#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <cstring>
#include <unordered_map>
#include <unordered_set>

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
#define IOS std::ios::sync_with_stdio(false)
#define pb push_back
#define inf 0x3f3f3f3f
#define YES cout << "YES" << endl;
#define yes cout << "yes" << endl;
#define no cout << "no" << endl;
#define NO cout << "NO" << endl;
#define int long long
#define x first
#define y second
#define cmp [&](PII a, PII b){ return a.y < b.y; }

const int N = 5e5+10, mod = 1e9+7, M = 1e6+5, K = 1e5+10, Z = 2e5+7;

using namespace std;
typedef long long LL;
typedef priority_queue<int> PQI; 
typedef priority_queue <int, vector<int>, greater<>> PQGI;
typedef pair<int, int> PII;

bool st[55][55];
char g[55][55];
int n, m;
int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};
vector<PII> a, b;

int get(int x1, int y1, int x2, int y2)
{
    return abs(x1 - x2) + abs(y1 - y2);
}

void bfs(int x, int y, int k)
{
    queue<PII> q;
    if(!k) a.pb({x, y});
    else b.pb({x, y});

    q.push({x, y});
    st[x][y] = true;

    while(q.size())
    {
        PII t = q.front();
        q.pop();
        int sx = t.first, sy = t.second;
        for(int i = 0; i < 4; i ++)
        {
            int nx = sx + dx[i], ny = sy + dy[i];
            if(nx >= 0 && nx <= n && ny >= 0 && ny <= m && g[nx][ny] == 'X' && !st[nx][ny])
            {
                q.push({nx, ny});
                if(!k) a.pb({nx, ny});
                else b.pb({nx, ny});
                st[nx][ny] = true;
            }
        }
    }
}

void solve() 
{
    cin >> n >> m;
    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= m; j ++)
            cin >> g[i][j];

    int t = 0;
    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= m; j ++)
            if(g[i][j] == 'X' && !st[i][j])
            {
                bfs(i, j, t);
                t = 1;
            }

    int res = inf;

    for(int i = 0; i < a.size(); i ++)
        for(int j = 0; j < b.size(); j ++)
            res = min(res, get(a[i].first, a[i].second, b[j].first, b[j].second));

    cout << res - 1 << endl;
    return;
}

signed main()
{
    IOS; cin.tie(nullptr), cout.tie(nullptr); 
    int T = 1;
    // cin >> T;
    while( T -- ) solve();
    return 0;
}
//这里填你的代码^^
//注意代码要放在两组三个点之间，才可以正确显示代码高亮哦~