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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
int main(){
  int n;
  cin>>n;
  vector<int64_t> a(n);
  int64_t s=0;
  for(int i=0;i<n;i++){
    cin>>a.at(i);
    s+=a.at(i);
  }
  int64_t k=s/((int64_t)n*(n+1)/2);
  for(int i=0;i<n;i++)
    a.at(i)-=k*(i+1);
  int64_t x=0;
  for(int i=1;i<n;i++){
    if(a.at(i)>a.at(i-1)||(a.at(i-1)-a.at(i))%n!=0){
      cout<<"NO"<<endl;
      return 0;
    }
    x+=(a.at(i-1)-a.at(i))/n;
  }
  cout<<(x<=k?"YES":"NO")<<endl;
}