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

  public static void main(String[] args) {
    int N = sc.nextInt();
    long[] A = new long[N + 1];
    long sum = 0;
    long single = 0;
    for (int i = 0; i < N; i++) {
      A[i] = sc.nextLong();
      sum += A[i];
      single += i + 1;
    }
    if (sum % single != 0) {
      System.out.println("NO");
      return;
    }
    A[N] = A[0];
    long count = sum / single;
    long use = 0;
    boolean ok = true;
    for (int i = 0; i < N; i++) {
      long diff = A[i + 1] - A[i] - count;
      if (diff > 0 || -diff % N != 0) {
        ok = false;
        break;
      }
      use += -diff / N;
    }
    ok &= use == count;
    System.out.println(ok ? "YES" : "NO");
  }

}
