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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;

public class A {

	static ArrayList<String> moves=new ArrayList<>();
	static boolean[][] board;
	static int w, h;
	
	public static void main(String[] args) {
		FastScanner fs=new FastScanner();
		int T=fs.nextInt();
		for (int tt=0; tt<T; tt++) {
			moves.clear();
			h=fs.nextInt(); w=fs.nextInt();
			board=new boolean[w][h];
			for (int y=0; y<h; y++) {
				char[] line=fs.next().toCharArray();
				for (int x=0; x<w; x++)
					board[x][y]=line[x]=='1';
			}
			for (int y=0; y+2<h; y++) {
				for (int x=0; x<w-1; x++) {
					if (board[x][y]) {
						topLeft(x, y);
					}
				}
				if (board[w-1][y]) {
					bottomRight(w-2, y);
				}
			}
			for (int x=0; x+2<w; x++) {
				if (board[x][h-2]) {
					topLeft(x, h-2);
				}
				if (board[x][h-1]) {
					bottomRight(x, h-2);
				}
			}
			//TODO: solve last 4x4
			boolean tl=board[w-2][h-2], tr=board[w-1][h-2], bl=board[w-2][h-1], br=board[w-1][h-1];
			if (!tl&&!tr&&!bl&&!br) {
				//nothing
			}
			//just one
			else if (!tl&&!tr&&!bl&&br) {
				tr();
				br();
				bl();
			}
			else if (!tl&&!tr&&bl&&!br) {
				tl();
				bl();
				br();
			}
			else if (!tl&&tr&&!bl&&!br) {
				tl();
				tr();
				br();
			}
			else if (tl&&!tr&&!bl&&!br) {
				tl();
				tr();
				bl();
			}
			//two oppisite, pick them
			else if (tl&&!tr&&!bl&&br) {
				tl();
				br();
			}
			else if (!tl&&tr&&bl&&!br) {
				tr();
				bl();
			}
			//two adjacent, pick others
			else if (tl&&tr&&!bl&&!br) {
				bl();
				br();
			}
			else if (!tl&&tr&&!bl&&br) {
				tl();
				bl();
			}
			else if (!tl&&!tr&&bl&&br) {
				tl();
				tr();
			}
			else if (tl&&!tr&&bl&&!br) {
				tr();
				br();
			}
			//three
			else if (!tl&&tr&&bl&&br) {
				br();
			}
			else if (tl&&!tr&&bl&&br) {
				bl();
			}
			else if (tl&&tr&&!bl&&br) {
				tr();
			}
			else if (tl&&tr&&bl&&!br) {
				tl();
			}
			// four
			else if (tl&&tr&&bl&&br) {
				tl();
				tr();
				bl();
				br();
			}
			else throw null;
			for (boolean[] a:board)
				for (boolean b:a) if (b) throw null;
			System.out.println(moves.size());
			for (String s:moves) {
				System.out.println(s);
			}
		}
	}
	
	static void tl() {
		topLeft(w-2, h-2);
	}
	static void tr() {
		topRight(w-2, h-2);
	}
	static void bl() {
		bottomLeft(w-2, h-2);
	}
	static void br() {
		bottomRight(w-2, h-2);
	}
	
	static void topLeft(int x, int y) {
		add(x, y, x+1, y, x, y+1);
	}
	static void topRight(int x, int y) {
		add(x, y, x+1, y, x+1, y+1);
	}
	static void bottomLeft(int x, int y) {
		add(x, y, x, y+1, x+1, y+1);
	}
	static void bottomRight(int x, int y) {
		add(x+1, y, x, y+1, x+1, y+1);
	}
	
	static void add(int x1, int y1, int x2, int y2, int x3, int y3) {
		board[x1][y1]^=true;
		board[x2][y2]^=true;
		board[x3][y3]^=true;
		x1++;
		y1++;
		x2++;
		y2++;
		x3++;
		y3++;
		moves.add(y1+" "+x1+" "+y2+" "+x2+" "+y3+" "+x3);
	}
	
	

	static void sort(int[] a) {
		ArrayList<Integer> l=new ArrayList<>();
		for (int i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}
	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}

}