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 scanner = new Scanner(System.in);

    int N = scanner.nextInt();
    int[] A = new int[N];
    for (int i = 0; i < N; i++) A[i] = scanner.nextInt();

    if (possible(N, A)) {
      System.out.println("YES");
    } else {
      System.out.println("NO");
    }
  }

  private static boolean possible(int N, int[] A) {
    long sum = 0;
    for (int i = 0; i < N; i++) sum += A[i];
    long NS = (long) N * (N + 1) / 2;
    if (sum % NS != 0) return false;
    long K = sum / NS;
    for (int i = 0; i < N; i++) {
      int j = i == 0 ? N - 1 : i - 1;
      long d = K - (A[i] - A[j]);
      if (d < 0 || d % N != 0) return false;
    }
    return true;
  }
}
