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<cmath>
#define MN 100005
using namespace std;
typedef long long ll;
int n,a[MN],b[MN];ll sum;
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]),sum+=a[i];
	if(sum%(1ll*n*(n+1)/2))return 0*puts("NO");ll round=sum/(1ll*(n+1)*n/2);
	for(int i=1;i<n;i++)b[i]=a[i+1]-a[i];b[n]=a[1]-a[n];
	for(int i=1;i<=n;i++){
		b[i]-=round;
		if(b[i]>0||b[i]%n)return 0*puts("NO");
	}puts("YES");
}
