description: Mr. Chanek has an array a of n integers. The prettiness value of a is denoted as:
  
  $$$∑_{i=1}^{n} {∑_{j=1}^{n} {\gcd(a_i, a_j) ⋅ \gcd(i, j)}}$$$
  
  where \gcd(x, y) denotes the greatest common divisor (GCD) of integers x and y.
  
  In other words, the prettiness value of an array a is the total sum of \gcd(a_i, a_j) ⋅ \gcd(i, j) for all pairs (i, j).
  
  Help Mr. Chanek find the prettiness value of a, and output the result modulo 10^9 + 7!
  
  Input
  
  The first line contains an integer n (2 ≤ n ≤ 10^5).
  
  The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^5).
  
  Output
  
  Output an integer denoting the prettiness value of a modulo 10^9 + 7.
  
  Example
  
  Input
  
  
  5
  3 6 2 1 4
  
  
  Output
  
  
  77
solution: #include <bits/stdc++.h>
  using namespace std;
  int phi[100001];
  vector<int> v[100001];
  int a[100001], cnt[100001];
  const int MOD = 1000000007;
  int main() {
    iota(phi, phi + 100001, 0);
    for (int i = 1; i <= 100000; i++) {
      v[i].push_back(i);
      for (int j = i + i; j <= 100000; j += i) {
        phi[j] -= phi[i];
        v[j].push_back(i);
      }
    }
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d", a + i);
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      long long t = 0;
      for (int j = i; j <= n; j += i) {
        for (int x : v[a[j]]) {
          t = (t + phi[x] * cnt[x]) % MOD;
          cnt[x]++;
        }
      }
      ans = (ans + t * phi[i]) % MOD;
      for (int j = i; j <= n; j += i) {
        for (int x : v[a[j]]) cnt[x]--;
      }
    }
    ans = (ans * 2) % MOD;
    for (int i = 1; i <= n; i++) {
      ans = (ans + 1ll * i * a[i]) % MOD;
    }
    printf("%lld\n", ans);
    return 0;
  }
  