#include<bits/stdc++.h>
#define MAX 100005
using namespace std;
 
int main(){
  long long n, m, a[MAX], sum = 0;
  scanf("%lld", &n);
  for(int i = 0; i < n; i++){
    scanf("%lld", &a[i]);
    sum += a[i];
  }
  if(sum%(n*(n+1)/2)) return 0*puts("NO");
  m = sum/(n*(n+1)/2);
  a[n] = a[0];
  for(int i = 0; i < n; i++){
    long long k = m + a[i] - a[i+1];
    if(k < 0 || k % n) return 0*puts("NO");
  }
  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>
#define MAX 100005
using namespace std;
 
int main(){
  long long n, m, a[MAX], sum = 0;
  scanf("%lld", &n);
  for(int i = 0; i < n; i++){
    scanf("%lld", &a[i]);
    sum += a[i];
  }
  if(sum%(n*(n+1)/2)) return 0*puts("NO");
  m = sum/(n*(n+1)/2);
  a[n] = a[0];
  for(int i = 0; i < n; i++){
    long long k = m + a[i] - a[i+1];
    if(k < 0 || k % n) return 0*puts("NO");
  }
  puts("YES");
}
```