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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<math.h>
int a[100010],b[10010];
int main()
{
	int i,n,m,sum,t1,t2,t;
	while(scanf("%d",&n)!=EOF)
	{
		sum=0;
		for(i=1;i<=n;i++)
		{
		scanf("%d",&a[i]);
		sum=sum+a[i];
     	}
    	t1=n*(n+1)/2;
		if(sum%t1!=0)
		{
			printf("NO\n");
			continue;
		}
		t2=sum/t1;
		for(i=1;i<n;i++)
		{
			b[i]=a[i+1]-a[i];
		}
		b[n]=a[1]-a[n];
		for(i=1;i<=n;i++)
		{
			t=t2*(n-1);
			t=t+b[i];
			if(t%n!=0)
			break;
		}
		if(i!=n+1)
		printf("NO\n");
		else
		printf("YES\n");
	}
	return 0;
}