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;
  vector<int> v[100001];
  const int skc = 1e9 + 7;
  int n, a[100001], i, j, k, x, u[100001], p[100001];
  long long ans, sum, cnt[100001];
  int main() {
    scanf("%d", &n);
    for (i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      for (j = 1; j * j <= a[i]; j++)
        if (a[i] % j == 0) {
          v[i].push_back(j);
          if (j * j != a[i]) v[i].push_back(a[i] / j);
        }
    }
    for (i = 1; i <= 100000; i++) {
      x = i;
      u[i] = i;
      for (j = 2; j * j <= x; j++)
        if (x % j == 0) {
          u[i] = u[i] / j * (j - 1);
          while (x % j == 0) x /= j;
        }
      if (x != 1) u[i] = u[i] / x * (x - 1);
    }
    for (i = 1; i <= n; i++) {
      sum = 0;
      for (j = i; j <= n; j += i) p[a[j]]++;
      for (j = i; j <= n; p[a[j]] = 0, j += i)
        if (p[a[j]])
          for (k = 0; k < v[j].size(); k++)
            x = v[j][k],
            sum = (sum -
                   (cnt[x] * cnt[x] - (cnt[x] + p[a[j]]) * (cnt[x] + p[a[j]])) %
                       skc * u[x]) %
                  skc,
            cnt[x] += p[a[j]];
      for (j = i; j <= n; j += i) p[a[j]]++;
      for (j = i; j <= n; p[a[j]] = 0, j += i)
        for (k = 0; k < v[j].size(); k++) x = v[j][k], cnt[x] -= p[a[j]];
      ans = (ans + sum * u[i]) % skc;
    }
    printf("%lld\n", (ans + skc) % skc);
  }
  