#include<bits/stdc++.h>
using namespace std;
#define IOS {cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);}
#define rep(i,j,k) for(int i=j;i<=k;++i)
#define per(i,j,k) for(int i=j;i>=k;--i)
#define P pair<int,int>
#define ll long long
#define vi vector<int>
const int N = 2e5+5, mod = 998244353, G = 3;
ll qp(ll x, ll y) {
    ll res = 1;
    while (y) {
        if (y & 1)res = res * x % mod;
        x = x * x % mod; y >>= 1;
    }return res;
}
const int Gi = qp(G, mod - 2);
typedef vi poly;
namespace Poly {
    int limit = 1, L = 0; int r[N * 4];
    void NTT(poly& A, int type) { //下标在[0,limit)范围内,数组开四倍即可
        A.resize(limit);
        for (int i = 0; i < limit; i++)
            if (i < r[i]) swap(A[i], A[r[i]]);
        for (int mid = 1; mid < limit; mid <<= 1) {
            ll Wn = qp(type == 1 ? G : Gi, (mod - 1) / (mid << 1)); //G是模数的原根,Gi是逆元
            for (int j = 0; j < limit; j += (mid << 1)) {
                ll w = 1; //ll不一定够
                for (int k = 0; k < mid; k++, w = (w * Wn) % mod) {
                    int x = A[j + k], y = w * A[j + k + mid] % mod; //int不一定够
                    A[j + k] = (x + y) % mod, A[j + k + mid] = (x - y + mod) % mod;
                }
            }
        }
    }
    poly operator + (poly a, poly b) {
        int n = max(a.size(), b.size());
        a.resize(n); b.resize(n);
        rep(i, 0, n - 1)a[i] = (a[i] + b[i]) % mod;
        return a;
    }
    poly operator - (poly a, poly b) {
        int n = max(a.size(), b.size());
        a.resize(n); b.resize(n);
        rep(i, 0, n - 1)a[i] = (a[i] - b[i] + mod) % mod;
        return a;
    }
    void poly_mul_init(poly& a, poly& b) {
        limit = 1; L = 0;
        int N = a.size() - 1, M = b.size() - 1;
        while (limit <= N + M) limit <<= 1, L++;
        for (int i = 0; i < limit; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (L - 1));
    }
    poly poly_mul(poly a, poly b) { //原先的a,b不需要维持原状的话,可以加&
        int n = a.size() + b.size() - 1;
        poly_mul_init(a, b);
        NTT(a, 1); NTT(b, 1);
        for (int i = 0; i < limit; i++) a[i] = 1ll * a[i] * b[i] % mod; //a[i]为ll不一定够
        NTT(a, -1);
        ll INV = qp(limit, mod - 2);
        rep(i, 0, limit - 1)a[i] = a[i] * INV % mod;
        a.resize(n); return a;
    }
    poly poly_inv(poly& a, int n) { //n为答案阶数, %(x^(n+1))意义下
        if (a.size() < n + 1)a.resize(n + 1); //!!!!
        if (n == 0)return { (int)qp(a[0],mod - 2) };
        int n2 = n / 2;
        poly b = poly_inv(a, n2);
        poly ta(n + 1); rep(i, 0, n)ta[i] = a[i];
        /*等价于:
        poly ans=poly_mul(ta,b); ans.resize(n+1);
        rep(i,0,n)ans[i]=(((i==0)?2:0)-ans[i]+mod)%mod;
        ans=poly_mul(b,ans); ans.resize(n+1); return ans;*/
        b.resize(n + 1); //ta和b的阶数一定要相同
        poly_mul_init(ta, b);
        NTT(ta, 1); NTT(b, 1);
        rep(i, 0, limit)b[i] = (2 - 1ll * ta[i] * b[i] % mod + mod) * b[i] % mod;  //蝴蝶变换这里每一项2-,然而多项式乘法时还是{2,0,0...}
        NTT(b, -1);
        ll inv = qp(limit, mod - 2);
        rep(i, 0, n)b[i] = b[i] * inv % mod;
        b.resize(n + 1); ta.clear(); ta.shrink_to_fit();
        return b;
    }
    poly poly_derivate(poly& a) { //求导
        int n = (int)a.size() - 1; poly da(n + 1);
        rep(i, 1, n)da[i - 1] = 1ll * a[i] * i % mod;
        return da;
    }
    poly poly_integral(poly& a) { //积分
        int n = (int)a.size(); poly ia(n + 1);
        rep(i, 1, n)ia[i] = 1ll * a[i - 1] * qp(i, mod - 2) % mod;
        return ia;
    }
    poly poly_ln(poly a) {
        int n = (int)a.size() - 1;
        assert(a[0] == 1);
        poly da = poly_derivate(a);
        a = poly_inv(a, n);
        poly b = poly_mul(a, da); b.resize(n + 1);
        vi res = poly_integral(b); res.resize(n + 1); return res;
    }
    poly poly_exp(poly& a, int n) {  //牛顿迭代, nxtg = g*(1-ln(g)+A)
        if (n == 0) {
            assert(a[0] == 0); return { 1 };
        }
        int n2 = n / 2;
        poly g = poly_exp(a, n2); g.resize(n + 1);//这里求逆要扩展g
        poly lng = poly_ln(g);
        rep(i, 0, n)lng[i] = ((i == 0) + a[i] - lng[i] + mod) % mod;
        poly res = poly_mul(g, lng); res.resize(n + 1);
        return res;
    }
    //多项式ln+exp可以用来优化多项式快速幂：f(x)^m = exp(ln(f(x))*m)
    poly poly_pow(poly a, int y) { //a[0]=1. a[0]=0的时候直接移位; 非0的时候整个多项式/a[0],最终结果再*qp(a[0],y)
        assert(a[0] == 1);
        int n = (int)a.size() - 1;
        poly lna = poly_ln(a);
        rep(i, 0, n)lna[i] = 1ll * lna[i] * y % mod;
        return poly_exp(lna, n);
    }
    void cdq_fft(poly& f, poly& g, int l, int r) { //f=F(f)*g形式,考虑[l,mid]对[mid+1,r]的贡献,一次poly_mul即可
        if (l == r)return;
        int mid = l + r >> 1;
        cdq_fft(f, g, l, mid);
        poly b(r - l + 1); rep(i, 0, r - l)b[i] = g[i];
        poly a(mid - l + 1); rep(i, 0, mid - l)a[i] = f[i + l];
        poly res = poly_mul(a, b);
        rep(i, mid + 1, r)f[i] = (f[i] + res[i - l]) % mod;
        cdq_fft(f, g, mid + 1, r);
    }
    pair<poly, poly>poly_divison(poly f, poly g) { //f=q*g+r, f(n),g(m),q(n-m),r(m-1)
        int m = (int)g.size() - 1;
        if (f.size() < m + 1)f.resize(m + 1); int n = (int)f.size() - 1;
        vi fR = f; reverse(fR.begin(), fR.end()); fR.resize(n - m + 1);//快了很多!!
        vi gR = g; reverse(gR.begin(), gR.end());
        auto qR = poly_mul(poly_inv(gR, n - m), fR); qR.resize(n - m + 1);
        auto q = qR; reverse(q.begin(), q.end());
        auto r = f - poly_mul(q, g); r.resize(m);
        return { q,r };
    }
    int linear_recurrence_poly(vi a, vi f, int n) { //常系数齐次线性递推,求a_n.  
        //a_i=sigma(a_{i-j}*f_j),1<=j<=k. 已知a的[0,k-1]项
        int k = a.size(); assert(f.size() == k + 1);
        if (n < k)return a[n];
        vi p(k + 1); p[k] = 1; rep(i, 1, k)p[k - i] = (mod - f[i]) % mod;
        vi res = { 1 }, x = { 0,1 };
        vi gR = p; reverse(gR.begin(), gR.end()); vi invgR = poly_inv(gR, k); //gR的逆预处理出来
        auto poly_division_r = [&](poly f, poly g)->poly {
            int m = (int)g.size() - 1;
            if (f.size() < m + 1)f.resize(m + 1); int n = (int)f.size() - 1;
            vi fR = f; reverse(fR.begin(), fR.end()); fR.resize(n - m + 1);
            auto qR = poly_mul(invgR, fR); qR.resize(n - m + 1);
            auto q = qR; reverse(q.begin(), q.end());
            auto r = f - poly_mul(q, g); r.resize(m);
            return r;
        };
        while (n) {
            int nn = x.size() * 2 - 1;
            poly_mul_init(x, x);
            NTT(x, 1);
            if (n & 1) {
                NTT(res, 1);
                for (int i = 0; i < limit; i++) res[i] = 1ll * res[i] * x[i] % mod;
            }
            for (int i = 0; i < limit; i++)x[i] = 1ll * x[i] * x[i] % mod;
            NTT(x, -1);
            ll INV = qp(limit, mod - 2);
            if (n & 1) {
                NTT(res, -1);
                for (int i = 0; i < limit; i++)res[i] = res[i] * INV % mod;
                res.resize(nn);
            }
            for (int i = 0; i < limit; i++)x[i] = x[i] * INV % mod;
            x.resize(nn);
            if (n & 1)res = poly_division_r(res, p); //四次NTT即可
            x = poly_division_r(x, p);
            n >>= 1;
        }
        ll ans = 0;
        rep(i, 0, k - 1)ans = (ans + 1ll * res[i] * a[i]) % mod; return ans;
    }
}
namespace multi_eval{//多点求值
    poly q[N*4]; //N=O(求值点)
    int xs[N],res[N]; //求值的点,答案
    void dfs(int i,int l,int r){
        if(l==r){
            q[i]={1,mod-xs[l]}; return;
        }
        int mid=l+r>>1;
        dfs(i<<1,l,mid); dfs(i<<1|1,mid+1,r);
        q[i]=Poly::poly_mul(q[i<<1],q[i<<1|1]);
    }
    poly MulT(poly a,poly b) {
        int n = a.size(), m = b.size(); 
        std::reverse(b.begin(),b.end()), b = Poly::poly_mul(a,b);
        for(int i = 0;i < n;i++) a[i] = b[i + m - 1];
        return a;
    }
    void slv(poly F,int i,int l,int r) {
        F.resize(r - l + 1);
        if(l == r) return void(res[l] = F[0]);
        int mid = (l + r) / 2;
        slv(MulT(F,q[i<<1|1]),i<<1,l,mid);
        slv(MulT(F,q[i<<1]),i<<1|1,mid + 1,r);
        return;
    }
    vi eval(poly a,vi ask){ //ask的下标[1,m]. 返回ans的下标[1,m]
        int m=(int)ask.size()-1; rep(i,1,m)xs[i]=ask[i];
        a.resize(m+1);
        dfs(1,1,m);  slv( MulT(a,Poly::poly_inv(q[1],m)) ,1,1,m);
        vi ans(m+1); rep(i,1,m)ans[i]=res[i]; return ans;
    }
};
ll fac[N],inv[N];
ll C(int i,int j){
    return fac[i]*inv[j]%mod*inv[i-j]%mod;
}
poly dfs(vi &a, int l,int r){
    if(l==r){
        return {a[l],mod-1};
    }
    int mid=l+r>>1;
    return Poly::poly_mul(dfs(a,l,mid),dfs(a,mid+1,r));
}
ll cal(int k,vi a,int v){ //[1,k]个数, 包含[1,v]
    vi ask(v+2);
    rep(i,1,v+1)ask[i]=i-1;
    poly A=dfs(a,1,k);
    auto res=multi_eval::eval(A,ask);
    int fl=1; ll ans=0;
    per(i,v,0){
        ans+=fl*res[v-i+1]*C(v,i)%mod;
        fl=-fl;
    }
    return (ans%mod+mod)%mod;
}
int a[N];
int main(){
    IOS
    const int M=2e5;
    fac[0]=1; rep(i,1,M)fac[i]=fac[i-1]*i%mod;
    inv[M]=qp(fac[M],mod-2); per(i,M,1)inv[i-1]=inv[i]*i%mod;
    int n;cin>>n;
    rep(i,1,n)cin>>a[i];
    int k;
    rep(i,1,n){
        if(a[n-i+1]>=i)k=i;
        else break;
    }
    vi aa(k+1);
    rep(i,1,k)aa[i]=a[n-i+1];
    ll ans=cal(k,aa,a[n-k]);
    int v=0;
    rep(i,1,k){
        if(a[n-i+1]>k)++v;
    }
    int r=1;
    rep(i,1,k){
        while(a[r]<i)++r;
        aa[i]=n-r+1;
    }
    ans=(ans+cal(k,aa,v))%mod;
    cout<<k<<' '<<(ans-fac[k]+mod)%mod<<'\n';
}