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 <cstring>
#include <algorithm>
#include <iostream>
#define ll long long
#define N 100005
#define inf 0x3f3f3f3f
using namespace std;
ll sum,a[N],t,tmp,n;
int main() {
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>a[i];
		sum+=a[i];
		t+=i;
	}

	if(sum%t)cout<<"NO";
	else{
		tmp=a[1];
		for(int i=1;i<n;i++)a[i]=a[i+1]-a[i];
		a[n]=tmp-a[n];
		ll d=sum/t,ok=1,ans=0;
		for(int i=1;i<=n&&ok;i++){
			a[i]-=d;
			if(a[i]%n||a[i]>0)ok=0;
			ans+=a[i]/n;
		}
		if(ok&&ans==-d)cout<<"YES";
		else cout<<"NO";
	}
	return 0;
}