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>
#define MN 100000
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
long long tot,per;
int n,a[MN+5];
int main()
{
	n=read();per=1LL*(n+1)*n/2;
	for(int i=1;i<=n;++i) a[i]=read(),tot+=a[i];
	if(tot%per) return 0*puts("NO");
	for(int i=1;i<=n;++i)
	{
		long long x=a[i%n+1]-a[i]-(tot/per);
		if(x>0||(-x)%n) return 0*puts("NO");
	}
	puts("YES");
	return 0;
}
