You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled with rock. It takes one minute to move one unit north, south, east, west, up or down. You cannot move diagonally and the maze is surrounded by solid rock on all sides. 

Is an escape possible? If yes, how long will it take? 

Input

The input consists of a number of dungeons. Each dungeon description starts with a line containing three integers L, R and C (all limited to 30 in size). 
L is the number of levels making up the dungeon. 
R and C are the number of rows and columns making up the plan of each level. 
Then there will follow L blocks of R lines each containing C characters. Each character describes one cell of the dungeon. A cell full of rock is indicated by a '#' and empty cells are represented by a '.'. Your starting position is indicated by 'S' and the exit by the letter 'E'. There's a single blank line after each level. Input is terminated by three zeroes for L, R and C.

Output

Each maze generates one line of output. If it is possible to reach the exit, print a line of the form 

Escaped in x minute(s).

where x is replaced by the shortest time it takes to escape. 
If it is not possible to escape, print the line 

Trapped!

Sample Input

3 4 5
S....
.###.
.##..
###.#

#####
#####
##.##
##...

#####
#####
#.###
####E

1 3 3
S##
#E#
###

0 0 0

Sample Output

Escaped in 11 minute(s).
Trapped!

#include "iostream"
#include "queue"
#include "cstring"
using namespace std;
queue<int> que;
char map[32][32][32];
int book[32][32][32];
struct point
{
	int x,y,z;
	int step;
};
int l,r,c;
point start,end;
int bfs()
{
	int next[6][3]={{0,0,1},{0,1,0},{1,0,0},{0,0,-1},{-1,0,0},{0,-1,0}};
//上面的next可以稍微的記憶下這種用法！！！

	queue<point> que;
	que.push(start);
	memset(book,0,sizeof(book));
//因為S 和 E是必定不可能在一個位置上的，所以這裡並不需要判斷
	book[start.x][start.y][start.z]=1;
	while(!que.empty())
	{
		point cur=que.front();
		que.pop();//刪除直接在這裡用就好了，放在迴圈結束很容易忘記
		for(int i=0;i<=5;i++)
		{
			point now=cur;
			now.x=cur.x+next[i][0];
			now.y=cur.y+next[i][1];
			now.z=cur.z+next[i][2];
			now.step=cur.step+1;
			if(now.x>=0&&now.y>=0&&now.z>=0&&now.x<l&&now.y<r&&now.z<c&&book[now.x][now.y][now.z]==0&&map[now.x][now.y][now.z]!='#')
//注意判斷條件，一定判斷x，y，z的合理性在前！！！否則陣列越界
			{
				que.push(now);
				book[now.x][now.y][now.z]=1;
				if(now.x==end.x&&now.y==end.y&&now.z==end.z)
				{
					return now.step;
				}
//在這裡就可以進行對值進行判斷了，演算法複雜度會變小，大概小6倍，判斷放在上面也可以
			}
			
		}
	}
	return -1;
}
int main()
{
	while(~scanf("%d%d%d",&l,&r,&c))
	{
		if(l==0&&r==0&&c==0)
		break;
		getchar();
		for(int i=0;i<l;i++)
		{
			for(int j=0;j<r;j++)
			{
				for(int k=0;k<c;k++)
				{
					scanf("%c",&map[i][j][k]);
					if(map[i][j][k]=='S')
					{
						start.x=i;
						start.y=j;
						start.z=k;
						start.step=0;
					}
					if(map[i][j][k]=='E')
					{
						end.x=i;
						end.y=j;
						end.z=k;
					}
				}
				getchar();
			}
			if(i!=(l-1))
			getchar();
		}
		if(bfs()==-1)
		{
			cout<<"Trapped!"<<endl;
		}
		else
		{
			cout<<"Escaped in "<<bfs()<<" minute(s)."<<endl;
		}
	}	
	return 0;
 } 

 直接用模板就好！！！