#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

long long MOD = 998244353;
static const int MAXN = 2500000;  // 必要に応じて大きく設定

// 階乗・階乗逆元の配列
static long long fact[MAXN+1], invFact[MAXN+1];

// 繰り返し二乗法 (a^b mod M)
long long modpow(long long a, long long b, long long M) {
    long long ret = 1 % M;
    a %= M;
    while (b > 0) {
        if (b & 1) ret = (ret * a) % M;
        a = (a * a) % M;
        b >>= 1;
    }
    return ret;
}

// 前処理: 階乗と逆元のテーブルを作る
void initFactorials() {
    // 階乗テーブル
    fact[0] = 1;
    for (int i = 1; i <= MAXN; i++) {
        fact[i] = fact[i-1] * i % MOD;
    }
    // 階乗逆元テーブル
    invFact[MAXN] = modpow(fact[MAXN], MOD - 2, MOD);  // フェルマーの小定理で逆元を計算
    for (int i = MAXN; i >= 1; i--) {
        invFact[i-1] = invFact[i] * i % MOD;
    }
}

// 組み合わせ数 C(n, r)
long long comb(int n, int r) {
    if (r < 0 || r > n) return 0;
    return fact[n] * invFact[r] % MOD * invFact[n - r] % MOD;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll n, k;
    cin >> n >> k;
    if (k > n - k) k = n - k;
    if (k < 100000000) {
        ll ans = 1;
        const ll mod = 998244353;
        ll temp = 1;
        if (k > n - k) k = n - k;
        for (ll i = 1; i <= k; ++i) {
            ans = ans * ((n - i + 1) % mod) % mod;
            temp = temp * i % mod;
        }
        ans = ans * modpow(temp, mod - 2, mod) % mod;
        cout << ans << '\n';
        return 0;
    }
    vector<ll> prev(11);
    prev[0] = 1;
    prev[1] = 808258749;
    prev[2] = 117153405;
    prev[3] = 761699708;
    prev[4] = 573994984;
    prev[5] = 62402409;
    prev[6] = 511621808;
    prev[7] = 242726978;
    prev[8] = 887890124;
    prev[9] = 875880304;
    prev[10] = 0;
    ll ans = prev[n / 100000000];
    const ll mod = 998244353;
    for (ll i = (n / 100000000) * 100000000 + 1; i <= n; ++i) {
        ans = ans * i % mod;
    }
    ll temp = prev[k / 100000000];
    for (ll i = (k / 100000000) * 100000000 + 1; i <= k; ++i) {
        temp = temp * i % mod;
    }
    ans = ans * modpow(temp, mod - 2, mod) % mod;
    temp = prev[(n - k) / 100000000];
    for (ll i = ((n - k) / 100000000) * 100000000 + 1; i <= n - k; ++i) {
        temp = temp * i % mod;
    }
    ans = ans * modpow(temp, mod - 2, mod) % mod;
    cout << ans << '\n';
}