$$\begin{gathered} \sum _{i=1}^n{i}^m[\gcd (i,n)=1] \\ \sum _{d\mid n}\mu (d)d^m\sum _{i=1}^{\frac{n}{d}}{i}^m \\ 由伯努利数可知\sum _{i=0}^n{i}^m=\frac{1}{m+1}\sum _{i=0}^m{C}_{m+1}^i{B}_i(n+1{)}^{m-i+1} \\ 设f_i=\frac{1}{m+1}{B}_{m-i+1}{C}_{m+1}^i,则有 \\ \sum _{d\mid n}\mu (d)d^m\left(\sum _{i=1}^{m+1}{f}_i(\frac{n}{d}{)}^i+(\frac{n}{d}{)}^m\right) \\ \sum _{i=1}^{m+1}{f}_i{n}^i\sum _{d\mid n}\mu (d)d^{m-i}+(n^m\sum _{d\mid n}\mu (d)) \\ 考虑后项\sum _{d\mid n}\mu (d)d^{m-i}，迪利克雷卷积乘积，为积性函数 \\ F(n)=\sum _{d\mid n}\mu (d)d^{m-i},F(1)=1,F(p^k)=1-p^{m-i} \end{gathered}$$

由于$m$较小，且模数是$10^9+7$，所以可以考虑$O(m^2)$递推得到伯努利数，之后可以枚举$i$，然后每次$O(w)$做一次线性筛，整体复杂度$O(mw)$。

#include <bits/stdc++.h>using namespace std;const int N = 1e3 + 10, mod = 1e9 + 7;int C[N][N], B[N], inv[N], f[N], p[N], n, m;void init() {for (int i = 0; i < N; i++) {C[i][0] = C[i][i] = 1;for (int j = 1; j < i; j++) {C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;}}inv[1] = 1;for (int i = 2; i < N; i++) {inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;}B[0] = 1;for (int i = 1; i < N; i++) {int cur = 0;for (int j = 0; j < i; j++) {cur = (cur + 1ll * C[i + 1][j] * B[j] % mod) % mod;}cur = 1ll * cur * inv[i + 1] % mod;B[i] = (mod - cur) % mod;}
}int quick_pow(int a, int n) {int ans = 1;while (n) {if (n & 1) {ans = 1ll * ans * a % mod;}a = 1ll * a * a % mod;n >>= 1;}return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);init();scanf("%d %d", &m, &n);for (int i = 1; i <= m + 1; i++) {f[i] = 1ll * inv[m + 1] * B[m - i + 1] % mod * C[m + 1][i] % mod;}int base = 1;for (int i = 1, w; i <= n; i++) {scanf("%d %d", &p[i], &w);base = 1ll * base * quick_pow(p[i], w) % mod;}int ans = 0, now = 1;for (int i = 1; i <= m + 1; i++) {now = 1ll * now * base % mod;int cur = 1ll * f[i] * now % mod, res = 1, po = m - i;if (po < 0) {po += mod - 1;}for (int j = 1; j <= n; j++) {res = 1ll * res * (1 - quick_pow(p[j], po) + mod) % mod;}cur = 1ll * cur * res % mod;ans = (ans + cur) % mod;}printf("%d\n", ans);return 0;
}