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 <iostream>
#include <vector>
#include <set>

using namespace std;

int main(){
	int N, sum=0, k;
	cin>>N;
	vector<int> A(N), A2(N);
	for(int i=0; i<N; i++){
		cin>>A[i];
		sum+=A[i];
	}
	k=sum/(N*(N+1)/2);

	for(int i=0; i<N; i++){
		A2[i]=A[(i+1)%N]-A[i];
	}

	set<int> s;
	for(int i=0; i<N; i++){
		if(s.find(A2[i]-k)==s.end()){
			s.insert((A2[i]-k)*-1);
		}
	}

	const set<int>::const_iterator ite=s.end();
	for(set<int>::const_iterator it=s.begin(); it!=ite; ++it){
		if((*it)%N!=0){
			cout<<"NO"<<endl;
			return 0;
		}
	}

	cout<<"YES"<<endl;

	return 0;
}