#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 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);
ll fac[N], inv[N];
namespace Poly {
    typedef vi 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) {
        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);
    }
}
using namespace Poly;
char s[N];
signed main() {
    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 v=1; 
    vi lg(M+1),b(M+2),c(M+1);
    rep(i,0,mod-2){
        if(v<=M)lg[v]=i;
        v=1ll*v*G%mod;
    }
    // cerr<<clock(); return 0;
    int n;cin>>n;
    vi a(n+1); rep(i,1,n)cin>>a[i];
    cin>>s+1;
    vi sgn(n+1); sgn[0]=1;
    rep(i,1,n-1)sgn[i]=(s[i]=='+'?1:-1),c[i]=i+1-sgn[i],++b[c[i]];
    vi res=poly_mul(lg,b);
    ll ans=0;
    rep(i,1,n){
        ll prc=1;
        rep(j,1,i-1)prc=prc*(i-c[j])%mod;
        prc=(prc+mod)%mod;
        // ans=(ans+a[i]*inv[i-1]%mod*prc)%mod;
        ll val=qp(G,res[i]);
        rep(j,max(1,i-2),i-1){
            if(c[j]==i)val=0;
            else if(c[j]==i+1)val=mod-val;
        }
        assert(prc==val);
        // cout<<val<<'\n';
        ans=(ans+a[i]*inv[i-1]%mod*val)%mod;
    }
    cout<<ans*fac[n-1]%mod<<'\n';
}
/*
4
9 1 4 1
-+-

5
1 2 3 4 5
+-+-

*/