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.Scanner;
public class Main {
	public static void main(String[] args){
		Scanner as = new Scanner(System.in);
		int n = as.nextInt();
		int[] a;
		a = new int[n];
		long t = 0;
		for(int i=0;i<n;i++){
			a[i] = Integer.parseInt(as.next());
			t+=a[i];
		}
		long l=0;
		boolean asa = true;
		if(t%((n+1)*(long)n/2)==0){
			long k=t/((n+1)*(long)n/2);
			for(int i=0;i<n;i++){
				l = a[(i+1)%n]-a[i];
				if(l>k||(k-l)%n!=0){
					asa = false;
					break;
				}
			}
		}else{
			asa = false;
		}
		if(asa){
			System.out.println("YES");
		}else{
			System.out.println("NO");
		}
	}
}