Problem: Vivek has encountered a problem. He has a maze that can be represented as an n × m grid. Each of the grid cells may represent the following:

  * Empty — '.' 
  * Wall — '#' 
  * Good person — 'G' 
  * Bad person — 'B' 



The only escape from the maze is at cell (n, m).

A person can move to a cell only if it shares a side with their current cell and does not contain a wall. Vivek wants to block some of the empty cells by replacing them with walls in such a way, that all the good people are able to escape, while none of the bad people are able to. A cell that initially contains 'G' or 'B' cannot be blocked and can be travelled through.

Help him determine if there exists a way to replace some (zero or more) empty cells with walls to satisfy the above conditions.

It is guaranteed that the cell (n,m) is empty. Vivek can also block this cell.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers n, m (1 ≤ n, m ≤ 50) — the number of rows and columns in the maze.

Each of the next n lines contain m characters. They describe the layout of the maze. If a character on a line equals '.', the corresponding cell is empty. If it equals '#', the cell has a wall. 'G' corresponds to a good person and 'B' corresponds to a bad person.

Output

For each test case, print "Yes" if there exists a way to replace some empty cells with walls to satisfy the given conditions. Otherwise print "No"

You may print every letter in any case (upper or lower).

Example

Input


6
1 1
.
1 2
G.
2 2
#B
G.
2 3
G.#
B#.
3 3
#B.
#..
GG.
2 2
#B
B.


Output


Yes
Yes
No
No
Yes
Yes

Note

For the first and second test cases, all conditions are already satisfied.

For the third test case, there is only one empty cell (2,2), and if it is replaced with a wall then the good person at (1,2) will not be able to escape.

For the fourth test case, the good person at (1,1) cannot escape.

For the fifth test case, Vivek can block the cells (2,3) and (2,2).

For the last test case, Vivek can block the destination cell (2, 2).

Solution (in JAVA):

Solution:
/**
 * @author vivek
 * programming is thinking, not typing
 */

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class Main {
    static HashMap<Pair, Boolean> maap;

    private static void solveTC(int __) {

        /* For Google */
//		ans.append("Case #").append(__).append(": ");

        //code start

        maap = new HashMap<>();

        int n = scn.nextInt();
        int m = scn.nextInt();

        char[][] array = new char[n][];

        HashSet<Pair> god = new HashSet<>();
        HashSet<Pair> bad = new HashSet<>();
        HashSet<Pair> wall = new HashSet<>();

        for (int i = 0; i < n; i++) {
            array[i] = scn.next().toCharArray();
            for (int j = 0; j < m; j++) {
                if (array[i][j] == 'B') {
                    bad.add(new Pair(i, j));
                } else if (array[i][j] == 'G') {
                    god.add(new Pair(i, j));
                } else if (array[i][j] == '#') {
                    wall.add(new Pair(i, j));
                }
            }
        }

        for (Pair p : bad) {
            int a = p.a;
            int b = p.b;
            int x = a - 1, y = b;
            if (x >= 0) {
                if (array[x][y] == 'G') {
                    print("No\n");
                    return;
                }
                array[x][y] = '#';
            }
            x = a + 1;
            if (x < n) {
                if (array[x][y] == 'G') {
                    print("No\n");
                    return;
                }
                array[x][y] = '#';
            }
            x = a;
            y = b - 1;
            if (y >= 0) {
                if (array[x][y] == 'G') {
                    print("No\n");
                    return;
                }
                array[x][y] = '#';
            }
            y = b + 1;
            if (y < m) {
                if (array[x][y] == 'G') {
                    print("No\n");
                    return;
                }
                array[x][y] = '#';
            }
        }

        for (Pair p : god) {
            if (!isPath(p.a, p.b, array)) {
                print("No\n");
                return;
            }
        }

        print("Yes");

        //code end

        print("\n");
    }

    private static boolean isPath(int a, int b, char[][] array) {
        if (a == array.length - 1 && b == array[a].length - 1) {
            return true;
        }
        HashSet<Pair> set = new HashSet<>();
        set.add(new Pair(a, b));
        return isPath(a + 1, b, array, set) ||
                isPath(a - 1, b, array, set) ||
                isPath(a, b + 1, array, set) ||
                isPath(a, b - 1, array, set);
    }

    private static boolean isPath(int a, int b, char[][] array, HashSet<Pair> visited) {
        Pair paa = new Pair(a, b);
        if (maap.containsKey(paa)) {
            if (maap.get(paa))
                return true;
        }
        if (a < 0 || b < 0) {
            return false;
        }
        if (a == array.length - 1 && b == array[a].length - 1) {
            return array[a][b] == '.' || array[a][b] == 'G';
        }
        if (a >= array.length || b >= array[a].length) {
            return false;
        }
        if (array[a][b] != '.' && array[a][b] != 'G') {
            return false;
        }
        Pair p = new Pair(a, b);
        if (visited.contains(p)) {
            return false;
        } else {
            visited.add(p);
        }

        maap.put(paa, isPath(a - 1, b, array, visited) ||
                isPath(a + 1, b, array, visited) ||
                isPath(a, b - 1, array, visited) ||
                isPath(a, b + 1, array, visited));
        return maap.get(paa);

    }

    static class Pair {
        int a, b;

        public Pair(int a, int b) {
            this.a = a;
            this.b = b;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Pair pair = (Pair) o;
            return a == pair.a &&
                    b == pair.b;
        }

        @Override
        public int hashCode() {
            return Objects.hash(a, b);
        }
    }

    public static void main(String[] args) {

//        int limit=               ;
//        sieve(limit);

        scn = new Scanner();
        ans = new StringBuilder();
        int t = scn.nextInt();
//        int t = 1;
        for (int i = 1; i <= t; i++) {
            solveTC(i);
        }
        System.out.print(ans);
    }
    //Stuff for prime start
    /**
     * List containing prime numbers <br>
     * <b>i<sup>th</sup></b> position contains <b>i<sup>th</sup></b> prime number <br>
     * 0th index is <b>null</b>
     */
    private static ArrayList<Integer> listOfPrimes;

    /**
     * query <b>i<sup>th</sup></b> element to get if its prime of not
     */
    private static boolean[] isPrime;

    /**
     * Performs Sieve of Erathosnesis and initialise isPrime array and listOfPrimes list
     *
     * @param limit the number till which sieve is to be performed
     */
    private static void sieve(int limit) {
        listOfPrimes = new ArrayList<>();
        listOfPrimes.add(null);
        boolean[] array = new boolean[limit + 1];

        Arrays.fill(array, true);
        array[0] = false;
        array[1] = false;

        for (int i = 2; i <= limit; i++) {
            if (array[i]) {
                for (int j = i * i; j <= limit; j += i) {
                    array[j] = false;
                }
            }
        }
        isPrime = array;
        for (int i = 0; i <= limit; i++) {
            if (array[i]) {
                listOfPrimes.add(i);
            }
        }
    }
    //stuff for prime end

    /**
     * Calculates the Least Common Multiple of two numbers
     *
     * @param a First number
     * @param b Second Number
     * @return Least Common Multiple of <b>a</b> and <b>b</b>
     */
    private static long lcm(long a, long b) {
        return a * b / gcd(a, b);
    }

    /**
     * Calculates the Greatest Common Divisor of two numbers
     *
     * @param a First number
     * @param b Second Number
     * @return Greatest Common Divisor of <b>a</b> and <b>b</b>
     */
    private static long gcd(long a, long b) {
        return (b == 0) ? a : gcd(b, a % b);
    }

    static void print(Object obj) {
        ans.append(obj.toString());
    }


    static Scanner scn;
    static StringBuilder ans;

    //Fast Scanner
    static class Scanner {
        BufferedReader br;
        StringTokenizer st;

        public Scanner() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        int[] nextIntArray(int n) {
            int[] array = new int[n];
            for (int i = 0; i < n; i++) {
                array[i] = nextInt();
            }
            return array;
        }

        Integer[] nextIntegerArray(int n) {
            Integer[] array = new Integer[n];
            for (int i = 0; i < n; i++) {
                array[i] = nextInt();
            }
            return array;
        }

        long[] nextLongArray(int n) {
            long[] array = new long[n];
            for (int i = 0; i < n; i++) {
                array[i] = nextLong();
            }
            return array;
        }

        String[] nextStringArray() {
            return nextLine().split(" ");
        }

        String[] nextStringArray(int n) {
            String[] array = new String[n];
            for (int i = 0; i < n; i++) {
                array[i] = next();
            }
            return array;
        }
    }
}