Problem: This is the easy version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.

You are given a binary table of size n × m. This table consists of symbols 0 and 1.

You can make such operation: select 3 different cells that belong to one 2 × 2 square and change the symbols in these cells (change 0 to 1 and 1 to 0).

Your task is to make all symbols in the table equal to 0. You are allowed to make at most 3nm operations. You don't need to minimize the number of operations.

It can be proved that it is always possible.

Input

The first line contains a single integer t (1 ≤ t ≤ 5000) — the number of test cases. The next lines contain descriptions of test cases.

The first line of the description of each test case contains two integers n, m (2 ≤ n, m ≤ 100).

Each of the next n lines contains a binary string of length m, describing the symbols of the next row of the table.

It is guaranteed that the sum of nm for all test cases does not exceed 20000.

Output

For each test case print the integer k (0 ≤ k ≤ 3nm) — the number of operations.

In the each of the next k lines print 6 integers x_1, y_1, x_2, y_2, x_3, y_3 (1 ≤ x_1, x_2, x_3 ≤ n, 1 ≤ y_1, y_2, y_3 ≤ m) describing the next operation. This operation will be made with three cells (x_1, y_1), (x_2, y_2), (x_3, y_3). These three cells should be different. These three cells should belong into some 2 × 2 square.

Example

Input


5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101


Output


1
1 1 2 1 2 2
2 
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2 
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2 
1 4 1 5 2 5 
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2

Note

In the first test case, it is possible to make only one operation with cells (1, 1), (2, 1), (2, 2). After that, all symbols will be equal to 0.

In the second test case:

  * operation with cells (2, 1), (3, 1), (3, 2). After it the table will be: 
    
          
    011  
    001  
    000  
    

  * operation with cells (1, 2), (1, 3), (2, 3). After it the table will be: 
    
          
    000  
    000  
    000  
    




In the fifth test case:

  * operation with cells (1, 3), (2, 2), (2, 3). After it the table will be: 
    
          
    010  
    110  
    

  * operation with cells (1, 2), (2, 1), (2, 2). After it the table will be: 
    
          
    000  
    000  
    

Solution (in JAVA):

Solution:
import java.io.BufferedOutputStream;
import java.io.Closeable;
import java.io.DataInputStream;
import java.io.Flushable;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.util.ArrayList;
import java.util.InputMismatchException;
import java.util.Objects;
import java.util.function.Function;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	static class TaskAdapter implements Runnable {
		@Override
		public void run() {
			long startTime = System.currentTimeMillis();
			InputStream inputStream = System.in;
			OutputStream outputStream = System.out;
			FastReader in = new FastReader(inputStream);
			Output out = new Output(outputStream);
			A1BinaryTableEasyVersion solver = new A1BinaryTableEasyVersion();
			int testCount = in.nextInt();
			for(int i = 1; i<=testCount; i++)
				solver.solve(i, in, out);
			out.close();
			System.err.println(System.currentTimeMillis()-startTime+"ms");
		}
	}

	public static void main(String[] args) throws Exception {
		Thread thread = new Thread(null, new TaskAdapter(), "", 1<<28);
		thread.start();
		thread.join();
	}

	static class A1BinaryTableEasyVersion {
		ArrayList<Pair<Integer, Integer>> ans;
		int[][] arr;

		public A1BinaryTableEasyVersion() {
		}

		public void solve(int x, int y) {
			int ox = 0, oy = 0, zx = 0, zy = 0, cnt = 0;
			for(int i = x; i<=x+1; i++) {
				for(int j = y; j<=y+1; j++) {
					if(arr[i][j]==1) {
						ox = i;
						oy = j;
						cnt++;
					}else {
						zx = i;
						zy = j;
					}
				}
			}
			if(cnt==0) {
				return;
			}
			int ax, ay;
			if(cnt==1) {
				ax = zx;
				ay = zy;
			}else if(cnt==2) {
				ax = ox;
				ay = oy;
			}else if(cnt==3) {
				ax = zx;
				ay = zy;
			}else {
				ax = ox;
				ay = oy;
			}
			for(int i = x; i<=x+1; i++) {
				for(int j = y; j<=y+1; j++) {
					if(i!=ax||j!=ay) {
						ans.add(new Pair<>(i, j));
						arr[i][j] ^= 1;
					}
				}
			}
			if(cnt!=3) {
				solve(x, y);
			}
		}

		public void solve(int kase, InputReader in, Output pw) {
			ans = new ArrayList<>();
			int n = in.nextInt(), m = in.nextInt();
			arr = in.nextIntChar(n, o -> o-'0');
			for(int i = 0; i<n-1; i++) {
				for(int j = 0; j<m-1; j++) {
					solve(i, j);
				}
			}
			pw.println(ans.size()/3);
			for(var v: ans) {
				pw.println(v.a+1, v.b+1);
			}
		}

	}

	static interface InputReader {
		String next();

		int nextInt();

		default int[] nextIntChar(Function<Character, Integer> f) {
			String s = next();
			int[] ret = new int[s.length()];
			for(int i = 0; i<s.length(); i++) {
				ret[i] = f.apply(s.charAt(i));
			}
			return ret;
		}

		default int[][] nextIntChar(int n, Function<Character, Integer> f) {
			int[][] ret = new int[n][];
			for(int i = 0; i<n; i++) {
				ret[i] = nextIntChar(f);
			}
			return ret;
		}

	}

	static class Pair<T1, T2> implements Comparable<Pair<T1, T2>> {
		public T1 a;
		public T2 b;

		public Pair(Pair<T1, T2> p) {
			this(p.a, p.b);
		}

		public Pair(T1 a, T2 b) {
			this.a = a;
			this.b = b;
		}

		public String toString() {
			return a+" "+b;
		}

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

		public boolean equals(Object o) {
			if(o instanceof Pair) {
				Pair p = (Pair) o;
				return a.equals(p.a)&&b.equals(p.b);
			}
			return false;
		}

		public int compareTo(Pair<T1, T2> p) {
			int cmp = ((Comparable<T1>) a).compareTo(p.a);
			if(cmp==0) {
				return ((Comparable<T2>) b).compareTo(p.b);
			}
			return cmp;
		}

	}

	static class FastReader implements InputReader {
		final private int BUFFER_SIZE = 1<<16;
		private DataInputStream din;
		private byte[] buffer;
		private int bufferPointer;
		private int bytesRead;

		public FastReader(InputStream is) {
			din = new DataInputStream(is);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public String next() {
			StringBuilder ret = new StringBuilder(64);
			byte c = skip();
			while(c!=-1&&!isSpaceChar(c)) {
				ret.appendCodePoint(c);
				c = read();
			}
			return ret.toString();
		}

		public int nextInt() {
			int ret = 0;
			byte c = skipToDigit();
			boolean neg = (c=='-');
			if(neg) {
				c = read();
			}
			do {
				ret = ret*10+c-'0';
			} while((c = read())>='0'&&c<='9');
			if(neg) {
				return -ret;
			}
			return ret;
		}

		private boolean isSpaceChar(byte b) {
			return b==' '||b=='\r'||b=='\n'||b=='\t'||b=='\f';
		}

		private byte skip() {
			byte ret;
			while(isSpaceChar((ret = read()))) ;
			return ret;
		}

		private boolean isDigit(byte b) {
			return b>='0'&&b<='9';
		}

		private byte skipToDigit() {
			byte ret;
			while(!isDigit(ret = read())&&ret!='-') ;
			return ret;
		}

		private void fillBuffer() {
			try {
				bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
			}catch(IOException e) {
				e.printStackTrace();
				throw new InputMismatchException();
			}
			if(bytesRead==-1) {
				buffer[0] = -1;
			}
		}

		private byte read() {
			if(bytesRead==-1) {
				throw new InputMismatchException();
			}else if(bufferPointer==bytesRead) {
				fillBuffer();
			}
			return buffer[bufferPointer++];
		}

	}

	static class Output implements Closeable, Flushable {
		public StringBuilder sb;
		public OutputStream os;
		public int BUFFER_SIZE;
		public String lineSeparator;

		public Output(OutputStream os) {
			this(os, 1<<16);
		}

		public Output(OutputStream os, int bs) {
			BUFFER_SIZE = bs;
			sb = new StringBuilder(BUFFER_SIZE);
			this.os = new BufferedOutputStream(os, 1<<17);
			lineSeparator = System.lineSeparator();
		}

		public void print(String s) {
			sb.append(s);
			if(sb.length()>BUFFER_SIZE >> 1) {
				flushToBuffer();
			}
		}

		public void println(int i) {
			println(String.valueOf(i));
		}

		public void println(Object... o) {
			for(int i = 0; i<o.length; i++) {
				if(i!=0) {
					print(" ");
				}
				print(String.valueOf(o[i]));
			}
			println();
		}

		public void println(String s) {
			sb.append(s);
			println();
		}

		public void println() {
			sb.append(lineSeparator);
		}

		private void flushToBuffer() {
			try {
				os.write(sb.toString().getBytes());
			}catch(IOException e) {
				e.printStackTrace();
			}
			sb = new StringBuilder(BUFFER_SIZE);
		}

		public void flush() {
			try {
				flushToBuffer();
				os.flush();
			}catch(IOException e) {
				e.printStackTrace();
			}
		}

		public void close() {
			flush();
			try {
				os.close();
			}catch(IOException e) {
				e.printStackTrace();
			}
		}

	}
}