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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class Main {
	static long N;
	static long[] A;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		N = sc.nextLong();
		A = new long[(int) N];
		long s = 0;
		for (int i = 0; i < N; ++i) {
			A[i] = sc.nextLong();
			s += A[i];
		}
		long all = N * (N + 1) / 2;
		if (s % all != 0) {
			System.out.println("NO");
			return;
		}
		long K = (s / all);
		long[] diff = new long[(int) N];
		for (int i = 0; i < N; ++i) {
			diff[i] = A[(int) ((i + 1) % N)] - A[i];
		}
		for (int i = 0; i < N; ++i) {
			diff[i] -= K;
			if (diff[i] > 0 || (-diff[i]) % N != 0) {
				System.out.println("NO");
				return;
			}
		}

		System.out.println("YES");

	}

	static void tr(Object... objects) {
		System.out.println(Arrays.deepToString(objects));
	}

}
