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>
# define rep(i,a,b) for(int i=(a);i<=(b);++i)
typedef long long ll;
using namespace std;
const int maxn = 1e5+10;
int n;
int a[maxn];
int main(){
	scanf("%d",&n);ll sum=0;
	rep(i,1,n){ scanf("%d",a+i);sum+=a[i]; }
	if(sum % (1ll*n*(n+1)/2)!=0){puts("NO");return 0;}
	ll k=sum / (1ll*n*(n+1)/2);
	a[0]=a[n];
	ll ans = 0;
	rep(i,1,n){
		ll now=k-a[i]+a[i-1];
		if (now%n ||now<0){puts("NO");return 0;}
		ans+=now/n;
	}
	if (ans!=k) puts("NO"); else printf("YES\n");
	return 0;
}