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>
using namespace std;

bool solve(void) {
	long long int i,n,sum=0;
	cin>>n;
	long int a[n+1],b[n];
	for(i=0;i<n;i++){
		cin>>a[i];
		sum+=a[i];
	}
	a[n]=a[0];
	long long int k=n*(n+1)/2;
	
	if(sum%k!=0)return false;
	long long int max=sum/k;
	
	for(i=0;i<n;i++){
		if((max-(a[i+1]-a[i]))%n!=0)return false;
		b[i]=(max-(a[i+1]-a[i]))/n;
		if(b[i]<0||b[i]>max)return false;
	}
	return true;
}

int main(){
	if(solve())cout<<"YES"<<endl;
	else{
		cout<<"NO"<<endl;
	}
}