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 F(i,j,k) for(int i=(j);i<=(k);++i)
using namespace std;
#define ll long long
const int N =1e5+5;
int n;
ll w[N],sum,num;
int main(){
	//freopen("txt.in","r",stdin);freopen("txt.out","w",stdout);
	scanf("%d\n",&n);
	F(i,1,n)scanf("%lld",&w[i]),sum+=w[i];
	if(sum%((ll)n*(n+1)/2))printf("NO\n");
	else{
		num=sum/((ll)n*(n+1)/2);
		w[0]=w[n];
		F(i,1,n)if((w[i]-w[i-1]-num)%n||(w[i]-w[i-1]-num)>0){printf("NO\n");return 0;}
		printf("YES\n");return 0;
	}
	return 0;
}
//w(i)-w(i-1)=x*1+y(1-n)=x+y-yn=sum-yn,yn=w[i-1]-w[i]+sum 
//x+y=sum