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 <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long ll;
const int maxn=1e5+5;

ll a[maxn],n,sum,k,t,d[maxn];

int main()
{
	scanf("%lld",&n);
	for(int i=0;i<n;i++)
	{
	    scanf("%lld",a+i);
	    sum+=a[i];
	}
	k=(1+n)*n/2;
	if(sum%k!=0)
	{
		printf("NO\n");
		return 0; 
	}
	k=sum/k;
	ll s=a[0];
	for(int i=0;i<n-1;i++)
	    d[i]=a[i+1]-a[i];
	d[n-1]=a[0]-a[n-1];
	if(k==1)
	{
		printf("YES\n");
		return 0; 
	}
	for(int i=0;i<n;i++)
	{
		if(k==d[i])
		    continue;
		t=k-d[i];
		if(t<0||t%n!=0)
		{
			printf("NO\n");
			return 0;
		}
	}
	printf("YES\n");
	return 0;
}