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

用一个三维数组去标记，注意求的是两个人在路上用的总时间

import java.util.*;
public class Main {
    static int n,m;
    static String[] map;
    static int[][][] vis;
    static int[][] move = {{0,1},{0,-1},{1,0},{-1,0}};
    static Queue<Node> q;
    
    static int bfs(Node a,Node b) {
        q.add(a);q.add(b);
        int res = Integer.MAX_VALUE;
        vis[a.x][a.y][a.op] = 1;vis[b.x][b.y][b.op] = 1;
        Node cur;
        while(!q.isEmpty()) {
            cur = new Node(q.peek().x,q.peek().y,q.poll().op);
            if(map[cur.x].charAt(cur.y) == '@' && vis[cur.x][cur.y][0] != 0 && vis[cur.x][cur.y][1] != 0)
                res = Math.min(res,vis[cur.x][cur.y][0] + vis[cur.x][cur.y][1]);
            for(int i = 0;i < 4;i++) {
                Node new_cur = new Node(cur.x + move[i][0],cur.y + move[i][1],cur.op);
                if(check(new_cur.x,new_cur.y,new_cur.op)) {
                    vis[new_cur.x][new_cur.y][new_cur.op] = vis[cur.x][cur.y][cur.op] + 1;
                    q.add(new_cur);
                }
            }
        }
        return res - 2;
    }
    
    static boolean check(int x,int y,int i) {
        return x >= 0 && x < n && y >= 0 && y < m  && map[x].charAt(y) != '#' && vis[x][y][i] == 0;
    }
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        while(cin.hasNext()) {
            q = new LinkedList<Node>();
            n = cin.nextInt();
            m = cin.nextInt();
            vis = new int[205][205][2];
            map = new String[205];
            Node a = new Node();
            Node b = new Node();
            for(int i = 0;i < n;i++) {
                map[i] = cin.next();
                for(int j = 0;j < m;j++) {
                    if(map[i].charAt(j) == 'Y') 
                        a = new Node(i,j,0);
                    if(map[i].charAt(j) == 'M') 
                        b = new Node(i,j,1);
                }
            }
            int ans = bfs(a,b);
            System.out.println(ans * 11);
        }
    }
}
class Node {
    int x,y,op;//op等于0代表y,op等于1代表m
    Node(){}
    Node(int x,int y,int op) {
        this.x = x;
        this.y = y;
        this.op = op;
    }
}