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<iostream>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int A[110000],n;
long long ans[110000];
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d",&A[i]);
	long long tot=0;
	for (int i=1;i<=n;i++) tot+=A[i];
	long long one=1ll*n*(n+1)/2;
	if (tot%one){
		printf("NO\n"); return 0;
	}
	long long t=tot/one; A[n+1]=A[1];
	for (int i=2;i<=n+1;i++){
		long long w=A[i-1]-A[i]+t;
		if (w<0||w%n){
			printf("NO\n"); return 0; 
		}
	}
	printf("YES\n");
	return 0;
}