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.io.*;
import java.util.*;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        long n=Long.parseLong(br.readLine());
        StringTokenizer st=new StringTokenizer(br.readLine());
        long[] a=new long[(int)n];
        long sum=0;
        long count=0;
        for(int i=0;i<n;i++){ a[i]=Long.parseLong(st.nextToken()); sum+=a[i]; }
        if(sum%((n*(n+1))/2)!=0){
            System.out.println("NO");
            System.exit(0);
        }
        long times=sum/((n*(n+1))/2);
        for(int i=0;i<n;i++){
            count=(a[(i+1)%(int)n]-a[i]);
            if((count>times)||(times-count)%n!=0){
                System.out.println("NO");
                System.exit(0);
            }
        }System.out.println("YES");
    }
}