#include<bits/stdc++.h>

 using namespace std;

 typedef unsigned int ui;

 typedef long long ll;

 typedef unsigned long long ull;

 #define pf printf

 #define mem(a,b) memset(a,b,sizeof(a))

 #define prime1 1e9+7

 #define prime2 1e9+9

 #define pi 3.14159265

 #define lson l,mid,rt<<1

 #define rson mid+1,r,rt<<1|1

 #define scand(x) scanf("%llf",&x)

 #define f(i,a,b) for(int i=a;i<=b;i++)

 #define scan(a) scanf("%d",&a)

 #define dbg(args) cout<<#args<<":"<<args<<endl;

 #define inf 0x3f3f3f3f

 #define maxn 205

 int n,m,t;

 char Map[maxn][maxn];

 bool vis[maxn][maxn];

 int a[][];

 struct node{

     int x,y,step;

 };

 node st,cur,nxt;

 vector<node>p1;

 vector<node>p2;

 int dir[][]={,,,-,,,-,};

 void bfs(int x,int y,vector<node> &p)

 {

     mem(vis,false);//记住放在bfs里面，该函数将会多次调用

     queue<node>q;

     st.x=x,st.y=y,st.step=;

     q.push(st);

     vis[x][y]=true;

     while(!q.empty())

     {

         cur=q.front();

         q.pop();

         if(Map[cur.x][cur.y]=='@')

         {

             p.push_back(cur);//每个点只有一次机会进入队列，故vector中只有不同的'@'位置

         }

         f(i,,)

         {

             nxt=cur;

             nxt.x+=dir[i][];

             nxt.y+=dir[i][];

             nxt.step++;

             if(nxt.x<||nxt.x>n||nxt.y<||nxt.y>m||Map[nxt.x][nxt.y]=='#')continue;

             if(vis[nxt.x][nxt.y])continue;

             vis[nxt.x][nxt.y]=true;

             q.push(nxt);

         }

     }

 }

 int main()

 {

     //freopen("input.txt","r",stdin);

     //freopen("output.txt","w",stdout);

     std::ios::sync_with_stdio(false);

     while(scanf("%d%d",&n,&m)==)

     {

         char c;

         f(i,,n)

             f(j,,m)

             {

                 scanf(" %c",&Map[i][j]);

                 if(Map[i][j]=='Y')a[][]=i,a[][]=j;

                 if(Map[i][j]=='M')a[][]=i,a[][]=j;

             }

             p1.clear();

             p2.clear();

             bfs(a[][],a[][],p1);

             bfs(a[][],a[][],p2);

             int ans=inf;

             f(i,,p1.size()-)

                 f(j,,p2.size()-)

                 {

                     if((p1[i].x==p2[j].x)&&(p1[i].y==p2[j].y))//到达同一个点

                     {

                         ans=min(ans,p1[i].step+p2[j].step);

                     //    dbg(ans);

                     }

                 }

             pf("%d\n",ans*);

      }

  }