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>
  #pragma GCC target("avx2")
  #pragma GCC optimize("O3")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  template <class T, class S>
  ostream &operator<<(ostream &o, const pair<T, S> &p) {
    return o << '(' << p.first << ", " << p.second << ')';
  }
  template <template <class, class...> class T, class... A>
  typename enable_if<!is_same<T<A...>, string>(), ostream &>::type operator<<(
      ostream &o, T<A...> V) {
    o << '[';
    for (auto a : V) o << a << ", ";
    return o << ']';
  }
  int n, a[100010], b[100010];
  long long int ansg[100010], ct[100010];
  vector<int> divs[100010];
  int v[100010 * 150];
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    for (int i = 1; i < (100010); ++i)
      for (int j = 1; j * j <= i; ++j)
        if (!(i % j)) {
          divs[i].push_back(j);
          if (j * j - i) divs[i].push_back(i / j);
        }
    cin >> n;
    for (int i = 1; i < (n + 1); ++i) cin >> a[i];
    for (int g = 1; g < (n + 1); ++g) {
      long long int vsz = 0, bsz = 0;
      for (int i = g; i <= n; i += g) b[bsz++] = a[i];
      sort(b, b + bsz);
      long long int c = 0;
      for (int i = 0; i < (bsz); ++i) {
        c++;
        if (i == bsz - 1 || b[i] - b[i + 1]) {
          for (long long int z : divs[b[i]]) {
            v[vsz++] = z;
            ct[z] += c;
          }
          c = 0;
        }
      }
      sort(v, v + vsz);
      vsz = unique(v, v + vsz) - v;
      for (int i = 0; i < (vsz); ++i) ct[v[i]] *= ct[v[i]];
      reverse(v, v + vsz);
      for (int i = 0; i < (vsz); ++i) {
        long long int x = v[i];
        for (int y : divs[x])
          if (y - x) ct[y] -= ct[x];
        ct[x] %= 1000000007;
        ansg[g] = (ansg[g] + x * ct[x]) % 1000000007;
      }
      for (int i = 0; i < (vsz); ++i) ct[v[i]] = 0;
    }
    long long int ans = 0;
    for (int i = n; i > (0); --i) {
      for (int j = 2 * i; j <= n; j += i)
        ansg[i] = (ansg[i] + 1000000007 - ansg[j]) % 1000000007;
      ans = (ans + ansg[i] * i) % 1000000007;
    }
    cout << ans << '\n';
  }
  