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;
using i64 = long long;
#define rep(i,s,e) for(int (i) = (s);(i) <= (e);(i)++)

i64 N;
vector<i64> a;
int main(){
  cin >> N;
  a.resize(N);
  i64 sum = 0;
  rep(i,0,N - 1){
    cin >> a[i];
    sum += a[i];
  }

  if(sum % (N * (N + 1) / 2) != 0){
    cout << "NO" << endl;
    return 0;
  }

  i64 need = sum / (N * (N + 1) / 2);

  rep(i,0,N - 1){
    int j = (i + 1) % N;
    i64 d = a[j] - a[i] - need;
    if(0 < d || d % N != 0){
      cout << "NO" << endl;
      return 0;
    }
  }

  cout << "YES" << endl;
  return 0;
}
