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>
using namespace std;
typedef long long ll;
const int inf=2000000000;
ll m,j,k;
int i,n,a[200001],d[200001],tot,b[200001];
int main(){
	scanf("%d",&n);m=(ll)(n)*(ll)(n+1)/2LL;
	for (i=1;i<=n;i++) scanf("%d",&a[i]),a[i]%=m;
	for (a[0]=a[n],i=1,k=-inf;i<=n;i++) d[i]=a[i]-a[i-1],k=max(k,(ll)d[i]);
	for (i=1;i<=n;i++)
	 if ((k-d[i])%n){
	 	puts("NO");return 0;
	 }else b[i]=(k-d[i])/n;
	for (i=2,j=a[1]-b[1];i<=n;i++) j-=(ll)(n-i+2)*(ll)b[i];
	puts(j?"NO":"YES");
	return 0;
}