设$f[i]$为$\gcd$为$i$，还需要多少个数，那么有$f[i]=1+\frac{\sum _{j=1}^{m}f[\gcd (i,j)]}{m}$，

$f[1]=0$，考虑化简$\sum _{j=1}^{m}f[\gcd (i,j)]$，
$$\begin{gathered} \sum _{d\mid i}f[d]\sum _{j=1}^m[\gcd (i,j)=d] \\ 给定i,d，求\sum _{j=1}^m[\gcd (i,j)=d] \\ \sum _{j=1}^{\frac{m}{d}}[\gcd (\frac{i}{d},j)=1] \\ \sum _{k\mid \frac{i}{d}}\mu (k)\frac{m}{kd} \\ 则原式为\sum _{d\mid i}f[d]\sum _{k\mid \frac{i}{d}}\mu (k)\frac{m}{kd} \end{gathered}$$

#include <bits/stdc++.h>using namespace std;const int N = 1e5 + 10, mod = 1e9 + 7;int f[N], prime[N], mu[N], m, cnt;vector<int> fac[N];bool st[N];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;
}inline int inv(int a) {return quick_pow(a, mod - 2);
}void init() {mu[1] = 1;for (int i = 2; i < N; i++) {if (!st[i]) {prime[++cnt] = i;mu[i] = mod - 1;}for (int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;if (i % prime[j] == 0) {break;}mu[i * prime[j]] = (mod - mu[i]) % mod;}}for (int i = 1; i < N; i++) {for (int j = i; j < N; j += i) {fac[j].push_back(i);}}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);init();scanf("%d", &m);int inv_m = inv(m);for (int i = 1; i <= m; i++) {for (auto &d : fac[i]) {if (d == i) {continue;}int cur = 0;for (auto &k : fac[i / d]) {cur = (cur + 1ll * mu[k] * (m / (k * d)) % mod) % mod;}f[i] = (f[i] + 1ll * f[d] * cur % mod) % mod;}f[i] = 1ll * (m + f[i]) * inv(m - m / i) % mod;}int ans = 0;for (int i = 1; i <= m; i++) {ans = (ans + f[i]) % mod;}cout << (1 + 1ll * ans * inv_m) % mod << "\n";return 0;
}