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(){
  ll n,a[100000],s=0;
  cin>>n;
  for(int i=0;i<n;i++)cin>>a[i],s+=a[i];
  if(s%(n*(n+1)/2))cout<<"NO"<<endl;
  else{
    s/=n*(n+1)/2;
    ll ans=1,b[100000]={};
    for(int i=0;i<n;i++){
      if(-a[(i+1)%n]+a[i]+s>=0)
      b[(i+1)%n]=(-a[(i+1)%n]+a[i]+s)/n;
    }
    ll ss=0;
    for(int i=0;i<n;i++)ss+=b[i];
    if(ss!=s)ans=0;
    if(ans)cout<<"YES"<<endl;
    else cout<<"NO"<<endl;
  }
  return 0;
}