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 fi first
#define se second
#define mp make_pair
#define int long long
int n;
int a[100007],b[100007];
pair<int,int> sv[100007];
long long sum=0,sum2=0;
signed main(){
	ios_base::sync_with_stdio(0);
	cin>>n;
	for(int i=1;i<=n;i++) cin>>a[i],sum2+=i,sum+=a[i];
	for(int i=1;i<n;i++){
		b[i]=a[i+1]-a[i];
	}
	b[n]=a[1]-a[n];
	if(sum%sum2!=0){
		cout<<"NO";
		return 0;
	}
	sum2=sum/sum2;
	for(int i=1;i<=n;i++){
		b[i]-=sum2;
		if(b[i]>0 || b[i]%n!=0){
			cout<<"NO";
			return 0;
		}
	}
	cout<<"YES";
}
