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 <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <map>
#include <vector>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;

const int MaxN = 1e5;
int n;
LL a[MaxN + 5], b[MaxN + 5];

int main()
{
	bool flag = 1;
	scanf("%d", &n);
	LL sum = 0, tot = 0;
	for(int i = 1; i <= n; i++) 
	{
		scanf("%lld", &a[i]);
		sum += a[i];
		tot += (LL)i;
	}
	if(sum % tot != (LL)0) flag = 0;
	LL num = sum / tot;
	a[n + 1] = a[1];
	sum = 0;
	for(int i = 1; i <= n; i++)
		b[i] = a[i + 1] - a[i], sum += b[i];
	if(sum != (LL)0) flag = 0;
	tot = 0;
	for(int i = 1; i <= n; i++) 
		if(b[i] != num) tot++;
	if(num < tot) flag = 0; 
	if(!flag) printf("NO\n");
	else printf("YES\n");
}