#include <iostream>
#include <vector>

using namespace std;

int main(){
	long N, k;
	long sum=0;
	cin>>N;
	vector<int> A(N);
	for(int i=0; i<N; i++){
		cin>>A[i];
		sum+=A[i];
	}
	if(sum%(N*(N+1)/2)!=0){
		cout<<"NO"<<endl;
		return 0;
	}
	k=sum/(N*(N+1)/2);

	for(int i=0; i<N; i++){
		if(abs(A[(i+1)%N]-A[i]-k)%N!=0 || (A[(i+1)%N]-A[i]-k)>0){
			cout<<"NO"<<endl;
			return 0;
		}
	}

	cout<<"YES"<<endl;

	return 0;
}

### Prompt

Generate a cpp solution to the following problem:

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`.

Examples

Input

5
4 5 1 2 3


Output

YES


Input

5
6 9 12 10 8


Output

YES


Input

4
1 2 3 1


Output

NO

### Response

```cpp
#include <iostream>
#include <vector>

using namespace std;

int main(){
	long N, k;
	long sum=0;
	cin>>N;
	vector<int> A(N);
	for(int i=0; i<N; i++){
		cin>>A[i];
		sum+=A[i];
	}
	if(sum%(N*(N+1)/2)!=0){
		cout<<"NO"<<endl;
		return 0;
	}
	k=sum/(N*(N+1)/2);

	for(int i=0; i<N; i++){
		if(abs(A[(i+1)%N]-A[i]-k)%N!=0 || (A[(i+1)%N]-A[i]-k)>0){
			cout<<"NO"<<endl;
			return 0;
		}
	}

	cout<<"YES"<<endl;

	return 0;
}

```