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, MAGIC = 100;
  vector<int> fac[N];
  vector<int> elem[N];
  long long g[N];
  int cnt[N];
  long long solve(vector<int>& val) {
    vector<int> all;
    vector<int> distinct;
    for (int x : val) {
      if (cnt[x] == 0) {
        distinct.push_back(x);
      }
      ++cnt[x];
    }
    sort(distinct.begin(), distinct.end());
    all = distinct;
    for (int x : distinct) {
      for (int z : fac[x]) {
        if (cnt[z] == 0) {
          all.push_back(z);
        }
        cnt[z] += cnt[x];
      }
    }
    sort(all.begin(), all.end());
    reverse(all.begin(), all.end());
    for (int x : all) {
      cnt[x] = 1LL * cnt[x] * cnt[x] % mod;
    }
    long long ans = 0;
    for (int x : all) {
      int cur = cnt[x];
      ans = (ans + 1LL * x * cur) % mod;
      for (int z : fac[x]) {
        cnt[z] -= cur;
        if (cnt[z] < 0) cnt[z] += mod;
      }
      cnt[x] = 0;
    }
    return ans;
  }
  int main() {
    for (int i = 1; i < N; ++i) {
      for (int j = i + i; j < N; j += i) {
        fac[j].push_back(i);
      }
    }
    int maxi = 0, s = 0;
    for (int i = 2; i < N; ++i) {
      maxi = max(maxi, (int)fac[i].size());
      s += fac[i].size();
    }
    int n;
    scanf("%d", &n);
    vector<int> a(n + 1);
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &a[i]);
    }
    for (int i = 1; i <= n; ++i) {
      for (int j = i; j <= n; j += i) {
        elem[i].push_back(a[j]);
      }
    }
    long long ans = 0;
    for (int i = n; i >= 1; --i) {
      g[i] = solve(elem[i]);
      for (int j = i + i; j <= n; j += i) {
        g[i] -= g[j];
        if (g[i] < 0) g[i] += mod;
      }
      ans = (ans + g[i] * i) % mod;
    }
    printf("%lld\n", ans);
    return 0;
  }
  