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;
  long long get_time() {
    return chrono::duration_cast<chrono::nanoseconds>(
               chrono::steady_clock::now().time_since_epoch())
        .count();
  }
  template <typename T1, typename T2>
  string print_iterable(T1 begin_iter, T2 end_iter, int counter) {
    bool done_something = false;
    stringstream res;
    res << "[";
    for (; begin_iter != end_iter and counter; ++begin_iter) {
      done_something = true;
      counter--;
      res << *begin_iter << ", ";
    }
    string str = res.str();
    if (done_something) {
      str.pop_back();
      str.pop_back();
    }
    str += "]";
    return str;
  }
  template <typename S, typename T>
  ostream& operator<<(ostream& out, const pair<S, T>& p) {
    out << "{" << p.first << ", " << p.second << "}";
    return out;
  }
  template <typename T>
  ostream& operator<<(ostream& out, const vector<T>& v) {
    out << "[";
    for (int i = 0; i < (int)v.size(); i++) {
      out << v[i];
      if (i != (int)v.size() - 1) out << ", ";
    }
    out << "]";
    return out;
  }
  template <class TH>
  void _dbg(const char* name, TH val) {
    clog << name << ": " << val << endl;
  }
  template <class TH, class... TA>
  void _dbg(const char* names, TH curr_val, TA... vals) {
    while (*names != ',') clog << *names++;
    clog << ": " << curr_val << ", ";
    _dbg(names + 1, vals...);
  }
  const int MAXN = 100100;
  int ppp[MAXN];
  int phi[MAXN];
  vector<int> divisors[MAXN];
  const long long mod = 1e9 + 7;
  int cnt[MAXN];
  long long solve(const vector<int>& a) {
    vector<int> used;
    for (int x : a) {
      for (int d : divisors[x]) {
        cnt[d] += 1;
        used.push_back(d);
      }
    }
    long long res = 0;
    for (int d : used) {
      if (cnt[d] == 0) continue;
      long long c = cnt[d];
      res += (phi[d] * c * c) % mod;
      cnt[d] = 0;
    }
    return res % mod;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    for (int p = 2; p < MAXN; p++) {
      if (ppp[p]) continue;
      for (int d = p; d < MAXN; d += p) ppp[d] = p;
    }
    for (int n = 1; n < MAXN; n++) {
      int m = n;
      phi[n] = 1;
      while (m != 1) {
        int p = ppp[m];
        m /= p;
        phi[n] *= p - 1;
        while (m % p == 0) m /= p, phi[n] *= p;
      }
    }
    for (int d = 1; d < MAXN; d++) {
      for (int n = d; n < MAXN; n += d) divisors[n].push_back(d);
    }
    int n;
    cin >> n;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++) cin >> a[i];
    long long res = 0;
    for (int d = 1; d < MAXN; d++) {
      vector<int> vec;
      for (int i = d; i <= n; i += d) vec.push_back(a[i]);
      res += phi[d] * solve(vec) % mod;
    }
    cout << res % mod << "\n";
  }
  