#include <bits/stdc++.h>
using namespace std;

int main(){
  long long N;
  cin >> N;
  vector<long long> A(N);
  long long k=0;
  for (int i=0; i<N; i++){
    cin >> A[i];
    k+=A[i];
  }
  if (k%(N*(N+1)/2) != 0){
    cout << "NO" << endl;
    return 0;
  }
  k/=N*(N+1)/2;
  vector<long long> B(N);
  for (int i=0; i<N; i++){
    B[i] = A[(i+1)%N] - A[i];
    B[i] -= k;
  }
  for (int i=0; i<N; i++){
    if (B[i] > 0){
      cout << "NO" << endl;
      return 0;
    }
    if ((-B[i])%N != 0){
      cout << "NO" << endl;
      return 0;
    }
  }
  cout << "YES" << endl;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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;

int main(){
  long long N;
  cin >> N;
  vector<long long> A(N);
  long long k=0;
  for (int i=0; i<N; i++){
    cin >> A[i];
    k+=A[i];
  }
  if (k%(N*(N+1)/2) != 0){
    cout << "NO" << endl;
    return 0;
  }
  k/=N*(N+1)/2;
  vector<long long> B(N);
  for (int i=0; i<N; i++){
    B[i] = A[(i+1)%N] - A[i];
    B[i] -= k;
  }
  for (int i=0; i<N; i++){
    if (B[i] > 0){
      cout << "NO" << endl;
      return 0;
    }
    if ((-B[i])%N != 0){
      cout << "NO" << endl;
      return 0;
    }
  }
  cout << "YES" << endl;
}

```