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 MOD = 1e9 + 7;
  int add(int x, int y) { return x + y >= MOD ? x + y - MOD : x + y; }
  int sub(int x, int y) { return x >= y ? x - y : x + MOD - y; }
  int gcd(int x, int y) { return y ? gcd(y, x % y) : x; }
  long long fast_pow(long long b, long long p) {
    long long ans = 1;
    while (p) {
      if (p & 1) ans = ans * b % MOD;
      b = b * b % MOD;
      p >>= 1;
    }
    return ans;
  }
  long long INV(long long x) { return fast_pow(x, MOD - 2); }
  const int B = 400;
  bool prm[200000];
  int prm_lst[200000], pcnt;
  long long phi[200000];
  void init_mu(int n) {
    for (int i = 2; i <= n; i++) prm[i] = 1;
    phi[1] = 1;
    for (int i = 2; i <= n; i++) {
      if (prm[i]) prm_lst[++pcnt] = i, phi[i] = i - 1;
      for (int j = 1; j <= pcnt && prm_lst[j] * i <= n; j++) {
        prm[prm_lst[j] * i] = 0;
        if (i % prm_lst[j] == 0) {
          phi[i * prm_lst[j]] = phi[i] * prm_lst[j];
          break;
          ;
        }
        phi[i * prm_lst[j]] = phi[i] * (prm_lst[j] - 1);
      }
    }
  }
  int a[200000];
  long long cnt[200000];
  int main() {
    init_mu(100000);
    int n = 0;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    long long ans = 0;
    for (int d = 1; d <= n; d++) {
      long long cur = 0;
      if (d <= B) {
        memset(cnt, 0, sizeof(cnt));
        for (int j = 1; j * d <= n; j++) cnt[a[j * d]]++;
        for (int i = 1; i <= pcnt; i++) {
          for (int j = 100000 / prm_lst[i]; j >= 1; j--)
            cnt[j] = add(cnt[j], cnt[j * prm_lst[i]]);
        }
        for (int i = 1; i <= 100000; i++)
          cur = add(cur, cnt[i] * cnt[i] % MOD * phi[i] % MOD);
      } else {
        for (int j = 1; j * d <= n; j++)
          for (int k = 1; k * d <= n; k++)
            cur = add(cur, gcd(a[j * d], a[k * d]));
      }
      ans = add(ans, cur * phi[d] % MOD);
    }
    printf("%lld\n", ans);
  }
  