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<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
#define N 100010
using namespace std;
int n, w[N];
ll s, m, x, y, a[N];
int main(){
	scanf("%d", &n);
	for(int i=1; i<=n; i++)scanf("%d", &w[i]);
	if(n==1){printf("YES"); return 0;}
	s=0; for(int i=1; i<=n; i++)s+=w[i]; m=(ll)n*(n+1)/2;
	if(s%m){printf("NO"); return 0;}
	x=s/m; w[n+1]=w[1];
	for(int i=1; i<=n; i++){
		y=w[i+1]-w[i]-x;
		if(y>=1||y%n){printf("NO"); return 0;}
		a[i+1]=-y/n;
	}
	a[1]=a[n+1];
	s=0; for(int i=1; i<=n; i++)s+=a[i];
	if(s!=x)printf("NO"); else printf("YES");
	return 0;
}