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 + 1e3 + 7, P = 1e9 + 7;
  int n, a[N];
  int vis[N], p[N], ptot, phi[N];
  vector<int> v;
  vector<int> g[N];
  void pre(int n) {
    vis[1] = 1;
    phi[1] = 1;
    for (int i = 2; i <= n; i++) {
      if (!vis[i]) p[++ptot] = i, phi[i] = i - 1;
      for (int j = 1; j <= ptot && i * p[j] <= n; j++) {
        phi[i * p[j]] = phi[i] * (p[j] - 1);
        vis[i * p[j]] = 1;
        if (i % p[j] == 0) {
          phi[i * p[j]] = phi[i] * p[j];
          break;
        }
      }
    }
    for (int i = 1; i <= n; i++)
      for (int j = i; j <= n; j += i) g[j].push_back(i);
  }
  int s[N];
  int calc(vector<int>& v) {
    int ret = 0, dlt = 0;
    for (auto x : v)
      for (auto v : g[x]) {
        ret = (ret + 1ll * s[v] * phi[v]) % P;
        dlt = (dlt + phi[v]) % P;
        s[v]++;
      }
    for (auto x : v)
      for (auto v : g[x]) s[v]--;
    return (ret * 2ll + dlt) % P;
  }
  int main() {
    pre(1e5);
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    int ans = 0;
    for (int d = 1; d <= n; d++) {
      v.clear();
      for (int i = 1; i * d <= n; i++) v.push_back(a[i * d]);
      ans = (ans + 1ll * phi[d] * calc(v)) % P;
    }
    printf("%d\n", ans);
  }
  