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>
#define N 100001
using namespace std;
typedef pair<int,int> P;
typedef long long ll;
int n;
ll A[N],B[N];

bool check(){
  ll sum=0;
  for(int i=0;i<n;i++) sum+=i+1;
  for(int i=0;i<n;i++) if(A[0]!=A[i])return 0;
  return A[0]%sum==0;
}

bool calc(){
  int I=0;
  ll Sum=0;
  for(int i=0;i<n;i++) if(A[i]<0)return 0;
  for(int i=0;i<n;i++)B[i]=A[(i+1)%n]-A[i],Sum+=B[i];

  for(int i=0;i<n;i++)if(B[i]!=B[(i+1)%n]&&B[i]<0)I = (i+1)%n;
  for(int i=0;i<n;i++)A[(I+i)%n]-=i+1;
  return Sum==0;
}
 
int solve(){
  if(n==1)return 1;
  if(check())return 1;
  for(int i=0;i<100;i++) {
    if(!calc()) return check();
    if(check())return 1;
  }
  return 1;
}
 
int main(){
  cin>>n;
  for(int i=0;i<n;i++)cin>>A[i];
  cout<<(solve()? "YES":"NO")<<endl;
  return 0;
}