使用BFS算法完成以下问题：Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison. Angel's friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards. You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.) Input First line contains two integers stand for N and M. Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend. Process to the end of the file. Output For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing "Poor ANGEL has to stay in the prison all his life."

解题思路：

这道题可以使用 BFS 算法来解决。首先需要找到起点和终点，并标记出所有的墙和警卫。使用 BFS 算法搜索整个迷宫，每次向四个方向扩展，如果扩展到的位置是墙或者警卫则跳过，否则加入到队列中继续扩展。如果扩展到终点则直接返回步数。最后如果没有找到终点，则说明无法到达，输出 "Poor ANGEL has to stay in the prison all his life."。

具体实现可以按照以下步骤进行：

1. 读入输入数据，找到起点和终点，并标记出所有的墙和警卫。

2. 使用 BFS 算法搜索整个迷宫，每次向四个方向扩展，如果扩展到的位置是墙或者警卫则跳过，否则加入到队列中继续扩展。

3. 如果扩展到终点则直接返回步数。

4. 最后如果没有找到终点，则输出 "Poor ANGEL has to stay in the prison all his life."。

代码实现：