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(){
	long long n;
	scanf("%lld",&n);
	long long arr[n+1],ops=0;
	for(int i=0;i<n;i++){
		scanf("%lld",&arr[i]);ops+=arr[i];
	}
	arr[n]=arr[0];
	if(ops%(n*(n+1)/2)!=0){printf("NO");return 0;}
	ops/=(n*(n+1)/2);
	for(int i=0;i<n;i++){
		if((ops+arr[i]-arr[i+1])%n==0&&ops+arr[i]-arr[i+1]>=0){
		}else{
			printf("NO");return 0;
		}
	}
	printf("YES");
}
