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 N = 1e5 + 7;
  const int mod = 1e9 + 7;
  int phi[N];
  int check[N];
  int a[N];
  vector<int> v[N];
  int cnt[N], prime[N];
  void init() {
    phi[1] = 1;
    check[0] = 1;
    for (int i = 1; i < N; i++)
      for (int j = i; j < N; j += i) {
        v[j].push_back(i);
      }
    for (int i = 2; i < N; i++) {
      if (!check[i]) {
        prime[++prime[0]] = i;
        phi[i] = i - 1;
      }
      for (int j = 1; j <= prime[0]; j++) {
        if ((long long)i * prime[j] >= N) break;
        check[i * prime[j]] = 1;
        if (i % prime[j] == 0) {
          phi[i * prime[j]] = phi[i] * prime[j];
          break;
        } else {
          phi[i * prime[j]] = phi[i] * phi[prime[j]];
        }
      }
    }
  }
  int main() {
    init();
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      vector<int> vv;
      for (int j = i; j <= n; j += i) {
        for (int d : v[a[j]]) {
          vv.push_back(d);
          cnt[d]++;
        }
      }
      long long tmp = 0;
      for (int d : vv) {
        tmp = (tmp + (((long long)cnt[d] * cnt[d] * phi[d]) % mod)) % mod;
        cnt[d] = 0;
      }
      ans = (ans + tmp * phi[i] % mod) % mod;
    }
    printf("%lld\n", ans);
    return 0;
  }
  