/*
	The technique on "constexpr" is derived from:
	https://atcoder.jp/contests/agc023/submissions/2435844
*/

#include <iostream>

constexpr int MOD = 998244353;

template<int N> class precalculation {
public:
	int inv[N];
	int d[N];
	constexpr precalculation() : inv(), d() {
		inv[1] = 1;
		for (int i = 2; i < N; i++) {
			inv[i] = (long long)(inv[MOD % i]) * (MOD - MOD / i) % MOD;
		}
		for (int i = 1; i < N; i++) {
			d[i] = 1;
		}
		for (int i = 2; i < N; i++) {
			if (d[i] == 1) {
				for (int j = i * 2; j < N; j += i) {
					if (d[j] == 1) {
						d[j] = i;
					}
				}
			}
		}
	}
};

constexpr precalculation precalc = precalculation<200001>();

int modpow(int a, int b) {
	int answer = 1;
	while (b != 0) {
		if ((b & 1) == 1) {
			answer = (long long)(answer) * a % MOD;
		}
		a = (long long)(a) * a % MOD;
		b >>= 1;
	}
	return answer;
}

int p[200009];
int main() {
	int N, K;
	std::cin >> N >> K;
	N -= 1;
	for (int i = 1; i <= N; i++) {
		if (precalc.d[i] == 1) {
			p[i] = modpow(i, K);
		}
		else {
			p[i] = (long long)(p[precalc.d[i]]) * p[i / precalc.d[i]] % MOD;
		}
	}
	int mult = 1;
	int answer = 0;
	for (int i = 0; i <= N; i++) {
		answer = (answer + (long long)(p[i]) * mult) % MOD;
		mult = (long long)(mult) * (N - i) % MOD;
		mult = (long long)(mult) * precalc.inv[i + 1] % MOD;
	}
	std::cout << answer << '\n';
	return 0;
}