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 = sc.nextInt();
    long[] a = new long[n];
    long sum = 0;
    for(int i = 0; i < n; i++) {
      a[i] = sc.nextLong();
      sum += a[i];
    }
    long t = ((long)n * (long)(n + 1)) / (long)2;
    String ans = "NO";
    if((sum % t) == 0) {
      long k = sum / t;
      int p = 0;
      for(int i = 0; i < n; i++) {
        long c = 0;
        if(i == 0) {
          c = a[i] - a[n - 1] - k;
        } else {
          c = a[i] - a[i - 1] - k;
        }
        if(!((c <= 0) && ((c % (long)n) == 0))) p++;
      }
      if(p == 0) ans = "YES";
    }
    System.out.println(ans);
  }
}