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;
typedef pair<int,int> pii;
typedef vector<int> VI;
typedef vector<VI> VVI;
const double PI = 3.14159265358979323846;
const double EPS = 1e-12;
const int INF = numeric_limits<int>::max() / 2;
const int NEG_INF = numeric_limits<int>::min() / 2;
const int MOD = 1e9+7;

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  ll n,asum=0,nsum;cin>>n;
  std::vector<ll> dif;
  ll a[100010];
  nsum=n*(n+1)/2;
  for(int i=0;i<n;i++){
      cin>>a[i];
      asum+=a[i];
  }
  for(int i=0;i<n;i++) dif.push_back(a[(i+1)%n]-a[i]);
  ll change=0;//dif_ma=dif[0];
  for(int i=1;i<n;i++){
      if(dif[i]!=dif[i-1]) change++;
      //dif_ma=max(dif_ma,dif[n]);
  }
  if(asum%nsum!=0){
      cout<<"NO"<<endl;
      return 0;
  }
  else{
     // ll k=asum/nsum;
      if(change>2) cout<<"NO"<<endl;
      else cout <<"YES"<<endl;
      return 0;
  }
}