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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>

long a[100001];

int main(int argc, char const *argv[]) {
	long n,i,sum=0,num;
	scanf("%ld",&n);
	for(i=0;i<n;i++){
		scanf("%ld",&a[i]);
		sum+=a[i];
	}
	a[n]=a[0];
	for(i=0;i<n;i++){
		a[i]=a[i+1]-a[i];
	}
	num=n*(n+1)/2;
	if(sum%num==0){
		num=sum/num;
		for(i=0;i<n;i++){
			a[i]-=num;
			a[i]=-a[i];
			if(a[i]%5!=0) break;
		}
		if(i==n) printf("YES\n");
		else printf("NO\n");
	}else printf("NO\n");
	return 0;
}