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 long long ma = 1e5, dv = 1e9 + 7;
  long long n, mb[100069], pc[100069], a[100069], vi[100069], dp[100069];
  bitset<100069> vtd;
  vector<long long> fz[100069];
  int main() {
    long long i, j, r, k, sz, z = 0;
    for (i = 1; i <= ma; i++) {
      mb[i] = 1;
    }
    for (i = 2; i <= ma; i++) {
      if (!vtd[i]) {
        for (j = i; j <= ma; j += i) {
          vtd[j] = 1;
          mb[j] = (mb[j] * (dv - (j % (i * i) != 0))) % dv;
        }
      }
    }
    for (i = 1; i <= ma; i++) {
      for (j = i; j <= ma; j += i) {
        pc[j] = (pc[j] + mb[j / i] * i) % dv;
        fz[j].push_back(i);
      }
    }
    scanf("%lld", &n);
    for (i = 1; i <= n; i++) {
      scanf("%lld", a + i);
    }
    for (i = 1; i <= n; i++) {
      for (j = i; j <= n; j += i) {
        sz = fz[a[j]].size();
        for (r = 0; r < sz; r++) {
          k = fz[a[j]][r];
          if (vi[k] < i * 2 - 1) {
            vi[k] = i * 2 - 1;
            dp[k] = 0;
          }
          dp[k]++;
        }
      }
      for (j = i; j <= n; j += i) {
        sz = fz[a[j]].size();
        for (r = 0; r < sz; r++) {
          k = fz[a[j]][r];
          if (vi[k] < i * 2) {
            vi[k] = i * 2;
            z = (z + dp[k] * dp[k] % dv * pc[i] % dv * pc[k]) % dv;
          }
        }
      }
    }
    printf("%lld\n", z);
  }
  