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;
long long n,a[100005];
int main(void)
{
	int flag=1;
	long long sum=0;
	cin>>n;
	for (int i = 0; i < n; i++) {
		cin>>a[i];
		sum+=a[i];
	}
	if(sum%((1+n)*n/2)!=0)flag=0;
	long long m=sum/((1+n)*n/2);
	for (int i = 0; i < n; i++) {
		if((a[(i+n-1)%n]-a[i]+m)%n!=0)flag=0;
		if((a[(i+n-1)%n]-a[i]+m)/n <0)flag=0;

	}
	if(flag==1)cout<<"YES"<<endl;
	else cout<<"NO"<<endl;
	return 0;
}
