#include <bits/stdc++.h>
using namespace std;
int phi[100001];
vector<int> v[100001];
int a[100001], cnt[100001];
const int MOD = 1000000007;
int main() {
  iota(phi, phi + 100001, 0);
  for (int i = 1; i <= 100000; i++) {
    v[i].push_back(i);
    for (int j = i + i; j <= 100000; j += i) {
      phi[j] -= phi[i];
      v[j].push_back(i);
    }
  }
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long t = 0;
    for (int j = i; j <= n; j += i) {
      for (int x : v[a[j]]) {
        t = (t + phi[x] * cnt[x]) % MOD;
        cnt[x]++;
      }
    }
    ans = (ans + t * phi[i]) % MOD;
    for (int j = i; j <= n; j += i) {
      for (int x : v[a[j]]) cnt[x]--;
    }
  }
  ans = (ans * 2) % MOD;
  for (int i = 1; i <= n; i++) {
    ans = (ans + 1ll * i * a[i]) % MOD;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
int phi[100001];
vector<int> v[100001];
int a[100001], cnt[100001];
const int MOD = 1000000007;
int main() {
  iota(phi, phi + 100001, 0);
  for (int i = 1; i <= 100000; i++) {
    v[i].push_back(i);
    for (int j = i + i; j <= 100000; j += i) {
      phi[j] -= phi[i];
      v[j].push_back(i);
    }
  }
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long t = 0;
    for (int j = i; j <= n; j += i) {
      for (int x : v[a[j]]) {
        t = (t + phi[x] * cnt[x]) % MOD;
        cnt[x]++;
      }
    }
    ans = (ans + t * phi[i]) % MOD;
    for (int j = i; j <= n; j += i) {
      for (int x : v[a[j]]) cnt[x]--;
    }
  }
  ans = (ans * 2) % MOD;
  for (int i = 1; i <= n; i++) {
    ans = (ans + 1ll * i * a[i]) % MOD;
  }
  printf("%lld\n", ans);
  return 0;
}

```