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<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#define ll long long
using namespace std;

const int N=100005;

int n;
ll sum,d;
ll a[N];

int main(){
	int i;
	scanf("%d",&n);
	for (i=1;i<=n;i++){
		scanf("%lld",&a[i]);
		sum+=a[i];
	}
	a[n+1]=a[1];
	for (i=1;i<=n;i++) a[i]=a[i+1]-a[i];
	if (sum%(1ll*n*(n+1)/2)) printf("NO\n");
	else{
		d=sum/(1ll*n*(n+1)/2);
		for (i=1;i<=n;i++){
			a[i]-=d;
			if (a[i]%n!=0||a[i]>0){
				printf("NO\n");
				return 0;
			}
		}
		printf("YES\n");
	}
	return 0;
}