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 inf = 1 << 30;
  const long long Inf = 1ll << 60ll;
  namespace io {
  char buf[(1 << 23)], *p1 = buf, *p2 = buf, c;
  int f;
  template <typename T>
  T read() {
    T x = 0;
    f = 0;
    c = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, (1 << 23), stdin), p1 == p2)
             ? EOF
             : *p1++);
    while (!isdigit(c))
      (c == '-') && (f = 1),
          c = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, (1 << 23), stdin),
                            p1 == p2)
                   ? EOF
                   : *p1++);
    while (isdigit(c))
      x = x * 10 + (c & 15),
      c = (p1 == p2 &&
                   (p2 = (p1 = buf) + fread(buf, 1, (1 << 23), stdin), p1 == p2)
               ? EOF
               : *p1++);
    return f ? -x : x;
  }
  }  // namespace io
  const int N = 100005;
  const long long mod = 1000000007;
  void reduce(int& x) { (x >= mod) && (x -= mod); }
  int n, a[N], cnt[N], phi[N], ans;
  vector<int> v, fac[N];
  int main() {
    for (int i = 1; i <= 100000; ++i) phi[i] = i;
    for (int i = 1; i <= 100000; ++i)
      for (int j = 2; i * j <= 100000; ++j) phi[i * j] -= phi[i];
    for (int i = 1; i <= 100000; ++i)
      for (int j = 1; i * j <= 100000; ++j) fac[i * j].push_back(i);
    n = io::read<int>();
    for (int i = 1; i <= n; ++i) a[i] = io::read<int>();
    for (int d = 1; d <= n; ++d) {
      v.clear();
      for (int i = d; i <= n; i += d) {
        for (int f : fac[a[i]]) {
          if (cnt[f] == 0) v.push_back(f);
          ++cnt[f];
        }
      }
      for (int f : v) {
        reduce(ans += 1ll * phi[d] * phi[f] % mod * cnt[f] % mod * cnt[f] % mod);
        cnt[f] = 0;
      }
    }
    printf("%d\n", ans);
    return 0;
  }
  