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;
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
int n,a[111111],b[111111],k,x,cnt;
signed main(){
	cin>>n;
	rep(i,n){
		cin>>a[i];
		k+=a[i];
	}
	if(k%((n+1)*n/2)){
		puts("NO");
		return 0;
	}
	k/=((n+1)*n/2);
	rep(i,n){
		if(i)b[i]=a[i]-a[i-1];
		else b[i]=a[0]-a[n-1];
		b[i]-=k;
		if(b[i]>0){
			puts("NO");
			return 0;
		}
	}
	rep(i,n){
		if(b[i]<0){
			if(-(b[i])%n==0)cnt+=-(b[i])/n;
		}
	}
	if(cnt==k)puts("YES");
	else puts("NO");
	return 0;
}
