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.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long sum = 0;

		int n = sc.nextInt();
		// int num = 0;
		int[] arr = new int[n];
		long check = (long) n * (long) (n + 1) / 2L;

		for (int i = 0; i < n; i++) {
			arr[i] = sc.nextInt();
			sum += arr[i];
		}

		if (sum % check != 0) {
			System.out.println("NO");
			return;
		}

		long k = sum / check;
		long s = 0;
		for (int i = 0; i < n; i++) {
			int current = arr[i];
			int next = arr[(i + 1) % n];
			long diff = next - current;
			if ((k - diff) % n != 0 || diff > k) {
				System.out.println("NO");
				return;
			}
			s += (k - diff) / n;
		}

		if (s != k) {

			System.out.println("NO");
			return;
		}

		System.out.println("YES");
	}
}
