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 maxn = 100000;
  const int mo = 1000000007;
  int n, a[100010], phi[100010], num[100010], po[100010], tot;
  long long res, ans;
  vector<int> ys[100010];
  void init() {
    for (int i = 1; i <= maxn; i++)
      for (int j = i; j <= maxn; j += i) ys[j].push_back(i);
    phi[1] = 1;
    for (int i = 2; i <= maxn; i++) {
      int tmp = 1, tmpp = i;
      while (tmpp % ys[i][1] == 0) {
        tmpp /= ys[i][1];
        tmp *= ys[i][1];
      }
      if (tmpp == 1)
        phi[i] = (ys[i][1] - 1) * (i / ys[i][1]);
      else
        phi[i] = phi[tmp] * phi[tmpp];
    }
  }
  void ins(int x, int y) {
    for (int i = 0; i < ys[x].size(); i++) {
      res = (res - 1ll * phi[ys[x][i]] * num[ys[x][i]] * num[ys[x][i]]) % mo;
      num[ys[x][i]] += y;
      res = (res + 1ll * phi[ys[x][i]] * num[ys[x][i]] * num[ys[x][i]]) % mo;
    }
  }
  int main() {
    init();
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= n; i++) {
      tot = 0;
      for (int j = i; j <= n; j += i) po[tot++] = a[j];
      sort(po, po + tot);
      int k = 0;
      for (int j = 0; j < tot; j = k) {
        while (k < tot && po[k] == po[j]) k++;
        ins(po[j], k - j);
      }
      ans = (ans + res * phi[i]) % mo;
      k = 0;
      for (int j = 0; j < tot; j = k) {
        while (k < tot && po[k] == po[j]) k++;
        ins(po[j], j - k);
      }
    }
    ans = (ans + mo) % mo;
    printf("%lld\n", ans);
  }
  