Chosen by god FZU - 2301

Chosen by god

 

Everyone knows there is a computer game names "hearth stone", recently xzz likes this game very much. If you want to win, you need both capacity and good luck.

There is a spell card names "Arcane Missiles", which can deal 3 damages randomly split among all enemies.

xzz have a "Arcane Missiles Plus" can deal n damage randomly split among all enemies. The enemy battle field have only two characters: the enemy hero and enemy minion.

Now we assume the enemy hero have infinite health, the enemy minion has m health.

xzz wants to know the probability of the "Arcane Missiles Plus" to kill the enemy minion (the minion has not more than 0 health after "Arcane Missiles Plus"), can you help him?

 

Input

 

The first line of the input contains an integer T, the number of test cases. T test cases follow. (1 ≤ T ≤ 100000)

Each test case consists of a single line containing two integer n and m. (0 ≤ m ≤ n ≤ 1000)

output

For each test case, because the probability can be written as x / y, please output x * y^-1 mod 1000000007. .(y * y^-1 ≡ 1 mod 10^9 + 7)

题目大意：

有n次攻击回合，每个回合的攻击有一点伤害，但是可能攻击到敌方无限血的英雄，也可能攻击到敌方的一个血量为m的小兵，问在n次攻击回合内把敌方小兵杀死的概率为多少？答案对1e9+7取膜。

分析:

求概率，那就是需要牵扯到除法和模数的问题，一看见答案那么大，那就是需要求逆元了，关于逆元的求法我就不在这里赘述了，这题我要提出的是组合数的，可以先打一个表，这样省时间，剩下的就按照求概率的方法做就好了

下面是AC的参考代码

#include 
#include 
#include 
#include