#include<bits/stdc++.h>
#define int i64 /* R.I.P */
#define all(x) std::begin(x), std::end(x)
#define rall(x) std::rbegin(x), std::rend(x)
using i64 = long long;
using u64 = unsigned long long;
using ld = long double;

constexpr int M = 998244353;

i64 powp(i64 a, i64 n) {
    a %= M;
    i64 ans = 1;
    while (n) {
        if (n & 1) (ans *= a) %= M;
        (a *= a) %= M;
        n >>= 1;
    }

    return ans;
}

i64 inv(i64 x) {
    return powp(x, M - 2);
}

constexpr int N = 3e5;
i64 fac[N], invfac[N];

void pre() {//      #######组合数预处理时不能超过模数########
    fac[0] = invfac[0] = 1;
    for (int i = 1; i < N; i++) fac[i] = fac[i-1] * i % M;
    invfac[N - 1] = inv(fac[N - 1]);
    for (int i = N - 2; i >= 1; i--) invfac[i] = invfac[i + 1] * (i + 1) % M;
}
 
i64 C(i64 n,i64 m) {
    if (m > n) return 0;
    return fac[n] * invfac[n - m] % M * invfac[m] % M;
}

/*
4 5 9 10 2 7 8 3 6 1 

A 中必须有 1
5 2 8 4 6 3 7 1 10


9 7 [3 6 5 1] 2

C(10, )



2 [1 3 4 5]

2 5 4 [3 2 1 6]

看数组中的第一个数，是不是恰好是数组的 mex - 1
如果不是，输出0

否则，所有值域在数组中的元素只能放在数组前面，其它元素随意放，答案为 ((max - min + 1) - m)! * C(n, (max - min + 1))
*/

void solve() {
    i64 n, k;
    std::cin >> k >> n;
    std::vector<int> a(n);
    for (auto &i : a) std::cin >> i;

    auto b = a;
    std::sort(all(b));

    int mex = 1;
    for (; mex <= n; mex++) {
        if (b[mex - 1] != mex) break;
    }

    if (b.back() != n && a[0] != mex - 1) {
        std::cout << 0;
        return;
    }

    int mx = 0, mi = k;
    for (auto i : a) {
        mx = std::max(mx, i);
        mi = std::min(mi, i);
    }

    i64 ans = fac[mx - mi + 1 - n];

    std::cout << ans * C(k, mx - mi + 1) * fac[k - (mx - mi + 1)] % M;
}

signed main() {
    std::cin.tie(0)->sync_with_stdio(false);
    int t = 1;
    pre();
    //std::cin >> t;
    while(t--) solve();
    return 0;
}