Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifenfei have many people to meet. Especially a good friend Merceki.
Yifenfei’s home is at the countryside, but Merceki’s home is in the center of city. So yifenfei made arrangements with Merceki to meet at a KFC. There are many KFC in Ningbo, they want to choose one that let the total time to it be most smallest.
Now give you a Ningbo map, Both yifenfei and Merceki can move up, down ,left, right to the adjacent road by cost 11 minutes.

Input

The input contains multiple test cases.
Each test case include, first two integers n, m. (2<=n,m<=200).
Next n lines, each line included m character.
‘Y’ express yifenfei initial position.
‘M’    express Merceki initial position.
‘#’ forbid road;
‘.’ Road.
‘@’ KCF

Output

For each test case output the minimum total time that both yifenfei and Merceki to arrival one of KFC.You may sure there is always have a KFC that can let them meet.

Sample Input


4 4 Y.#@ .... .#.. @..M 4 4 Y.#@ .... .#.. @#.M 5 5 Y..@. .#... .#... @..M. #...#


Sample Output


66 88 66


题意是 : 两个人约定要在最近的kfc见面 , 两个人分别从 Y ,M 出发 ,求最短时间。

思路 : 一开始思路是枚举每个 '@'然后 两遍 bfs  求'@' 分别到 Y 和 M 的最短时间,然后求和,结果T了.

另一个思路就是,两次bfs,用二维数组保存从起点到不同位置的时间,然后求出计算出两个人到kfc 的 时间和。

#include <iostream>
#include <cstring> 
#include <algorithm> 
#include <queue> 
using namespace std ;
const int MAX = 1005;
const int inf = 0x3f3f3f3f;
struct node{
    int x,y;
};
int m,n;
char map[MAX][MAX];
int st1[MAX][MAX],st2[MAX][MAX];
bool vis[MAX][MAX];
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
 
void bfs(int st[][MAX],node p)
{
    queue<node> q;
    q.push(p);
    vis[p.x][p.y] = 1;
    st[p.x][p.y] = 0;
    while(!q.empty())
    {
        node t = q.front();
        q.pop();
        for(int i=0;i<4;i++)
        {
            int tx=t.x+dx[i];
            int ty=t.y+dy[i];
            if(tx>=0&&tx<n&&ty>=0&&ty<m&&!vis[tx][ty]&&map[tx][ty]!='#')
            {
                st[tx][ty]=st[t.x][t.y]+1;
                vis[tx][ty]=true;
                node u ; 
                u.x = tx  ; 
                u.y = ty  ;
                q.push(u);
            }
        }
    }
}
 
int main()
{
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        memset(st1,-1,sizeof(st1));
        memset(st2,-1,sizeof(st2));
        memset(vis,0,sizeof(vis));
        node sty,stm;
        for(int i=0;i<n;i++)
        {
            scanf("%s",map[i]);
            for(int j=0;j<m;j++)
            {
                if(map[i][j]=='Y')
                {
                    sty.x=i;
                    sty.y=j;
                }
                if(map[i][j]=='M')
                {
                    stm.x=i;
                    stm.y=j;
                }
            }
        }
        bfs(st1,sty);
        memset(vis,0,sizeof(vis));
        bfs(st2,stm);
 
        int ans=inf;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(map[i][j]=='@'&&st1[i][j]!=-1&&st2[i][j]!=-1)
                {
                    ans=min(ans,st1[i][j]+st2[i][j]);
                }
            }
        }
 
        printf("%d\n",ans*11);
    }
    return 0;
}