#include <bits/stdc++.h>
using namespace std;
int main(){
	int n; cin >> n;
	vector<int> a(n);
	for(int i = 0; i < n; i++) cin >> a[i];
	long long sum = 0, dv = 0;
	for(int i = 1; i <= n; i++) sum += a[i - 1], dv += i;
	auto bad = [](){
		cout << "NO" << endl;
		exit(0);
	};
	if(sum%dv) bad();
	long long op = sum / dv;
	a.push_back(a[0]);
	for(int i = 0; i < n; i++){
		int fwd = a[i + 1] - a[i];
		if(op - fwd < 0) bad();
		if((op - fwd) % n) bad();
	}
	cout << "YES" << 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 <bits/stdc++.h>
using namespace std;
int main(){
	int n; cin >> n;
	vector<int> a(n);
	for(int i = 0; i < n; i++) cin >> a[i];
	long long sum = 0, dv = 0;
	for(int i = 1; i <= n; i++) sum += a[i - 1], dv += i;
	auto bad = [](){
		cout << "NO" << endl;
		exit(0);
	};
	if(sum%dv) bad();
	long long op = sum / dv;
	a.push_back(a[0]);
	for(int i = 0; i < n; i++){
		int fwd = a[i + 1] - a[i];
		if(op - fwd < 0) bad();
		if((op - fwd) % n) bad();
	}
	cout << "YES" << endl;
	return 0;
}

```