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;
  template <typename T>
  void _do(vector<T> x) {
    for (auto i : x) cerr << i << ' ';
    cerr << "\n";
  }
  template <typename T>
  void _do(set<T> x) {
    for (auto i : x) cerr << i << ' ';
    cerr << "\n";
  }
  template <typename T>
  void _do(unordered_set<T> x) {
    for (auto i : x) cerr << i << ' ';
    cerr << "\n";
  }
  template <typename T>
  void _do(T&& x) {
    cerr << x << endl;
  }
  template <typename T, typename... S>
  void _do(T&& x, S&&... y) {
    cerr << x << ", ";
    _do(y...);
  }
  template <typename T>
  void print(T&& x) {
    cout << x << "\n";
  }
  template <typename T, typename... S>
  void print(T&& x, S&&... y) {
    cout << x << ' ';
    print(y...);
  }
  void solve() {
    const int M = 1e5 + 1, MOD = 1e9 + 7;
    vector<int> mu(M, -2), f(M);
    vector<int> primes;
    mu[1] = 1;
    f[1] = 1;
    for (int i = 2; i < M; i++) {
      if (mu[i] == -2) {
        primes.push_back(i);
        mu[i] = -1;
        f[i] = i - 1;
      }
      for (int p : primes) {
        if (1LL * i * p >= M) break;
        if (i % p == 0) {
          mu[i * p] = 0;
          f[i * p] = f[i] * p;
          break;
        }
        mu[i * p] = -mu[i];
        f[i * p] = 1LL * f[i] * (p - 1) % MOD;
      }
    }
    int n;
    cin >> n;
    vector<vector<int>> fac(n + 1);
    for (int i = 1; i <= n; i++) {
      int a;
      cin >> a;
      for (int j = 1; 1LL * j * j <= a; j++)
        if (a % j == 0) {
          fac[i].push_back(j);
          if (j * j != a) fac[i].push_back(a / j);
        }
    }
    vector<int> cnt(M);
    int ans = 0;
    for (int T = 1; T <= n; T++) {
      int G = 0;
      vector<int> vec;
      for (int i = 1; i * T <= n; i++) {
        for (int p : fac[i * T])
          if (cnt[p]++ == 0) vec.push_back(p);
      }
      for (int p : vec) G = (G + 1LL * f[p] * cnt[p] % MOD * cnt[p]) % MOD;
      for (int i = 1; i * T <= n; i++) {
        for (int p : fac[i * T]) cnt[p]--;
      }
      ans = (ans + 1LL * G * f[T]) % MOD;
    }
    if (ans < 0) ans += MOD;
    cout << ans << endl;
  }
  int main() {
    cin.tie(0), cout.tie(0);
    ios::sync_with_stdio(false);
    solve();
  }
  