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;
  long long read() {
    long long x;
    scanf("%lld", &x);
    return x;
  }
  const long long mod = 1000000007;
  const long long N = 100010, p = 100000, K = 150;
  long long n, a[N], c[N], w[N], mu[N], pr[N], tot, q[N], ans;
  bool vis[N];
  inline long long gcd(long long x, long long y) {
    return y == 0 ? x : gcd(y, x % y);
  }
  long long calc() {
    long long ret = 0;
    for (long long i = 1; i <= p; ++i) {
      long long sum = 0;
      for (long long j = i; j <= p; j += i) sum += c[j];
      sum *= sum;
      ret += sum % mod * w[i] % mod;
    }
    return ret % mod;
  }
  int main() {
    n = read();
    for (long long i = (long long)(1); i <= (long long)(n); ++i) a[i] = read();
    mu[1] = 1;
    for (long long i = 2; i <= p; ++i) {
      if (!vis[i]) {
        pr[++tot] = i;
        mu[i] = -1;
      }
      for (long long j = 1; j <= tot; ++j) {
        if (i * pr[j] > p) break;
        vis[i * pr[j]] = 1;
        if (i % pr[j] == 0)
          mu[i * pr[j]] = 0;
        else
          mu[i * pr[j]] = -mu[i];
        if (i % pr[j] == 0) break;
      }
    }
    for (long long i = 1; i <= p; ++i) {
      for (long long j = 1; i * j <= p; ++j) {
        w[i * j] += mu[i] * j;
      }
    }
    for (long long i = (long long)(1); i <= (long long)(n); ++i) {
      long long m = 0;
      for (long long j = i; j <= n; j += i) {
        q[++m] = a[j];
      }
      if (m <= K) {
        long long sum = 0;
        for (long long x = 1; x <= m; ++x)
          for (long long y = 1; y <= m; ++y) sum += gcd(q[x], q[y]);
        ans += sum % mod * w[i] % mod;
      } else {
        memset(c, 0, sizeof c);
        for (long long x = 1; x <= m; ++x) ++c[q[x]];
        ans += calc() * w[i] % mod;
      }
    }
    printf("%lld\n", (ans % mod + mod) % mod);
    return 0;
  }
  