#include <iostream>
#include <vector>
using namespace std;

int main() {
	long N; cin >> N;
	vector<long> a(N);
	long sum=0, dd=((1+N)*N)/2;
	for(int i=0; i<N; ++i){
	    cin >> a[i];
	    sum += a[i];
	}
	if(sum%dd>0){
	    cout << "NO" << endl;
	    return 0;
	}
	sum /= dd;
	long cnt=0;
	for(int i=0; i<N; ++i){
	    if((a[i]-a[(i+1)%N]+sum)%N>0 || a[i]-a[(i+1)%N]+sum<0){
	        cout << "NO" << endl;
	        return 0;
	    }
	    cnt += (a[i]-a[(i+1)%N]+sum)/N;
	}
	if(cnt==sum) cout << "YES" << endl;
	else cout << "NO" << endl;
	return 0;
}

### Prompt

In CPP, your task is to solve 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; cin >> N;
	vector<long> a(N);
	long sum=0, dd=((1+N)*N)/2;
	for(int i=0; i<N; ++i){
	    cin >> a[i];
	    sum += a[i];
	}
	if(sum%dd>0){
	    cout << "NO" << endl;
	    return 0;
	}
	sum /= dd;
	long cnt=0;
	for(int i=0; i<N; ++i){
	    if((a[i]-a[(i+1)%N]+sum)%N>0 || a[i]-a[(i+1)%N]+sum<0){
	        cout << "NO" << endl;
	        return 0;
	    }
	    cnt += (a[i]-a[(i+1)%N]+sum)/N;
	}
	if(cnt==sum) cout << "YES" << endl;
	else cout << "NO" << endl;
	return 0;
}

```