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:
// Utilities
import java.io.*;
import java.util.*;

public class Main {
	static int T;
	static int N, M;
	static char[][] grid;
	static ArrayList<Pair> idxGood;
	static int goodCnt, goodCmp;
	static int idxGr, idxGc;
	static int[][] d = {{1,0},{0,1},{-1,0},{0,-1}};
	static boolean[][] vis;
	static int nr, nc;
	static boolean existGood, existBad, visEnd;
	
	public static void main(String[] args) throws IOException {
		T = in.iscan();
		outer : while (T-- > 0) {
			N = in.iscan(); M = in.iscan(); grid = new char[N][M];
			existGood = false; existBad = false; idxGood = new ArrayList<Pair>();
			goodCnt = 0; idxGr = -1; idxGc = -1;
			for (int i = 0; i < N; i++) {
				grid[i] = in.sscan().toCharArray();
				for (int j = 0; j < M; j++) {
					if (grid[i][j] == 'G') {
						existGood = true;
						idxGr = i; idxGc = j;
						goodCnt++;
					}
					if (grid[i][j] == 'B') existBad = true;
				}
			}
			for (int i = 0; i < N; i++) {
				for (int j = 0; j < M; j++) {
					if (grid[i][j] == 'B') {
						for (int k = 0; k < 4; k++) {
							nr = i + d[k][0]; nc = j + d[k][1];
							if (nr < 0 || nr == N || nc < 0 || nc == M) continue;
							if (grid[nr][nc] == 'G') {
								out.println("No");
								continue outer;
							}
							if (grid[nr][nc] == '.') grid[nr][nc] = '#';
						}
					}
				}
			}
			if (existGood) {
				vis = new boolean[N][M]; goodCmp = 0; visEnd = false;
				dfs(idxGr, idxGc);
				if (goodCmp == goodCnt && visEnd) out.println("Yes");
				else out.println("No");
			}
			else { // !existGood
				out.println("Yes");
			}
		}
		out.close();
	} 
	
	static void dfs(int r, int c) {
		vis[r][c] = true;
		if (grid[r][c] == 'G') goodCmp++;
		if (r == N-1 && c == M-1) visEnd = true;
		for (int j = 0; j < 4; j++) {
			nr = r + d[j][0]; nc = c + d[j][1];
			if (nr < 0 || nr == N || nc < 0 || nc == M) continue;
			if (grid[nr][nc] != '#' && !vis[nr][nc]) dfs(nr, nc);
		}
	}
	
	static class Pair {
		int r, c;
		Pair(int r, int c){ this.r = r; this.c = c; }
	}
	
	static INPUT in = new INPUT(System.in);
	static PrintWriter out = new PrintWriter(System.out);
	private static class INPUT {

		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar, numChars;

		public INPUT (InputStream stream) {
			this.stream = stream;
		}

		public INPUT (String file) throws IOException {
			this.stream = new FileInputStream (file);
		}

		public int cscan () throws IOException {
			if (curChar >= numChars) {
				curChar = 0;
				numChars = stream.read (buf);
			}
			
			if (numChars == -1)
				return numChars;

			return buf[curChar++];
		}

		public int iscan () throws IOException {
			int c = cscan (), sgn = 1;
			
			while (space (c))
				c = cscan ();

			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}

			int res = 0;

			do {
				res = (res << 1) + (res << 3);
				res += c - '0';
				c = cscan ();
			}
			while (!space (c));

			return res * sgn;
		}

		public String sscan () throws IOException {
			int c = cscan ();
			
			while (space (c))
				c = cscan ();

			StringBuilder res = new StringBuilder ();

			do {
				res.appendCodePoint (c);
				c = cscan ();
			}
			while (!space (c));

			return res.toString ();
		}

		public double dscan () throws IOException {
			int c = cscan (), sgn = 1;
			
			while (space (c))
				c = cscan ();

			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}

			double res = 0;

			while (!space (c) && c != '.') {
				if (c == 'e' || c == 'E')
					return res * UTILITIES.fast_pow (10, iscan ());
				
				res *= 10;
				res += c - '0';
				c = cscan ();
			}

			if (c == '.') {
				c = cscan ();
				double m = 1;

				while (!space (c)) {
					if (c == 'e' || c == 'E')
						return res * UTILITIES.fast_pow (10, iscan ());

					m /= 10;
					res += (c - '0') * m;
					c = cscan ();
				}
			}

			return res * sgn;
		}

		public long lscan () throws IOException {
			int c = cscan (), sgn = 1;
			
			while (space (c))
				c = cscan ();

			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}

			long res = 0;

			do {
				res = (res << 1) + (res << 3);
				res += c - '0';
				c = cscan ();
			}
			while (!space (c));

			return res * sgn;
		}

		public boolean space (int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
	}

	public static class UTILITIES {

		static final double EPS = 10e-6;

		public static int lower_bound (int[] arr, int x) {
			int low = 0, high = arr.length, mid = -1;

			while (low < high) {
				mid = (low + high) / 2;

				if (arr[mid] >= x)
					high = mid;
				else
					low = mid + 1;
			}

			return low;
		}

		public static int upper_bound (int[] arr, int x) {
			int low = 0, high = arr.length, mid = -1;

			while (low < high) {
				mid = (low + high) / 2;

				if (arr[mid] > x)
					high = mid;
				else
					low = mid + 1;
			}

			return low;
		}

		public static int gcd (int a, int b) {
			return b == 0 ? a : gcd (b, a % b);
		}

		public static int lcm (int a, int b) {
			return a * b / gcd (a, b);
		}

		public static long fast_pow_mod (long b, long x, int mod) {
			if (x == 0) return 1;
			if (x == 1) return b;
			if (x % 2 == 0) return fast_pow_mod (b * b % mod, x / 2, mod) % mod;

			return b * fast_pow_mod (b * b % mod, x / 2, mod) % mod;
		}

		public static int fast_pow (int b, int x) {
			if (x == 0) return 1;
			if (x == 1) return b;
			if (x % 2 == 0) return fast_pow (b * b, x / 2);

			return b * fast_pow (b * b, x / 2);
		}

		public static long choose (long n, long k) {
			k = Math.min (k, n - k);
			long val = 1;

			for (int i = 0; i < k; ++i)
				val = val * (n - i) / (i + 1);

			return val;
		}

		public static long permute (int n, int k) {
			if (n < k) return 0;
			long val = 1;

			for (int i = 0; i < k; ++i)
				val = (val * (n - i));

			return val;
		}
	}
}