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;
int main()
{
  ios_base::sync_with_stdio(false);
  int n , a[100002] , i ;
  bool f = true;
  long long int sum = 0 , x;
  cin >> n;
  for(i=0;i<n;i++)
  {
    cin >> a[i];
    sum = sum + a[i];
  }
  long long int ch = ((long long)n*(n+1))/2;
  if(sum%ch)
    f = false;
  x = sum/ch;
  for(i=0;i<n;i++)
  {
    int d  = a[(i+1)%n] - a[i];
    if(x - d < 0) f = false;
    if((x-d)%n) f = false;
  }
  if(f)
    cout << "YES\n";
  else cout << "NO\n";
  return 0;
}
