#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;

/*
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]

7 5 4 [3 2 6 1 9]

从小到大确定可以放的位置

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

二分一下前缀 max 
*/

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 && mex < a[0]) {
        std::cout << 0;
        return;
    }

    i64 ans = 1;

    std::set<int> c;
    for (int i = 1; i <= k; i++) c.insert(i);
    for (auto j : a) c.erase(j);

    std::vector<int> pre(n);
    pre[0] = a[0];
    for (int i = 1; i < n; i++) {
        pre[i] = std::max(pre[i - 1], a[i]);
    }

    int bonus = 0;
    for (auto i : c) {
        (ans *= (std::lower_bound(all(pre), i) - pre.begin()) + 1 + bonus) %= M;
        bonus++;
    }

    std::cout << ans;
}

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