#include <bits/stdc++.h>

using namespace std;

int n,i,a[100005],d[100005];

int main(){
	scanf("%d", &n);
	
	for(i = 0; i < n; i++)
	scanf("%d", &a[i]);
	
	long long s = 0;
	
	for(i = 0; i < n; i++)
	s += a[i];
	
	long long tmp = (long long)n * (n + 1) / 2;
	if(s % tmp != 0)
	return 0 * printf("NO\n");
	
	long long k = s / tmp;
	
	for(i = 0; i < n; i++){
		if(i == n - 1)
		d[i] = a[0] - a[i];
		else
		d[i] = a[i + 1] - a[i];
	}
	
	//(k - d[i]) % n == 0
	
	for(i = 0; i < n; i++){
		if((k - d[i]) % n != 0 || k - d[i] < 0)
		return 0 * printf("NO\n");
	}
	
	printf("YES\n");
}

### Prompt

Please create a solution in cpp 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 <bits/stdc++.h>

using namespace std;

int n,i,a[100005],d[100005];

int main(){
	scanf("%d", &n);
	
	for(i = 0; i < n; i++)
	scanf("%d", &a[i]);
	
	long long s = 0;
	
	for(i = 0; i < n; i++)
	s += a[i];
	
	long long tmp = (long long)n * (n + 1) / 2;
	if(s % tmp != 0)
	return 0 * printf("NO\n");
	
	long long k = s / tmp;
	
	for(i = 0; i < n; i++){
		if(i == n - 1)
		d[i] = a[0] - a[i];
		else
		d[i] = a[i + 1] - a[i];
	}
	
	//(k - d[i]) % n == 0
	
	for(i = 0; i < n; i++){
		if((k - d[i]) % n != 0 || k - d[i] < 0)
		return 0 * printf("NO\n");
	}
	
	printf("YES\n");
}

```