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;
  int n;
  int a[100005];
  int tmp[100005];
  int sl = 0;
  int dp[100005];
  int zs = 0;
  int sz[1200005];
  int dd[100005];
  int cl[100005];
  vector<int> g[100005];
  int sol() {
    zs = 0;
    for (int i = 1; i <= sl; ++i) {
      int u = tmp[i];
      for (auto &v : g[u]) {
        cl[v]++;
        if (cl[v] == 1) sz[++zs] = v;
      }
    }
    sort(sz + 1, sz + zs + 1);
    int cnt = 0;
    for (int i = zs; i >= 1; --i) {
      dd[sz[i]] =
          (dd[sz[i]] + (1ll * cl[sz[i]] * cl[sz[i]]) % 1000000007) % 1000000007;
      cnt = 0;
      for (auto &v : g[sz[i]])
        if (v != sz[i]) dd[v] = (dd[v] - dd[sz[i]] + 1000000007) % 1000000007;
    }
    int res = 0;
    for (int i = 1; i <= zs; ++i) {
      res = (res + (1ll * dd[sz[i]] * sz[i]) % 1000000007) % 1000000007;
      dd[sz[i]] = 0;
      cl[sz[i]] = 0;
    }
    return res;
  }
  int main() {
    ios ::sync_with_stdio(false);
    cin >> n;
    int ans = 0;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 1; i <= 100000; ++i)
      for (int j = 1; j * j <= i; ++j)
        if (i % j == 0) {
          g[i].push_back(j);
          if (j * j != i) g[i].push_back(i / j);
        }
    for (int i = 1; i <= n; ++i) {
      sl = 0;
      for (int j = 1; i * j <= n; ++j) tmp[++sl] = a[i * j];
      dp[i] = sol();
    }
    for (int i = n; i >= 1; --i)
      for (int j = 2; i * j <= n; ++j)
        dp[i] = (dp[i] - dp[i * j] + 1000000007) % 1000000007;
    for (int i = 1; i <= n; ++i)
      ans = (ans + (1ll * i * dp[i]) % 1000000007) % 1000000007;
    cout << ans;
  }
  