//多源最短路 两个源点的最短路
//存图和判断
#include <bits/stdc++.h>
using namespace std;

typedef pair<int, int> PII;
const int N = 210;
int n, m, res, d[N][N];
string g[N];
int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};
queue<PII> q;

int main()
{
    cin >> m >> n;
    n = n * 2 + 1;
    m = m * 2 + 1;

    string t;
    getline(cin, t);
    for (int i = 0; i < n; ++ i)
    {
        getline(cin, t);
        g[i] = t;
    }


    memset(d, -1, sizeof d);
    //找到两个出口
    for (int i = 0; i < n; ++ i)
        for (int j = 0; j < m; ++ j)
        {
            if (i != 0 && i != n-1 && j != 0 && j != m-1) continue;
            if (g[i][j] != '+' && g[i][j] != '-' && g[i][j] != '|') 
            {
                q.push({i, j});
                d[i][j] = 0;
            }
        }


    while (q.size())
    {
        auto t = q.front(); q.pop();
        for (int i = 0; i < 4; ++ i)
        {
            int x = t.first + dx[i], y = t.second + dy[i];
            if (x < 0 || x >= n || y < 0 || y >= m) continue;
            if (g[x][y] != ' ' || d[x][y] != -1) continue;
            d[x][y] = d[t.first][t.second] + 1;
            q.push({x, y});
            res = max(res, d[x][y]);
        }
    }
    cout << (res + 1) / 2; //迷宫内的移动，两步算真实的一步，最后出去的1步，算1步， 总共加起来除以2
    return 0;
}