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<cstdio>
#include<algorithm>
using namespace std;
const int N=100005;
typedef long long ll;
ll a[N],b[N];
int main ()
{
	ll n;
	while(scanf("%I64d",&n)!=EOF)
	{
		ll ans=0;
		for(int i=0;i<n;i++)
		{
			scanf("%I64d",&a[i]);
			ans+=a[i];
		}
		for(int i=0;i<n-1;i++)
			b[i]=a[i+1]-a[i];
		b[n-1]=a[0]-a[n-1];
		int flag=1;
		if(ans%((1+n)*n/2))
			flag=0;
		if(flag)
		{
			ll k=ans/((1+n)*n/2);
			for(int i=0;i<n;i++)
				if((k-b[i])%n||b[i]-k>0)
					flag=0;
		}
		printf("%s\n",flag?"YES":"NO");
	}
}