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 {
    static Scanner scanner = new Scanner(System.in);

    public static void main(String[]$) {
        int n = scanner.nextInt();
        long m = (long)n * (n + 1) / 2;
        long[] a = new long[n];
        long sum = 0;
        for (int i = 0; i < n; i++) {
            a[i] = scanner.nextInt();
            sum += a[i];
        }
        if (sum % m == 0) {
            long[] b = new long[n];
            b[0] = a[n - 1] - a[0] + sum / m;
            for (int i = 1; i < n; i++) {
                b[i] = a[i - 1] - a[i] + sum / m;
            }
            for (int i = 0; i < n; i++) {
                if (b[i] < 0 || b[i] % n != 0) {
                    System.out.println("NO");
                    return;
                }
            }
            System.out.println("YES");
        } else {
            System.out.println("NO");
        }
    }
}