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>
  template <class T>
  inline T read() {
    T r = 0, f = 0;
    char c;
    while (!isdigit(c = getchar())) f |= (c == '-');
    while (isdigit(c)) r = (r << 1) + (r << 3) + (c ^ 48), c = getchar();
    return f ? -r : r;
  }
  bool n_p[101101];
  int s_p, p[101101], phi[101101];
  std::vector<int> fact[101101];
  inline void init(int n) {
    phi[1] = 1;
    for (int i = 2; i <= n; i++) {
      if (!n_p[i]) p[++s_p] = i, phi[i] = i - 1;
      for (int j = 1; j <= s_p && p[j] <= n / i; j++) {
        n_p[i * p[j]] = 1;
        if (!(i % p[j])) {
          phi[i * p[j]] = phi[i] * p[j];
          break;
        }
        phi[i * p[j]] = phi[i] * (p[j] - 1);
      }
    }
    for (int i = 1; i <= n; i++)
      for (int j = i; j <= n; j += i) fact[j].push_back(i);
  }
  int n, a[101101], cnt[101101];
  int main() {
    n = read<int>();
    for (int i = 1; i <= n; i++) a[i] = read<int>();
    init(1e5);
    long long ans = 0;
    for (int d = 1; d <= n; d++) {
      long long s = 0;
      for (int i = d; i <= n; i += d)
        for (auto x : fact[a[i]]) {
          (s += 1ll * (2 * cnt[x] + 1) * phi[x]) %= 1000000007;
          cnt[x]++;
        }
      (ans += phi[d] * s) %= 1000000007;
      for (int i = d; i <= n; i += d)
        for (auto x : fact[a[i]]) cnt[x] = 0;
    }
    printf("%lld\n", ans);
    return 0;
  }
  