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<cmath>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
#define ll long long
ll n,num[1000005],sum,ans=0;
int main()
{
	scanf("%lld",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%lld",&num[i]);
		sum+=num[i];
	}
	if(sum%((n+1)*n/2)){printf("NO");return 0;}
	sum/=(n+1)*n/2;int tmp=num[1];
	for(int i=1;i<n;i++)
	{
		num[i]-=num[i+1];
	}num[n]-=tmp;
	for(int i=1;i<=n;i++)
	{
		if(num[i]<-sum||(num[i]+sum)%n)
		{printf("NO");return 0;}
		ans+=(sum+num[i])/n;
	}
	if(ans==sum)printf("YES");
	else printf("NO");
	return 0;
}