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>
  const int maxn = 1e5 + 10;
  const int MOD = 1e9 + 7;
  int phi[maxn];
  int n;
  int arr[maxn];
  std::vector<int> vec[maxn];
  int cnt[maxn];
  void init() {
    phi[1] = 1;
    for (int i = 2; i < maxn; i++) {
      if (!phi[i]) {
        for (int j = i; j < maxn; j += i) {
          if (!phi[j]) phi[j] = j;
          phi[j] = phi[j] / i * (i - 1);
        }
      }
    }
    for (int i = 1; i < maxn; i++) {
      for (int j = i; j < maxn; j += i) {
        vec[j].emplace_back(i);
      }
    }
    return;
  }
  int main() {
    init();
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &arr[i]);
    }
    long long ans = 0;
    for (int x = 1; x <= n; x++) {
      for (int i = x; i <= n; i += x) {
        for (const auto &val : vec[arr[i]]) {
          ++cnt[val];
        }
      }
      for (int i = x; i <= n; i += x) {
        for (const auto &val : vec[arr[i]]) {
          if (cnt[val]) {
            ans += (long long)cnt[val] * cnt[val] % MOD * phi[x] % MOD *
                   phi[val] % MOD;
            ans %= MOD;
            cnt[val] = 0;
          }
        }
      }
    }
    printf("%lld\n", ans);
    return 0;
  }
  