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>
  #pragma comment(linker, "/STACK:36777216")
  using namespace std;
  long long mod = 1e9 + 7;
  long double eps = 1e-12;
  long double pi = acosl(-1);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  ifstream in("input.txt");
  ofstream out("output.txt");
  #pragma GCC optimize("Ofast")
  #pragma GCC target("avx,avx2,fma")
  #pragma GCC optimization("unroll-loops")
  using ll = long long;
  using ull = unsigned long long;
  using ld = long double;
  using uc = unsigned char;
  using us = unsigned short;
  ll phi[100050];
  ll a[100050];
  vector<int> g[100050];
  int used[100050];
  int step = 1;
  void solve() {
    phi[1] = 1;
    g[1].push_back(1);
    for (int i = 2; i < 100050; i++) {
      phi[i] = i - 1;
      g[i].push_back(1);
      g[i].push_back(i);
    }
    for (ll i = 2; i < 100050; i++) {
      for (ll j = 2; j * i < 100050; j++) {
        phi[j * i] -= phi[i];
        g[i * j].push_back(j);
      }
    }
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    ll ans = 0;
    for (ll i = 1; i <= n; i++) {
      ll res = 0;
      ll p = 1;
      unordered_map<int, ll> m;
      for (ll t = 1; t * i <= n; t++) {
        for (auto x : g[a[t * i]]) m[x]++;
      }
      for (auto x : m) {
        res += (((x.second * x.second) % mod) * phi[x.first]) % mod;
        if (res >= mod) res -= mod;
      }
      ans += res * phi[i] % mod;
      if (ans >= mod) ans -= mod;
    }
    cout << ans;
  }
  int main() {
    cout << setprecision(20) << fixed;
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    clock_t t1 = clock();
    ;
    int t = 1;
    int cs = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  