Little penguin Polo loves his home village. The village has n houses, indexed by integers from 1 to n. Each house has a plaque containing an integer, the i-th house has a plaque containing integer pi (1 ≤ pi ≤ n).

Little penguin Polo loves walking around this village. The walk looks like that. First he stands by a house number x

. Then he goes to the house whose number is written on the plaque of house x (that is, to house px), then he goes to the house whose number is written on the plaque of house px (that is, to house ppx), and so on.

We know that:

1. When the penguin starts walking from any house indexed from 1 to k
   , inclusive, he can walk to house number 1.
2. When the penguin starts walking from any house indexed from k + 1 to n, inclusive, he definitely cannot walk to house number 1.
3. When the penguin starts walking from house number 1, he can get back to house number 1 after some non-zero number of walks from a house to a house.

You need to find the number of ways you may write the numbers on the houses' plaques so as to fulfill the three above described conditions. Print the remainder after dividing this number by 1000000007 (109 + 7).

Input

The single line contains two space-separated integers n and k (1 ≤ n ≤ 1000, 1 ≤ k ≤ min(8, n)) — the number of the houses and the number k from the statement.

Output

In a single line print a single integer — the answer to the problem modulo 1000000007 (109 + 7).

Examples

input

Copy

5 2

output

Copy

54

input

Copy

7 4

output

Copy

1728 

題意：

給定 n 和k，n 表示有n個房子，然後每個有一個編號，一隻鵝要從一個房間中開始走，下一站就是房間的編號，現在要你求出有多少

種方法編號並滿足下面的要求：

1.如果從1到k開始走，一定能走到 1。

2.如果從k+1到n 開始走，一定走不到 1.

3.如果從 1 開始走，那麼一定能回到1，並且走過房間數不為0.

思路：分類來看，

1、首先是從k+1到n，每個的p(i)只要∈(k+1,n)就行,所以一共有（n-k)^(n-k)種；

2、接下來是從1到k，這裡就要用到我們的purfer序列的結論了：n個點的 有標號有根樹的計數：n^(n-2)*n = n^(n-1)

https://www.cnblogs.com/dirge/p/5503289.html （原博主）

所以（1，k）有k^(k-1)種，

最後把這兩類的結果乘起來就是我們要的答案