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);

		int N = sc.nextInt();
		int[] A = new int[N];
		for(int i=0; i<N; i++)
			A[i] = sc.nextInt();
	
		System.out.println(check(A) ? "YES" : "NO");
		sc.close();
	}
	
	static boolean check(int[] A) {
		int N = A.length;
		long sum = 0;
		for(int i=0; i<N; i++)
			sum += A[i];
		
		if(sum%((long)N*(N+1)/2)!=0)
			return false;
		
		int op = (int)(sum/((long)N*(N+1)/2));
		
		for(int i=0; i<N; i++) {
			int diff = A[(i+1)%N]-A[i];
			if((diff>op))
				return false;
			if((op-diff)%N!=0)
				return false;
		}
		
		return true;
	}
}
