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<iostream>
#include<cstdio>
using namespace std;
int a[101000],b[101000];
int main()
{
	int n;scanf("%d",&n);long long sum=0,tot=1ll*n*(n+1)/2,tot2=0;
	for(int i=1;i<=n;i++)scanf("%d",&a[i]),sum+=a[i];
	if(sum%tot!=0)return 0*puts("NO");a[0]=a[n];
	int Max=0;for(int i=1;i<=n;i++)b[i]=a[i]-a[i-1],Max=max(Max,b[i]);
	for(int i=1;i<=n;i++)tot2+=(Max-b[i])/n,b[i]=(b[i]%n+n)%n;b[0]=b[n];
	for(int i=1;i<=n;i++)if(b[i]!=b[i-1])return 0*puts("NO");
	long long lst=sum-tot2*tot;
	if(lst/n%tot!=0)return 0*puts("NO");
	if(lst<0)return 0*puts("NO");
	return 0*puts("YES");
 } 