#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
const int M = 1e6 + 10;
const int mod = 1e9 + 7;
long long phi[N], vis[N], prime[N];
vector<int> vv[N];
int cnt;
void init() {
  phi[1] = 1;
  for (int i = 2; i < N; i++)
    if (!phi[i]) {
      for (int j = i; j < N; j += i) {
        if (!phi[j]) phi[j] = j;
        phi[j] = phi[j] / i * (i - 1);
      }
    }
}
long long a[N];
void run() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < N; i++) {
    for (int j = i; j < N; j += i) {
      vv[j].push_back(i);
    }
  }
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  long long ans = 0;
  unordered_map<long long, long long> mp;
  for (int i = 1; i <= n; i++) {
    long long tot = 0;
    mp.clear();
    for (int j = 1; j <= n / i; j++) {
      for (auto it : vv[a[j * i]]) {
        mp[it]++;
      }
    }
    for (auto it : mp) {
      tot = (tot + phi[it.first] * it.second % mod * it.second % mod) % mod;
    }
    ans = (ans + tot * phi[i] % mod) % mod;
  }
  printf("%lld\n", ans);
}
int main() {
  cnt = 0;
  init();
  run();
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
const int M = 1e6 + 10;
const int mod = 1e9 + 7;
long long phi[N], vis[N], prime[N];
vector<int> vv[N];
int cnt;
void init() {
  phi[1] = 1;
  for (int i = 2; i < N; i++)
    if (!phi[i]) {
      for (int j = i; j < N; j += i) {
        if (!phi[j]) phi[j] = j;
        phi[j] = phi[j] / i * (i - 1);
      }
    }
}
long long a[N];
void run() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < N; i++) {
    for (int j = i; j < N; j += i) {
      vv[j].push_back(i);
    }
  }
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  long long ans = 0;
  unordered_map<long long, long long> mp;
  for (int i = 1; i <= n; i++) {
    long long tot = 0;
    mp.clear();
    for (int j = 1; j <= n / i; j++) {
      for (auto it : vv[a[j * i]]) {
        mp[it]++;
      }
    }
    for (auto it : mp) {
      tot = (tot + phi[it.first] * it.second % mod * it.second % mod) % mod;
    }
    ans = (ans + tot * phi[i] % mod) % mod;
  }
  printf("%lld\n", ans);
}
int main() {
  cnt = 0;
  init();
  run();
  return 0;
}

```