Wand FZU

 Wand  FZU - 2282

错拍公式：有十本书放在书架上，问重新摆放之后没有书在自己原来的位置上的摆放方式有多少种？

         　　 设n本书的错拍方式为D[n]，那么把第n本书放在第k个位置，k有n-1种取值。对于第k个元素，要么放在第n个位置，此时剩下的就是n-2个元素的错拍                       D[n-2]，如果第k个元素不放在第n个位置，那么可以想想n对应的就是k，要求k不能放在n处，就是n-1个元素的错拍D[n-1]。于是D[n] = (n-1)*(D[n-                           1]+D[n-2])。

N wizards are attending a meeting. Everyone has his own magic wand. N magic wands was put in a line, numbered from 1 to n(Wand_i owned by wizard_i). After the meeting, n wizards will take a wand one by one in the order of 1 to n. A boring wizard decided to reorder the wands. He is wondering how many ways to reorder the wands so that at least k wizards can get his own wand.

For example, n=3. Initially, the wands are w1 w2 w3. After reordering, the wands become w2 w1 w3. So, wizard 1 will take w2, wizard 2 will take w1, wizard 3 will take w3, only wizard 3 get his own wand.

Input

First line contains an integer T (1 ≤ T ≤ 10), represents there are T test cases.

For each test case: Two number n and k.

1<=n <=10000.1<=k<=100. k<=n.

Output

For each test case, output the answer mod 1000000007(10^9 + 7).

Sample Input

2
1 1
3 1

Sample Output

1
4
题意：有n个巫师，每个巫师有一个魔法棒，现在要把魔法棒打乱顺序，巫师按照顺序领取魔法棒，问至少要有k个巫师领到自己的魔法棒的排列当时有多
少种？
题解：如果我们遍历有k、k+1、k+2...n个巫师领到自己的魔法棒，一定可以出来结果。
x个巫师领到自己的魔法棒，公式为 C(n,x)*D[n-x],预处理的话，需要把所有的C(n,x)都预处理出来，但是由于只有10组测试样例，所以不需要预处理，
于是对于每一组测试样例，我们需要把C(n , 1~n)都求出来，且O(n)的时间求出来D[n-x],这样的话，应该也不会超时，但是可以直接枚举不可以的减去就
行了