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;

const int maxn = 1e5+10;
long long a[maxn],sum,t,n;

int main(){
	cin >> n;
	for(int i=0; i<n; i++) {
		cin>>a[i];
		sum += a[i];
	}
	t = n*(n+1)/2;
	long long k = sum/t;
	int flag = 1;
	if(sum%t) flag = 0;
	else{
		long long tmp,tt=a[0];
		a[0] = a[0]-a[n-1]-k;
		for(int i=1; i<n; i++){
			tmp = a[i];
		 	a[i] = a[i]-tt;
		 	a[i] -= k;
		 	tt = tmp;
		}
		for(int i=0; i<n; i++){
			if(a[i]>0 || a[i]%n)
				flag = 0;
		}
	}


	if(flag) puts("YES");
	else puts("NO");
}
