$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^{i}f(j)[gcd(i,j)=1] \\ \sum _{i=1}^n\sum _{j=i}^{n}f(i)[gcd(i,j)=1] \\ \sum _{i=1}^n\sum _{j=1}^{n}f(i)[gcd(i,j)=1]-\sum _{i=1}^n\sum _{j=1}^{i}f(i)[gcd(i,j)=1]+f(1) \\ \sum _{d=1}^n\mu (d)\frac{n}{d}\sum _{i=1}^{\frac{n}{d}}f(id)-\sum _{i=1}^{n}f(i)\varphi (i)+f(1) \\ 然后就可以O(n\log n)求解了 \end{gathered}$$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e5 + 10;ll prime[N], phi[N], mu[N], f[N], g[N], cnt, n;bool st[N];void init() {phi[1] = mu[1] = f[1] = 1;for(int i = 2; i <= n; i++) {f[i] = f[i / 10] + i % 10;if(!st[i]) {prime[++cnt] = i;phi[i] = i - 1;mu[i] = -1;}for(int j = 1; j <= cnt && 1ll * i * prime[j] <= n; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) {phi[i * prime[j]] = phi[i] * prime[j];break;}phi[i * prime[j]] = phi[i] * (prime[j] - 1);mu[i * prime[j]] = -mu[i];}}for(int i = 1; i <= n; i++) {for(int j = i; j <= n; j += i) {g[i] += f[j];}}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);scanf("%lld", &n);init();ll ans1 = 0, ans2 = 0;for(int i = 1; i <= n; i++) {ans1 += mu[i] * (n / i) * g[i];ans2 += f[i] * phi[i];}printf("%lld\n", ans1 - ans2 + 1);return 0;
}