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 <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <set>
#include <map>
typedef long long ll;
using namespace std;

int main(){
  ll N;
  cin >> N;
  ll a[N];
  ll sum = 0;
  for(int i = 0; i < N; i++) {
    cin >> a[i];
    sum += a[i];
  }

  //N==1
  if(N == 1){
    cout << "YES" << endl;
    return 0;
  }
  
  //一回の操作で入れる石の数
  ll S = N * (N + 1) / 2;
  if(sum % S != 0){
    //cout << "hoge" << endl;
    cout << "NO" << endl;
    return 0;
  }
  //操作回数
  ll T = sum / S;

  //右の項との差分b
  ll b[N];
  for(int i = 0; i < N - 1; i++){
    b[i] = a[i+1] - a[i];
  }
  b[N-1] = a[0] - a[N-1];

  //iがm回選ばれるとすると、
  //-m(N-1)+(T-m)=b[i]
  //m=(T-b)/N;

  //選ばれる回数m
  ll m[N];
  for(int i = 0; i < N; i++){
    ll tmp = T - b[i];
    if(tmp % N != 0){
      //cout << i << endl;
      //cout << "Hoge1" << endl;
      cout << "NO" << endl;
      return 0;
    }
    m[i] = (T - b[i]) / N;
  }

  ll cnt = 0;
  for(int i = 0; i < N; i++) cnt += m[i];

  if(cnt != T){
    //cout << "Hoge12" << endl;
    cout << "NO" << endl;
    return 0;
  }

  cout << "YES" << endl;
  return 0;
}