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
#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

int N;
ll a[100001];
ll d[100001];
ll sum;
ll sum_a;

int main() {
  cin >> N;

  for(int i = 0; i < N; i++) {
    cin >> a[i];
    sum_a += a[i];
  }

  for(int i = 1; i <= N; i++) {
    sum += i;
    d[i%N] = a[i%N] - a[i-1];
  }

  if(sum_a % sum != 0) {
    cout <<"NO"<<endl;
    return 0;
  }

  ll k = sum_a / sum;

  bool is = true;
  for(int i = 0; i < N; i++) {
    if(d[i] > k || (k-d[i]) % N != 0) {
      is = false;
      break;
    }
  }

  if(is) cout << "YES" <<endl;
  else cout << "NO" <<endl;
}
    
