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>
#include<cstring>
inline void check_max(int a,int &b){if(a>b)b=a;}
typedef long long ll;
const int N=100100;
int a[N],b[N];
int n,m;
ll sum;
bool check()
{
	if(sum%((ll)n*(n+1)/2))return 0;
	sum/=(ll)n*(n+1)/2;
	
	int cnt=0;
	for(int i=1;i<=n;i++)
		if((m-b[i])%n)return 0;
		else cnt+=(m-b[i])/n;
	return cnt==m && cnt<=sum && ((sum-cnt)%n==0);
}
int main()
{
	scanf("%d",&n),sum=0;
	for(int i=1;i<=n;i++)scanf("%d",a+i),sum+=a[i];
	m=b[1]=a[1]-a[n];
	for(int i=2;i<=n;i++)check_max(b[i]=a[i]-a[i-1],m);
	printf("%s\n",check()?"YES":"NO");
	return 0;
}
