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 long double pi = acos(-1);
  const long long Mod = 1000000007LL;
  const int N = 1e5 + 10;
  const long long Inf = 2242545357980376863LL;
  const long long Log = 30;
  long long a[N], u[N], c[N], cnt[N], c2[N];
  vector<int> D[N];
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    for (int i = 1; i < N; i++)
      for (int j = i; j < N; j += i) D[j].push_back(i);
    int mx = 0;
    for (int i = 1; i < N; i++) {
      for (auto x : D[i])
        if (x < i) u[i] -= u[x];
      if (i == 1) u[i] = 1;
      mx = max(mx, (int)D[i].size());
    }
    for (int i = 1; i < N; i++)
      for (auto x : D[i]) c[i] += x * u[i / x];
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int g = 1; g <= n; g++) {
      for (int i = g; i <= n; i += g)
        for (auto x : D[a[i]]) cnt[x]++;
      for (int i = g; i <= n; i += g) {
        for (auto x : D[a[i]]) {
          c2[x] += cnt[x] * c[g];
          c2[x] %= Mod;
        }
      }
      for (int i = g; i <= n; i += g)
        for (auto x : D[a[i]]) cnt[x]--;
    }
    long long ans = 0;
    for (int i = N - 1; i >= 1; i--) {
      for (auto x : D[i]) {
        if (x < i) (c2[x] -= c2[i]) %= Mod;
      }
      c2[i] %= Mod;
      ans += (c2[i] * i);
      ans %= Mod;
    }
    ans += Mod;
    cout << ans % Mod << '\n';
    return 0;
  }
  