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: Wrong Answer)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		int a[]=new int [n+1];
		long sum=0;
		for (int i = 0; i < n; i++) {
			a[i]=sc.nextInt();
			sum+=a[i];
		}
		if(n==1){
			System.out.println("YES");
			return;
		}
		if(sum%((n*(n+1))/2)!=0)System.out.println("NO");
		else{
			long p=sum/((n*(n+1))/2);
			long q=0;
			a[n]=a[0];
			for (int i = 0; i < n; i++) {
				int b=a[i]-a[i+1];
				if((b+p)%n!=0){
					System.out.println("NO");
					return;
				}
				else{
					q+=(b+p)/n;
				}
			}
			if(p==q)System.out.println("YES");
			else System.out.println("NO");
			
			
		}
		
		}
		
	}