#include <bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef pair<int, int> pi;

lint n, a[100005], b[100005], sum;

int main(){
	scanf("%lld",&n);
	for(int i=1; i<=n; i++){
		scanf("%lld",&a[i]);
		sum += a[i];
	}
	if(sum % ((n * n + n) / 2) != 0){
		puts("NO");
		return 0;
	}
	sum /= (n * n + n) / 2;
	for(int i=1; i<=n; i++){
		if(i > 1) b[i] = a[i] - a[i-1];
		else b[i] = a[1] - a[n];
		b[i] = sum - b[i];
	}
	for(int i=1; i<=n; i++){
		if(b[i] < 0 || b[i] % n){
			puts("NO");
			return 0;
		}
	}
	puts("YES");
}

### 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 <bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef pair<int, int> pi;

lint n, a[100005], b[100005], sum;

int main(){
	scanf("%lld",&n);
	for(int i=1; i<=n; i++){
		scanf("%lld",&a[i]);
		sum += a[i];
	}
	if(sum % ((n * n + n) / 2) != 0){
		puts("NO");
		return 0;
	}
	sum /= (n * n + n) / 2;
	for(int i=1; i<=n; i++){
		if(i > 1) b[i] = a[i] - a[i-1];
		else b[i] = a[1] - a[n];
		b[i] = sum - b[i];
	}
	for(int i=1; i<=n; i++){
		if(b[i] < 0 || b[i] % n){
			puts("NO");
			return 0;
		}
	}
	puts("YES");
}

```