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;
  namespace PhiFunction {
  long long phi(long long n) {
    long long result = n;
    for (long long i = 2; i * i <= n; i++) {
      if (n % i == 0) {
        while (n % i == 0) n /= i;
        result -= result / i;
      }
    }
    if (n > 1) result -= result / n;
    return result;
  }
  vector<long long> phi_1_to_n(long long n) {
    vector<long long> phi(n + 1);
    phi[0] = 0;
    phi[1] = 1;
    for (long long i = 2; i <= n; i++) phi[i] = i;
    for (long long i = 2; i <= n; i++) {
      if (phi[i] == i) {
        for (long long j = i; j <= n; j += i) phi[j] -= phi[j] / i;
      }
    }
    return phi;
  }
  }  // namespace PhiFunction
  const long long MAX = 1e5 + 5, MOD = 1e9 + 7;
  long long n, a[MAX];
  vector<long long> g[MAX];
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    vector<long long> phi = PhiFunction::phi_1_to_n(100000);
    for (long long i = 1; i <= 1e5; i++)
      for (long long j = i; j <= 1e5; j += i) g[j].push_back(i);
    cin >> n;
    for (long long i = 1; i <= n; i++) cin >> a[i];
    long long rs = 0;
    for (long long d1 = 1; d1 <= n; d1++) {
      unordered_map<long long, long long> C;
      for (long long p = d1; p <= n; p += d1)
        for (long long d2 : g[a[p]]) C[d2]++;
      for (auto tmp : C) {
        long long d2 = tmp.first;
        long long val = tmp.second * tmp.second % MOD;
        rs = (rs + phi[d1] * phi[d2] % MOD * val) % MOD;
      }
    }
    cout << rs;
  }
  