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>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;
using lint=long long;

int main(){
	lint n; scanf("%lld",&n);
	vector<lint> a(n);
	rep(i,n) scanf("%lld",&a[i]);

	lint sum=0;
	rep(i,n) sum+=a[i];
	if(sum%(n*(n+1)/2)!=0) return puts("NO"),0;

	lint m=sum/(n*(n+1)/2),cnt=0;
	rep(i,n){
		lint d=a[(i+1)%n]-a[i];
		if(m-d<0 || (m-d)%n!=0) return puts("NO"),0;
		cnt+=(m-d)/n;
	}
	puts(cnt==m?"YES":"NO");

	return 0;
}
