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>
#include <cmath>
#include <ctime>
#include <cstdlib>
using namespace std;

int main()
{
    int n,i,j,tt=0;
    long long sum=0,a[100001],dis;
    cin >> n;
    for (i=1;i<=n;i++)
    {
        cin >> a[i];
        sum += a[i];
        tt += i;
    }

    if ( sum%tt!=0)
    {
      cout << "NO";
      return 0;
    }

    sum = sum / tt;
    long long fr;
    for (i=1;i<n;i++)
    {
        dis = a[i+1] - a[i];
        if ((sum-dis)%n!=0)
        {
            cout << "NO";
            return 0;
        }
        fr+= (sum-dis)/n;

    }

    if (fr=sum)
        cout << "YES";
    else cout << "NO";
}