A Simple Math Problem
/*
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;
} 
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