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 + 5, mod = 1e9 + 7;
  mt19937 gen(time(0));
  int rdint(int l, int r) { return uniform_int_distribution<int>(l, r)(gen); }
  int n, a[N], res[N], tot, b[N];
  vector<int> fac[N];
  int vis[N], d[N], vd[N];
  int sol(int n, int *a) {
    vector<int> vc;
    for (int i = 1; i <= n; i++)
      for (int x : fac[a[i]]) {
        if (!vis[x]) vc.push_back(x);
        vis[x] += vd[b[i]];
      }
    sort(vc.begin(), vc.end());
    reverse(vc.begin(), vc.end());
    int ans = 0;
    for (int x : vc) {
      int sum = 1ll * vis[x] * (vis[x] - 1) / 2 % mod;
      sum = (sum + mod - d[x]) % mod;
      for (int y : fac[x])
        if (y != x) d[y] = (d[y] + sum) % mod;
      ans = (ans + 1ll * sum * x % mod) % mod;
      vis[x] = d[x] = 0;
    }
    return ans;
  }
  int main() {
    for (int i = 1; i <= 100000; i++)
      for (int j = i; j <= 100000; j += i) fac[j].push_back(i);
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    int ans = 0;
    for (int d = n; d >= 1; d--) {
      tot = 0;
      for (int i = d; i <= n; i += d) {
        if (!vd[a[i]]) b[++tot] = a[i];
        vd[a[i]]++;
      }
      res[d] = sol(tot, b);
      for (int i = d; i <= n; i += d) vd[a[i]] = 0;
      for (int i = d + d; i <= n; i += d) res[d] = (res[d] + mod - res[i]) % mod;
      ans = (ans + 1ll * d * res[d]) % mod;
    }
    ans = ans * 2 % mod;
    for (int i = 1; i <= n; i++) ans = (ans + 1ll * a[i] * i % mod) % mod;
    printf("%d\n", ans);
  }
  