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;
typedef long long ll;
const int N=1e5+10;
int n;ll a[N],sum,cnt;
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%lld",&a[i]),sum+=a[i];
	a[n+1]=a[1];
	cnt=1ll*n*(n+1)/2;
	if (sum%cnt) return puts("NO"),0;
	cnt=sum/cnt;
	for (int i=1;i<=n;i++) a[i]=a[i+1]-a[i];
	for (int i=1;i<=n;i++) a[i]-=cnt;
	for (int i=1;i<=n;i++)
		if (a[i]>0||a[i]%n) return puts("NO"),0;
	puts("YES");
	return 0;
}