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;
  const int mod = 1000000007;
  vector<int> p;
  int gcd(int x, int y) { return y ? gcd(y, x % y) : x; }
  int a[100010], b[100010], c[100010], h[100010];
  long long f[100010];
  vector<int> m[100010], d[100010];
  void calc(int x) {}
  int main() {
    a[1] = 1;
    for (int i = 2; i <= 100000; i++) {
      if (a[i]) continue;
      p.push_back(i);
      for (int j = i + i; j <= 100000; j += i) {
        a[j] = 1;
        d[j].push_back(i);
      }
    }
    for (int i = 1; i <= 100000; i++) {
      h[i] = i;
    }
    for (int i = 1; i <= 100000; i++) {
      for (int j = i; j <= 100000; j += i) {
        m[j].push_back(i);
        if (!a[i]) h[j] = h[j] / i * (i - 1);
      }
    }
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &b[i]);
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      vector<int> w;
      for (int j = i; j <= n; j += i) {
        for (int k = 0; k < m[b[j]].size(); k++) {
          int x = m[b[j]][k];
          if (!c[x]++) w.push_back(x);
        }
      }
      for (int j = 0; j < w.size(); j++) {
        int x = w[j];
        f[i] = (f[i] + 1ll * h[x] * c[x] * c[x]) % mod;
      }
      for (int j = 0; j < w.size(); j++) {
        c[w[j]] = 0;
      }
    }
    for (int i = 1; i <= n; i++) {
      ans = (ans + f[i] * h[i]) % mod;
    }
    printf("%lld\n", ans);
    return 0;
  }
  