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: Time Limit Exceeded)

import java.util.*;
 
public class Main{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		long[] a = new long[n];
		long max = 0;
		int p=0;
		for(int i=0; i<n; i++){
			a[i] = sc.nextLong();
		}
		
		while( !check(a) ){
			max=0;
			p=0;
			for(int i=0; i<n; i++){
				if(a[i] > max){
					max = a[i];
					p=i;
				}
			}

			for(int i=1; i<=n; i++){
				if(i+p>=n){
					a[i+p-n] -= i;
					if(a[i+p-n]<0){
						System.out.println("NO");
						return;
					}
				}else{
					a[i+p] -= i;
					if(a[i+p]<0){
						System.out.println("NO");
						return;
					}
				}
			}
		}
				
		System.out.println("YES");
		return;
	}

	static boolean check(long[] a){
		int n = a.length;
		for(int i=0; i<n; i++){
			if(a[i] != 0){
				return false;
			}
		}
		return true;
	}
}