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.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;

public class Main {

  private void solve(Scanner in, PrintWriter out) {
    int N = in.nextInt();
    long[] a = new long[N];
    for (int i = 0; i < N; i++) {
      a[i] = in.nextInt();
    }

    long[] d = new long[N];
    for (int i = 0; i < N; i++) {
      d[i] = a[(i + 1) % N] - a[i];
    }

    long sum = Arrays.stream(a).sum();
    if (sum % ((long) N * ((long) N + 1) / 2) != 0) {
      out.println("NO");
      return;
    }

    long k = sum / ((long) N * ((long) N + 1) / 2);
    //1-N
    long count = 0;
    for (long e : d) {
      e -= k;

      if (e > 0 || (-e) % (long) N != 0) {
        out.println("NO");
        return;
      } else {
        count += (-e) / (long) N;
      }
    }
    if (count != k) {
      out.println("NO");
    } else {
      out.println("YES");
    }
  }

  public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    PrintWriter out = new PrintWriter(System.out);
    new Main().solve(in, out);
    in.close();
    out.close();
  }
}
