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;
int main(){
  int n,rm_num,rm_cnt,sum_stone=0;
  cin >> n;
  int a[n],dif[n];
  bool f=false;
  rm_num=(n*(n+1))/2;
  cin >> a[0];
  sum_stone+=a[0];
  for(int i=0;i<n-1;i++){
    cin >> a[i+1];
    dif[i] = a[i+1]- a[i];
    sum_stone+=a[i+1];
  }
  dif[n-1] = a[0]-a[n-1];
  rm_cnt = sum_stone/rm_num;
  if(rm_cnt*rm_num!=sum_stone){
    cout <<"NO\n";
    return 0;
  }
    for(int i=0;i<n;i++){
      dif[i]-=rm_cnt;
    }
    for(int i=0;i<n;i++){
      while(dif[i]<0){
        dif[i]+=n;
        rm_cnt--;
      }
      if(dif[i]!=0){
        f=true;
        break;
      }
    }
    if(f) cout<<"NO\n";
    else cout <<"YES\n";

}