#include<iostream>
#include<queue>
#include<stack>
#include<stdlib.h>
#include<iomanip>
using namespace std;

int n,m;
char map[101][101];

typedef struct point
{
	int x,y,step,time;//step用于记录到此点的方向，step=1,2,3,4 ->上下左右 
}*prt;//因为队中元素需要修改，用指针 
prt t[100][100];

bool cheak(int x,int y)//判断是否可达 
{
	if(x<0||y<0||x>=n||y>=m||map[x][y]=='X')
		return false;
	return true;
}
//广搜遍历所有可能 
void bfs()
{
	queue<prt> Q;//因为队中元素需要修改，用指针 
	prt corrent;
	int x,y,k,i;
	bool flag;
	t[0][0]->time=0;//出发点时间置0 
	Q.push(t[0][0]);//出发点进队 
	while(!Q.empty())
	{
		corrent=Q.front();
		x=corrent->x;
		y=corrent->y;
		Q.pop();
		//对于上下左右四个点进行查看：
		//	如果可以达到，并且从当前点走过去比以前方式到达他的时间小 
		//	就将这个点进队
		//	这里既有初始情况到达，也有事后修改，并且修改需要修改以后所有点，所以也需要进队（即最后样例所给） 
		if(cheak(x-1,y)&&corrent->time+map[x-1][y]-'0'+1<t[x-1][y]->time)
		{
				t[x-1][y]->time=corrent->time+map[x-1][y]-'0'+1;
				Q.push(t[x-1][y]);
				t[x-1][y]->step=4;
		}
		if(cheak(x+1,y)&&corrent->time+map[x+1][y]-'0'+1<t[x+1][y]->time)
		{
				t[x+1][y]->time=corrent->time+map[x+1][y]-'0'+1;
				Q.push(t[x+1][y]);
				t[x+1][y]->step=3;	
		}
		if(cheak(x,y-1)&&corrent->time+map[x][y-1]-'0'+1<t[x][y-1]->time)
		{
				t[x][y-1]->time=corrent->time+map[x][y-1]-'0'+1;
				Q.push(t[x][y-1]);
				t[x][y-1]->step=2;
		}
		if(cheak(x,y+1)&&corrent->time+map[x][y+1]-'0'+1<t[x][y+1]->time)
		{
				t[x][y+1]->time=corrent->time+map[x][y+1]-'0'+1;
				Q.push(t[x][y+1]);	
				t[x][y+1]->step=1;	
		}
	}
}

int main()
{
	int k,i,order,time;
	while(cin>>n>>m)
	{
		for(k=0;k<n;k++)
			for(i=0;i<m;i++)
			{
				t[k][i]=new point;
				cin>>map[k][i];
				if(map[k][i]=='.')
					map[k][i]='0';
				t[k][i]->x=k;
				t[k][i]->y=i;
				t[k][i]->time=10000000;
				t[k][i]->step=-1;
			}	
		t[0][0]->step=1;
		bfs();
		
		if(t[n-1][m-1]->time<10000000)//如果可以到达目的地 
		{
			cout<<"It takes "<<t[n-1][m-1]->time<<" seconds to reach the target position, let me show you the way."<<endl;
			stack<point> S;
			point p,last;
			k=n-1;
			i=m-1;
			while(k>=0&&i>=0)//由最后点按照每一点的step找到路线 ，找到的上一点都入栈 
			{
				p.x=k;
				p.y=i;
				S.push(p);
				order=t[k][i]->step;
				if(order==4)
					k+=1;
				if(order==3)
					k-=1;
				if(order==2)
					i+=1;
				if(order==1)
					i-=1;
			}
			p=S.top();
			S.pop();
			time=1;
			while(!S.empty())//输出栈 
			{
				last=p;
				p=S.top();
				cout<<(time++)<<"s:("<<last.x<<","<<last.y<<")->("<<p.x<<","<<p.y<<")"<<endl;
				for(k=0;k<map[p.x][p.y]-'0';k++)//输出在某点打怪消耗时间 
					cout<<(time++)<<"s:FIGHT AT ("<<p.x<<","<<p.y<<")"<<endl;
				S.pop();
			}
		}
		else
			cout<<"God please help our poor hero."<<endl;
			
		cout<<"FINISH"<<endl;
		
		for(k=0;k<n;k++)//删除申请空间 
			for(i=0;i<m;i++)
				delete t[k][i];
	}
	return 0;
}

测试样例：

5 6

.XX.1.

..X.9.

2...X.

...XX.

XXXXX.