推完式子就是去朴素地求就行了Orz
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
int m, mu[maxn], vis[maxn], primes[maxn], tot;
ll dp[maxn];
vector<int> factor[maxn];
ll ksm(ll a, ll b) {
ll ret = 1;
for (; b; b >>= 1) {
if (b & 1) ret = ret * a % mod;
a = a * a % mod;
}
return ret;
}
void pre(int n) {
rep(i, 1, n) {
for (int k = 1; k * i <= n; k++)
factor[k * i].push_back(i);
}
mu[1] = 1;
rep(i, 2, n) {
if (!vis[i]) {
primes[tot++] = i;
mu[i] = mod - 1;
}
for (int j = 0; j < tot && primes[j] * i <= n; j++) {
vis[primes[j] * i] = 1;
if (i % primes[j] == 0) break;
mu[primes[j] * i] = mod - mu[i];
}
}
}
ll calc(int x, int d) {
ll ret = 0;
for (int i : factor[x / d]) {
ret = (ret + (ll)mu[i] * (m / d / i) % mod) % mod;
}
return ret;
}
void DP() {
dp[1] = 0;
rep(i, 2, m) {
ll k = m - m / i;
ll tmp = m;
for (auto j : factor[i]) {
if (j == i) continue;
tmp = (tmp + dp[j] * calc(i, j) % mod) % mod;
}
dp[i] = tmp * ksm(k, mod - 2) % mod;
}
}
ll ANS(int m) {
ll ret = 0;
rep(i, 1, m) {
ret = (ret + dp[i] + 1) % mod;
}
return ksm(m, mod - 2) * ret % mod;
}
int main() {
read(m);
pre(m);
DP();
writeln(ANS(m));
return 0;
}