#include <bits/stdc++.h>
#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 uint unsigned int
#define ll long long
#define ull unsigned long long
#define db double
#define ldb long double
#define pii pair<int,int>
#define pll pair<ll,ll>
#define mkp make_pair
#define eb emplace_back
#define SZ(S) (int)S.size()
#define mod 998244353
//#define mod 1000000007
#define inf 0x3f3f3f3f
#define linf 0x3f3f3f3f3f3f3f3f
using namespace std;

template <class T>
void chkmax(T &x,T y){
    if(x < y) x = y;
}

template <class T>
void chkmin(T &x,T y){
    if(x > y) x = y;
}

inline int popcnt(int x){
    return __builtin_popcount(x);
}

inline int ctz(int x){
    return __builtin_ctz(x);
}

template <int P>
class mod_int
{
    using Z = mod_int;

private:
    static int mo(int x) { return x < 0 ? x + P : x; }

public:
    int x;
    int val() const { return x; }
    mod_int() : x(0) {}
    template <class T>
    mod_int(const T &x_) : x(x_ >= 0 && x_ < P ? static_cast<int>(x_) : mo(static_cast<int>(x_ % P))) {}
    bool operator==(const Z &rhs) const { return x == rhs.x; }
    bool operator!=(const Z &rhs) const { return x != rhs.x; }
    Z operator-() const { return Z(x ? P - x : 0); }
    Z pow(long long k) const
    {
        Z res = 1, t = *this;
        while (k)
        {
            if (k & 1)
                res *= t;
            if (k >>= 1)
                t *= t;
        }
        return res;
    }
    Z &operator++()
    {
        x < P - 1 ? ++x : x = 0;
        return *this;
    }
    Z &operator--()
    {
        x ? --x : x = P - 1;
        return *this;
    }
    Z operator++(int)
    {
        Z ret = x;
        x < P - 1 ? ++x : x = 0;
        return ret;
    }
    Z operator--(int)
    {
        Z ret = x;
        x ? --x : x = P - 1;
        return ret;
    }
    Z inv() const { return pow(P - 2); }
    Z &operator+=(const Z &rhs)
    {
        (x += rhs.x) >= P && (x -= P);
        return *this;
    }
    Z &operator-=(const Z &rhs)
    {
        (x -= rhs.x) < 0 && (x += P);
        return *this;
    }
    Z operator-()
    {
        return -x;
    }
    Z &operator*=(const Z &rhs)
    {
        x = 1ULL * x * rhs.x % P;
        return *this;
    }
    Z &operator/=(const Z &rhs) { return *this *= rhs.inv(); }
#define setO(T, o)                                  \
    friend T operator o(const Z &lhs, const Z &rhs) \
    {                                               \
        Z res = lhs;                                \
        return res o## = rhs;                       \
    }
    setO(Z, +) setO(Z, -) setO(Z, *) setO(Z, /)
#undef setO
    
    friend istream& operator>>(istream& is, mod_int& x)
    {
        long long tmp;
        is >> tmp;
        x = tmp;
        return is;
    }
    friend ostream& operator<<(ostream& os, const mod_int& x)
    {
        os << x.val();
        return os;
    }
};

using Z = mod_int<mod>;
Z power(Z p,ll k){
    Z ans = 1;
    while(k){
        if(k % 2 == 1) ans *= p;
        p *= p;
        k /= 2;
    }
    return ans;
}
int n,K;
int q[505];
Z dp[505][505],fact[505],C[505][505];

Z calc(int m,int s){//长为 m 的排列,可被划分为 s 个段
    Z answer = 0;
    rep(i,s,n){
        if(i % 2 == s % 2) answer += C[i - 1][s - 1] * dp[i][m];
        else answer -= C[i - 1][s - 1] * dp[i][m];
    }
//    printf("calc %d %d :%d\n",m,s,answer.val());
    return answer;
}

int main(){
//    freopen("test.in","r",stdin);
    scanf("%d%d",&n,&K);
    rep(i,1,n) scanf("%d",&q[i]);
    fact[0] = 1;
    rep(i,1,n) fact[i] = fact[i - 1] * i;
    rep(i,0,n){
        C[i][0] = 1;
        rep(j,1,i) C[i][j] = C[i - 1][j - 1] + C[i - 1][j]; 
    }

    dp[0][0] = 1;
    rep(i,1,n){
        rep(j,0,n){
            rep(k,1,n - j) dp[i][j + k] += dp[i - 1][j] * fact[k];
        }
    }
    int u = n + 1;
    rep(i,1,n){
        if(q[i] > q[i + 1]){
            u = i;
            break;
        }
    }
    if(u == n){
        Z output = 0;
        rep(i,K,n) output += calc(n,i);
        printf("%d\n",output.val());
        return 0;
    }
    Z output = 0;
    rep(i,u + 1,n){
        if(i > u + 1 && q[i] < q[u]) break;
        if(i > u + 2 && q[i] < q[i - 1]) break;
        output += calc(u,i - 1 - (n - K + 1) + 1) * fact[i - 1 - u];
    }
    printf("%d\n",output.val());
	return 0;
}