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 = Integer.parseInt(sc.next());
        int[] A = new int[N];
        long sum = 0L;
        boolean flag = true;
        for (int i=0;i<N;i++) {
            A[i] = Integer.parseInt(sc.next());
            sum += A[i];
        }

        if (sum%((long)N*(N+1)/2)!=0) flag=false;

        long val = sum/((long)N*(N+1)/2);
        for (int i=0;i<N-1;i++) {
            long tmp = val-(A[i+1]-A[i]);
            if (tmp<0 || tmp%N!=0) flag=false;
        }
        long tmp_last = val-(A[0]-A[N-1]);
        if (tmp_last<0 || tmp_last%N!=0) flag=false;

        if (flag) System.out.println("YES");
        else System.out.println("NO");
    }
}