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 optimize("Ofast")
  using namespace std;
  const int MOD = (int)1e9 + 7;
  const int N = (int)2e5;
  const long long INF = (long long)1e18;
  int T;
  int n;
  int v[100005], w[100005];
  int dp[100005], dp2[100005];
  int f[100005];
  vector<int> divs[100005];
  int lgput(int n, long long p) {
    int ans = 1, first = n;
    while (p) {
      if (p & 1) ans = 1LL * ans * first % MOD;
      first = 1LL * first * first % MOD;
      p >>= 1;
    }
    return ans;
  }
  void precalc() {
    for (int i = 1; i <= 100000; i++) {
      for (int j = i; j <= 100000; j += i) divs[j].push_back(i);
    }
  }
  int solve(int v[], int n) {
    int ans = 0;
    vector<int> vals;
    for (int i = 1; i <= n; i++) {
      for (auto &d : divs[v[i]]) {
        if (!f[d]) vals.push_back(d);
        f[d]++;
      }
    }
    for (int i = (int)vals.size() - 1; i >= 0; i--) {
      int nr = vals[i];
      dp2[nr] = (dp2[nr] + 1LL * f[nr] * f[nr]) % MOD;
      ans = (ans + 1LL * nr * dp2[nr]) % MOD;
      for (auto &d : divs[nr]) {
        dp2[d] = (dp2[d] - dp2[nr] + MOD) % MOD;
      }
    }
    for (auto &d : vals) f[d] = dp2[d] = 0;
    return ans;
  }
  void solve() {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> v[i];
    int ans = 0;
    for (int d = n; d >= 1; d--) {
      int m = 0;
      for (int i = d; i <= n; i += d) {
        w[++m] = v[i];
      }
      dp[d] = solve(w, m);
      for (int i = 2 * d; i <= n; i += d) {
        dp[d] = (dp[d] - dp[i] + MOD) % MOD;
      }
      ans = (ans + 1LL * d * dp[d]) % MOD;
    }
    cout << ans << "\n";
  }
  int main() {
    ios_base ::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    T = 1;
    precalc();
    for (; T; T--) {
      solve();
    }
    return 0;
  }
  