#include<cstdio>
#define MAXN 100005
#define LL long long
int n;
LL a[MAXN],b[MAXN],sum,cnt;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%lld",&a[i]);
        sum+=a[i];
    }
    LL tmp=1LL*n*(n+1)/2;
    if(sum%tmp)
    {
        printf("NO\n");
        return 0;
    }
    LL cnt=sum/tmp;
    b[n]=b[1]-a[n];
    b[n]=cnt-b[n];
    for(int i=1;i<n;i++)
    {
        b[i]=cnt-(a[i+1]-a[i]);
        if(b[i]<0||(b[i]%n))
        {
            printf("NO\n");
            return 0;
        }
    }
    printf("YES\n");
    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<cstdio>
#define MAXN 100005
#define LL long long
int n;
LL a[MAXN],b[MAXN],sum,cnt;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%lld",&a[i]);
        sum+=a[i];
    }
    LL tmp=1LL*n*(n+1)/2;
    if(sum%tmp)
    {
        printf("NO\n");
        return 0;
    }
    LL cnt=sum/tmp;
    b[n]=b[1]-a[n];
    b[n]=cnt-b[n];
    for(int i=1;i<n;i++)
    {
        b[i]=cnt-(a[i+1]-a[i]);
        if(b[i]<0||(b[i]%n))
        {
            printf("NO\n");
            return 0;
        }
    }
    printf("YES\n");
    return 0;
}
```