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.*;

class Main{

    public static void main(String args[]){
	Scanner sc=new Scanner(System.in);
	int n=sc.nextInt();
	long[] a=new long[n];
	for(int i=0;i<n;i++){
	    a[i]=sc.nextLong();
	}

	long sum=0;
	for(int i=0;i<n;i++){
	    sum+=a[i];
	}

	long consume=n*(n+1)/2;
	if(sum%consume!=0){
	    System.out.println("NO");
	    return;
	}

	long sump=sum/consume;
	long cnt=0;
	boolean isok=true;
	for(int i=0;i<n;i++){
	    //a[i-1]+(sump-p)=a[i]+(n-1)*p
	    long chil=(a[(i-1+n)%n]+sump-a[i]);
	    if(chil<0 || chil%n!=0) isok=false;
	    long p=chil/n;
	    if(a[(i-1+n)%n]+(sum-p)==a[i]+(n-1)*p) isok=false;
	    if(p<0) isok=false;
	    cnt+=p;
	}
	if(isok && cnt==sump){
	    System.out.println("YES");
	}
	else{
	    System.out.println("NO");
	}
	return;
    }
    

    
}