这两份代码是第二类斯特林数的两种求法，请帮我看看这两种求法有什么问题，对拍有答案不同的情况：

#include <iostream>
using namespace std;
const int N = 2e3, p = 1e9 + 7;
int dp[N][N];
signed main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr), cout.tie(nullptr);
	for (int i = 1; i <= N; i ++)
		dp[i][1] = 1;
	for (int i = 1; i <= N; i ++)
		for (int j = 2; j <= N; j ++)
			dp[i][j] = (1ll * dp[i - 1][j - 1] + 1ll * dp[i - 1][j] * j % p) % p;
	int x, y;
	cin >> x >> y;
	cout << dp[x][y];
    return 0;
}

#include <iostream>
using namespace std;
const int N = 2e6 + 5, p = 1e9 + 7;
int n, m, fac[N], inv[N];
int qpow (int a, int b) {
	int res = 1;
	while (b) {
		if (b & 1) res = 1ll * res * a % p;
		a = 1ll * a * a % p;
		b >>= 1; 
	}
	return res;
}
int C (int x, int y) {
	int z = 1ll * fac[x] * inv[y] % p;
	return 1ll * z * inv[x - y] % p;
}
int main () {
	int i;
	fac[0] = 1;
	cin >> n >> m;
	for (i = 1; i <= n + m; i ++) 
		fac[i] = 1ll * fac[i - 1] * i % p;
	inv[n + m] = qpow(fac[n + m], p - 2);
	for (i = n + m - 1; i >= 0; i --)
		inv[i] = 1ll * inv[i + 1] * (i + 1) % p;
	int ans = 0;
	for (int k = 0; k < m; k ++) {
		int cnt = (1ll * C(m, k) * qpow(m - k, n)) % p;
		ans += (k & 1 ? -1 : 1) * cnt;
		ans %= p;
	}
	cout << 1ll * ans * (1ll * inv[fac[m]] % p) % p;
	return 0;
}