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<iostream>
using namespace std;
main()
{
	long n,a[1<<17],m=1LL<<60,M=0;
	cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>a[i];
		m=min(m,a[i]);
		M=max(M,a[i]);
	}
	if(m*n<M)
	{
		cout<<"NO"<<endl;
		return 0;
	}
	for(int i=1;i<n-1;i++)
	{
		long x=a[i]-a[i-1];
		if(x<0)
		{
			x+=n*((-x)/n)+n;
		}
		x%=n;
		long y=a[i+1]-a[i];
		if(y<0)
		{
			y+=n*((-y)/n)+n;
		}
		y%=n;
		if(x!=y||!x||!y)
		{
			cout<<"NO"<<endl;
			return 0;
		}
	}
	if(n==2)
	{
		if(a[0]*2<a[1]||a[1]*2<a[0])
		{
			cout<<"NO"<<endl;
			return 0;
		}
		a[0]%=3;
		a[1]%=3;
		if((a[0]+a[1])%3)cout<<"NO"<<endl;
		else cout<<"YES"<<endl;
		return 0;
	}
	cout<<"YES"<<endl;
}