#include <iostream>
#include <cstdio>
#include <stdlib.h>
#define LL long long
using namespace std;
LL a[100010],n;
void priNo()
{
    printf("NO\n");
    exit(0);
}
int main()
{
    cin >> n;
    LL sum = 0;
    for (int i = 0;i < n;i++)
    {
        scanf("%lld",&a[i]);
        sum += a[i];
    }
    if (sum%((n*(n+1))/2) != 0) priNo();
    LL k = sum/((n*(n+1))/2);
    a[n] = a[0];
    for (int i = 1;i <= n;i++)
    {
        LL d = a[i]-a[i-1];
        if (k-d < 0||((k-d)%n) != 0) priNo();
    }
    printf("YES\n");
    return 0;
}
### Prompt

Create a solution in cpp for 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 <cstdio>
#include <stdlib.h>
#define LL long long
using namespace std;
LL a[100010],n;
void priNo()
{
    printf("NO\n");
    exit(0);
}
int main()
{
    cin >> n;
    LL sum = 0;
    for (int i = 0;i < n;i++)
    {
        scanf("%lld",&a[i]);
        sum += a[i];
    }
    if (sum%((n*(n+1))/2) != 0) priNo();
    LL k = sum/((n*(n+1))/2);
    a[n] = a[0];
    for (int i = 1;i <= n;i++)
    {
        LL d = a[i]-a[i-1];
        if (k-d < 0||((k-d)%n) != 0) priNo();
    }
    printf("YES\n");
    return 0;
}
```