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:
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStreamWriter;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.PriorityQueue;
import java.util.Map.Entry;


import java.util.Random;
import java.util.TreeSet;

public final class CF_648_D2_D
{


	static boolean verb=true;
	static void log(Object X){if (verb) System.err.println(X);}
	static void log(Object[] X){if (verb) {for (Object U:X) System.err.print(U+" ");System.err.println("");}}
	static void log(int[] X){if (verb) {for (int U:X) System.err.print(U+" ");System.err.println("");}}
	static void logWln(int[] X){if (verb) {for (int U:X) System.err.print(U+" ");}}
	static void log(int[] X,int L){if (verb) {for (int i=0;i<L;i++) System.err.print(X[i]+" ");System.err.println("");}}
	static void log(long[] X){if (verb) {for (long U:X) System.err.print(U+" ");System.err.println("");}}

	static void logWln(Object X){if (verb) System.err.print(X);}
	static void info(Object o){	System.out.println(o);}
	static void output(Object o){outputWln(""+o+"\n");	}
	static void outputFlush(Object o){try {out.write(""+ o+"\n");out.flush();} catch (Exception e) {}}
	static void outputWln(Object o){try {out.write(""+ o);} catch (Exception e) {}}


	static void logBin(int[] tm) {for (int x:tm) logWln(bin(8,x)+" ");log("");}

	static String bin(int L,int x) {
		String s=Integer.toBinaryString(x);
		while (s.length()<L) s="0"+s;
		return s;
	}

	static long powerMod(long b,long e,long m){
		long x=1;
		while (e>0) {
			if (e%2==1)
				x=(b*x)%m;
			b=(b*b)%m;
			e=e/2;
		}
		return x; 
	}


	static class Composite implements Comparable<Composite>{

		int val;
		int idx;


		public int compareTo(Composite X) {
			if (val!=X.val)
				return val-X.val;
			return idx-X.idx;
		}


		public Composite(int val, int idx) {
			this.val = val;
			this.idx = idx;
		}







	}

	// Global vars
	static BufferedWriter out;
	static InputReader reader;

	static int[][] dd={{-1,0},{0,1},{1,0},{0,-1}};



	static long mod=1000000007;

	static void process() throws Exception {

		out = new BufferedWriter(new OutputStreamWriter(System.out));
		reader = new InputReader(System.in);

		int MX=51;
		int[][] stack=new int[MX*MX][2];
		int[][] visited=new int[MX][MX];
		int time=0;
		int T=reader.readInt();
		for (int t=0;t<T;t++) {
			time++;
			int n=reader.readInt();
			int m=reader.readInt();
			char[][] map=new char[n][];
			for (int i=0;i<n;i++) {
				String s=reader.readString();
				map[i]=s.toCharArray();
			}
			boolean ok=true;
			int good=0;
			loop:for (int i=0;i<n;i++)
				for (int j=0;j<m;j++) {
					if (map[i][j]=='B') {
						for (int[] tm:dd) {
							int uu=i+tm[0];
							int vv=j+tm[1];
							if (uu>=0 && uu<n && vv>=0 && vv<m) {
								char c=map[uu][vv];
								if (c=='G') {
									ok=false;
									break loop;
								}
								if (c=='.') {
									map[uu][vv]='#';
								}
							}
						}
					} else if (map[i][j]=='G')
						good++;
				}
			String ans="No";
			if (ok) {
				// do flood fill from zero
				if (map[n-1][m-1]=='.') {
					int st=0;
					stack[st][0]=n-1;
					stack[st][1]=m-1;
					visited[n-1][m-1]=time;
					st++;
					while (st>0) {
						int u=stack[st-1][0];
						int v=stack[st-1][1];
						st--;
						for (int[] tm:dd) {
							int uu=u+tm[0];
							int vv=v+tm[1];
							if (uu>=0 && uu<n && vv>=0 && vv<m && visited[uu][vv]<time) {
								char c=map[uu][vv];
								if (c!='#') {
									stack[st][0]=uu;
									stack[st][1]=vv;
									st++;
									visited[uu][vv]=time;
									if (c=='G')
										good--;
								}
								
							}
						}
						
					}
					
					
				}
				
				if (good==0)
					ans="Yes";
				
			}
			output(ans);

		}

		try {
			out.close();
		} catch (Exception Ex) {
		}

	}





	public static void main(String[] args) throws Exception {
		process();

	}

	static final class InputReader {
		private final InputStream stream;
		private final byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;

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

		private int read() throws IOException {
			if (curChar >= numChars) {
				curChar = 0;
				numChars = stream.read(buf);
				if (numChars <= 0) {
					return -1;
				}
			}
			return buf[curChar++];
		}

		public final String readString() throws IOException {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.append((char) c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public final String readString(int L) throws IOException {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder(L);
			do {
				res.append((char) c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public final int readInt() throws IOException {
			int c = read();
			boolean neg = false;
			while (isSpaceChar(c)) {
				c = read();
			}
			char d = (char) c;
			// log("d:"+d);
			if (d == '-') {
				neg = true;
				c = read();
			}
			int res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			// log("res:"+res);
			if (neg)
				return -res;
			return res;

		}

		public final long readLong() throws IOException {
			int c = read();
			boolean neg = false;
			while (isSpaceChar(c)) {
				c = read();
			}
			char d = (char) c;
			// log("d:"+d);
			if (d == '-') {
				neg = true;
				c = read();
			}
			long res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			// log("res:"+res);
			if (neg)
				return -res;
			return res;

		}

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

}