Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

5
4 5 1 2 3

Sample Output 1

YES

All the stones can be removed in one operation by selecting the second box.

Sample Input 2

5
6 9 12 10 8

Sample Output 2

YES

Sample Input 3

4
1 2 3 1

Sample Output 3

NO

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main {
    static Scanner scanner = new Scanner(System.in);

    public static void main(String[]$) {
        int n = scanner.nextInt();
        long m = (long)n * (n + 1) / 2;
        long[] a = new long[n];
        long sum = 0;
        for (int i = 0; i < n; i++) {
            a[i] = scanner.nextInt();
            sum += a[i];
        }
        System.out.println(sum % m == 0 && f(a, n, sum / m) ? "YES" : "NO");
    }

    static boolean f(long[] a, int n, long k) {
        for (int i = 0; i < n; i++) {
            long b = a[i] - a[(i + 1) % n] + k;
            if (b < 0 || b % n != 0) {
                return false;
            }
        }
        return true;
    }
}