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
import java.util.*;

public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		// 整数の入力
		int N = sc.nextInt();
		long[] list=new long[N];
		for(int i=0;i<N;i++){
			list[i]=sc.nextInt();
		}
		//１回で取れる個数
		long pick=(long)N*(long)(N+1)/2;

		long sum=0;
		for(int i=0;i<N;i++){
			sum=sum+list[i];
		}
		if(sum-(long)(sum/pick)*pick!=0){
			System.out.println("NO");
			return;
		}
		//総ピック回数
		long count=sum/pick;
		if(((int)(list[0]-list[N-1])-count)%N!=0||((int)(list[0]-list[N-1])-count)>0){				
			System.out.println("NO");
			return;
		}

		for(int i=1;i<N;i++){
			if(((int)(list[i]-list[i-1])-count)%N!=0||((int)(list[i]-list[i-1])-count)>0){				
				System.out.println("NO");
				return;
			}
		}
		System.out.println("YES");

	}
}