#include <bits/stdc++.h>
using namespace std;
signed main() {
  cin.tie(0); ios::sync_with_stdio(false);
  int n; cin >> n;
  int a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  long long sum = accumulate(a, a + n, 0LL);
  if (sum % ((n + 1LL) * n / 2) != 0) {
    cout << "NO" << "\n";
    return 0;
  }
  int m = sum / ((n + 1LL) * n / 2), t = 0;
  for (int i = 0; i < n; i++) {
    int d = a[(i + 1) % n] - a[i % n];
    if ((m - d) % n != 0 or d > m) {
      cout << "NO" << "\n";
      return 0;
    }
    t += (m - d) / n;
  }
  cout << (t == m ? "YES" : "NO") << "\n";
  return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

There are N boxes arranged in a circle. The i-th box contains A_i stones.

Determine whether it is possible to remove all the stones from the boxes by repeatedly performing the following operation:

* Select one box. Let the box be the i-th box. Then, for each j from 1 through N, remove exactly j stones from the (i+j)-th box. Here, the (N+k)-th box is identified with the k-th box.



Note that the operation cannot be performed if there is a box that does not contain enough number of stones to be removed.

Constraints

* 1 ≦ N ≦ 10^5
* 1 ≦ A_i ≦ 10^9

Input

The input is given from Standard Input in the following format:


N
A_1 A_2 … A_N


Output

If it is possible to remove all the stones from the boxes, print `YES`. Otherwise, print `NO`.

Examples

Input

5
4 5 1 2 3


Output

YES


Input

5
6 9 12 10 8


Output

YES


Input

4
1 2 3 1


Output

NO

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
signed main() {
  cin.tie(0); ios::sync_with_stdio(false);
  int n; cin >> n;
  int a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  long long sum = accumulate(a, a + n, 0LL);
  if (sum % ((n + 1LL) * n / 2) != 0) {
    cout << "NO" << "\n";
    return 0;
  }
  int m = sum / ((n + 1LL) * n / 2), t = 0;
  for (int i = 0; i < n; i++) {
    int d = a[(i + 1) % n] - a[i % n];
    if ((m - d) % n != 0 or d > m) {
      cout << "NO" << "\n";
      return 0;
    }
    t += (m - d) / n;
  }
  cout << (t == m ? "YES" : "NO") << "\n";
  return 0;
}
```