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;

void out()
{
	puts("NO");
	exit(0);
}
typedef long long s64;
#define rep(i,l,r) for(int i=l;i<=r;++i)
const int N=1e5+5;
int a[N];

int main()
{
	//freopen("1.in","r",stdin);
	s64 s=0;
	int n;cin>>n;
	rep(i,1,n){scanf("%d",a+i);s+=a[i];}
	s64 one=(s64)n*(n+1)/2;
	if(s%one)out();
	s/=one;
	a[0]=a[n];
	s64 ans=0;
	rep(i,1,n)
	{
		s64 now=s-a[i]+a[i-1];
		if(now%n||now<0)out();
		ans+=now/n;
	}
	if(ans!=s)out();
	puts("YES");
}