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;
  using ll = long long;
  using ld = long double;
  using str = string;
  using pi = pair<int, int>;
  using pl = pair<ll, ll>;
  using vi = vector<int>;
  using vl = vector<ll>;
  using vpi = vector<pair<int, int>>;
  using vvi = vector<vi>;
  const int md = 1e9 + 7;
  int add(const int &a, const int &b) { return a + b >= md ? a + b - md : a + b; }
  int sub(const int &a, const int &b) { return a - b < 0 ? a - b + md : a - b; }
  int mul(const int &a, const int &b) { return (1ll * a * b) % md; }
  const int maxN = 1e5 + 5;
  int a[maxN];
  vi divs[maxN];
  int sm[maxN];
  int sumg[maxN];
  int cnt[maxN];
  ll too[maxN];
  void solve() {
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 1; i < maxN; ++i) {
      for (int j = i; j < maxN; j += i) {
        divs[j].push_back(i);
      }
    }
    for (int i = 1; i < maxN; ++i) {
      sm[i] = add(i, sm[i]);
      for (int j = i * 2; j < maxN; j += i) {
        sm[j] = sub(sm[j], sm[i]);
      }
    }
    int ans = 0;
    for (int g = n; g >= 1; --g) {
      int cur = 0;
      for (int d = g; d <= n; d += g) {
        for (auto &div : divs[a[d]]) {
          cur = add(cur, mul(cnt[div], sm[div]));
          cnt[div]++;
        }
        cur = sub(cur, sumg[d]);
      }
      for (int d = g; d <= n; d += g) {
        for (auto &div : divs[a[d]]) {
          cnt[div]--;
        }
      }
      sumg[g] = cur;
      ans = add(ans, mul(2, mul(cur, g)));
      ans = add(ans, mul(a[g], g));
    }
    cout << ans << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int t = 1;
    for (int i = 0; i < (t); ++i) {
      solve();
    }
    return 0;
  }
  