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<bits/stdc++.h>
using namespace std;
typedef long long ll;

int main(){
int N; cin >> N;
  //和がN　最大値がN以上　
  // 必要条件を集めて十分性をためしてみる方向で何通りかやってみる
  ll rem = N*(N+1); rem /= 2;
  vector<ll> A(N); ll sum = 0;
  map<ll,ll> mp; ll M = 0;
  for(int i = 0; i < N; i++){ cin >> A[i]; sum += A[i];mp[A[i]]++;}
  
  bool flag = true;
  for(auto ne : mp ){ 
    if( ne.second >= 2 ){ flag = false;}
    if( ne.first == 0 && ne.second == N ){ cout <<"YES" << endl;return 0;}
  }
  
  if( sum % rem == 0 && flag ){ cout << "YES" << endl;}
  else{ cout << "NO" << endl;}

  return 0;
}