Problem:
Score : 400 points

Problem Statement

Takahashi has a maze, which is a grid of H \times W squares with H horizontal rows and W vertical columns.

The square at the i-th row from the top and the j-th column is a "wall" square if S_{ij} is #, and a "road" square if S_{ij} is ..

From a road square, you can move to a horizontally or vertically adjacent road square.

You cannot move out of the maze, move to a wall square, or move diagonally.

Takahashi will choose a starting square and a goal square, which can be any road squares, and give the maze to Aoki.

Aoki will then travel from the starting square to the goal square, in the minimum number of moves required.

In this situation, find the maximum possible number of moves Aoki has to make.

Constraints

1 \leq H,W \leq 20

S_{ij} is . or #.

S contains at least two occurrences of ..

Any road square can be reached from any road square in zero or more moves.

Input

Input is given from Standard Input in the following format:

H W
S_{11}...S_{1W}
:
S_{H1}...S_{HW}

Output

Print the maximum possible number of moves Aoki has to make.

Sample Input 1

3 3
...
...
...

Sample Output 1

4

If Takahashi chooses the top-left square as the starting square and the bottom-right square as the goal square, Aoki has to make four moves.

Sample Input 2

3 5
...#.
.#.#.
.#...

Sample Output 2

10

If Takahashi chooses the bottom-left square as the starting square and the top-right square as the goal square, Aoki has to make ten moves.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {

    public static void main(String[] args) {

        Scanner in = new Scanner(System.in);
        String[] params = in.nextLine().split(" ");
        int h = Integer.parseInt(params[0]);
        int w = Integer.parseInt(params[1]);
        String[][] ss = new String[h][w];
        for (int i = 0; i < h; i++) {
            params = in.nextLine().split("");
            for (int j = 0; j < w; j++) {
                ss[i][j] = params[j];
            }
        }

        // 全ての場所をスタート地点として最長距離を回答とする
        int maxDistance = 0;
        String[][] copy = new String[h][];
        for (int i = 0; i < h; i++) {
            for (int j = 0; j < w; j++) {
                maxDistance = Math.max( maxDistance , getMaxDistanceWithBfs( ss , i , j ) );
            }
        }

        System.out.println(maxDistance);

    }

    /**
     * 幅優先探索で最長距離を返す
     * @param ss 迷路の配列
     * @param startY スタート地点のY座標
     * @param startX スタート地点のX座標
     * @return 最長距離
     */
    private static int getMaxDistanceWithBfs( String[][] ss , int startY , int startX ) {

        int[][] distances = new int[ss.length][ss[0].length];
        Queue<Integer> queue = new ArrayDeque<>();

        // 四方の定義 順に左上右下
        int[] dx = { -1 , 0 , 1 , 0 };
        int[] dy = { 0 , 1 , 0 , -1 };

        int maxDistance = 0;

        // スタート位置をキューに追加しておく
        queue.add( startX );
        queue.add( startY );

        // キューが空になった時 = 探索できる場所がなくなった時
        // ループ回数 = スタート位置からの距離
        while( !queue.isEmpty()  ) {

            int x = queue.remove();
            int y = queue.remove();

            // 今いる場所が壁の場合処理しない
            if( "#".equals( ss[y][x] ) ){
                continue;
            }

            maxDistance = Math.max( maxDistance , distances[y][x] );

            // 四方を確認
            for( int i = 0 ; i < 4 ; i++ ){

                // 確認先の座標
                int xx = x + dx[i];
                int yy = y + dy[i];

                // 確認先の座標が配列外にいないこと
                if( !( 0 <= yy && yy < ss.length && 0 <= xx && xx < ss[0].length ) ) {
                    continue;
                }

                // 確認先の座標が壁でないこと
                if( !".".equals( ss[yy][xx] ) ) {
                    continue;
                }

                // 既に通った道でないこと
                if( distances[yy][xx] != 0 ){
                    continue;
                }

                queue.add( xx );
                queue.add( yy );
                distances[yy][xx] = distances[y][x] + 1;

            }

        }

        return maxDistance;

    }

}