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 mod = 1000000007;
  const int N = 100010;
  int a[N];
  int miu[N];
  int dmiu[N];
  int primes[N];
  vector<int> fac[N];
  bool vis[N];
  int phi[N];
  void init() {
    int tot = 0;
    miu[1] = phi[1] = 1;
    for (int i = 2; i < N; i++) {
      if (!vis[i]) {
        miu[i] = -1;
        primes[tot++] = i;
        phi[i] = i - 1;
      }
      for (int j = 0; i * primes[j] < N && j < tot; j++) {
        int x = i * primes[j];
        vis[x] = true;
        if (i % primes[j] == 0)
          miu[x] = 0, phi[x] = phi[i] * primes[j];
        else
          miu[x] = -miu[i], phi[x] = phi[i] * (primes[j] - 1);
        if (i % primes[j] == 0) break;
      }
    }
    for (int i = 1; i < N; i++) {
      for (int j = i; j < N; j += i) {
        dmiu[j] += miu[i] * (j / i);
        fac[j].push_back(i);
      }
    }
  }
  int _gcd(int a, int b) { return b ? _gcd(b, a % b) : a; }
  long long sum[N];
  int occ[N];
  int times;
  int solve(vector<int> &v) {
    long long ans = 0;
    ++times;
    for (int i = 0; i < v.size(); i++) {
      int x = v[i];
      for (int j = 0; j < fac[x].size(); j++) {
        int y = fac[x][j];
        if (occ[y] != times) {
          occ[y] = times;
          sum[y] = phi[y];
        } else {
          sum[y] += phi[y];
        }
      }
    }
    for (int i = 0; i < v.size(); i++) {
      int x = v[i];
      for (int j = 0; j < fac[x].size(); j++) {
        int y = fac[x][j];
        ans += sum[y];
      }
    }
    return ans % mod;
  }
  int main() {
    init();
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      vector<int> v;
      for (int j = i; j <= n; j += i) {
        v.push_back(a[j]);
      }
      ans = (ans + 1LL * dmiu[i] * solve(v)) % mod;
    }
    if (ans < 0) ans += mod;
    cout << ans << endl;
    return 0;
  }
  