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 = 150000;
  const int mod = 1000000000 + 7;
  bool vis[N + 5];
  int prime[N + 5], phi[N + 5], a[N + 5], n, ans = 0, cnt[100005], bj[100005];
  vector<int> Div[N + 5];
  inline int add(const int &x, const int &y) { return (x + y) % mod; }
  inline int mul(const int &x, const int &y) { return (1ll * x * y) % mod; }
  void Euler() {
    for (int i = 2; i <= N; i++) {
      if (!vis[i]) {
        prime[++prime[0]] = i;
        phi[i] = i - 1;
      }
      for (int j = 1; j <= prime[0] && prime[j] * i <= N; j++) {
        vis[prime[j] * i] = 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() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      int x;
      scanf("%d", &a[i]);
      x = a[i];
      for (int j = 1; j * j <= x; j++) {
        if (x % j == 0) {
          Div[i].push_back(j);
          if (j != x / j) Div[i].push_back(x / j);
        }
      }
    }
    Euler();
    phi[1] = 1;
    for (int x = 1; x <= n; x++) {
      vector<int> box;
      for (int i = 1; i * x <= n; i++)
        for (auto d : Div[i * x]) {
          if (bj[d] != x) {
            bj[d] = x;
            cnt[d] = 1;
            box.push_back(d);
          } else
            cnt[d]++;
        }
      for (auto y : box)
        ans = add(ans, mul(phi[x], mul(phi[y], mul(cnt[y], cnt[y]))));
    }
    printf("%d\n", ans);
    return 0;
  }
  