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) {
		new Main().solve();
	}
	
	void solve(){
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		int[]a=new int[n];
		long sum=0;
		for(int i=0;i<n;i++){
			a[i]=sc.nextInt();
			sum+=a[i];
		}
		long nsum=n*(n+1)/2;
		if(sum%nsum!=0){
			System.out.println("NO");
			return;
		}
		
		for(int t=0;t<sum/nsum;t++){
			int id=0;
			int max=0;
			for(int i=0;i<n;i++){
				if(max<a[i]){
					max=a[i];
					id=i;
				}
			}
			for(int i=0;i<n;i++){
				if(i==id)a[i]-=n;
				else a[i]-=(i-id+n)%n;
			}
		}
		
		for(int i=0;i<n;i++){
			if(a[i]!=0){
				System.out.println("NO");
				return;
			}
		}
		System.out.println("YES");
	}

}