Problem: Pink Floyd are pulling a prank on Roger Waters. They know he doesn't like [walls](https://www.youtube.com/watch?v=YR5ApYxkU-U), he wants to be able to walk freely, so they are blocking him from exiting his room which can be seen as a grid.

Roger Waters has a square grid of size n× n and he wants to traverse his grid from the upper left (1,1) corner to the lower right corner (n,n). Waters can move from a square to any other square adjacent by a side, as long as he is still in the grid. Also except for the cells (1,1) and (n,n) every cell has a value 0 or 1 in it.

Before starting his traversal he will pick either a 0 or a 1 and will be able to only go to cells values in which are equal to the digit he chose. The starting and finishing cells (1,1) and (n,n) are exempt from this rule, he may go through them regardless of picked digit. Because of this the cell (1,1) takes value the letter 'S' and the cell (n,n) takes value the letter 'F'.

For example, in the first example test case, he can go from (1, 1) to (n, n) by using the zeroes on this path: (1, 1), (2, 1), (2, 2), (2, 3), (3, 3), (3, 4), (4, 4)

The rest of the band (Pink Floyd) wants Waters to not be able to do his traversal, so while he is not looking they will invert at most two cells in the grid (from 0 to 1 or vice versa). They are afraid they will not be quick enough and asked for your help in choosing the cells.  Note that you cannot invert cells (1, 1) and (n, n).

We can show that there always exists a solution for the given constraints.

Also note that Waters will pick his digit of the traversal after the band has changed his grid, so he must not be able to reach (n,n) no matter what digit he picks.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 50). Description of the test cases follows.

The first line of each test case contains one integers n (3 ≤ n ≤ 200).

The following n lines of each test case contain the binary grid, square (1, 1) being colored in 'S' and square (n, n) being colored in 'F'.

The sum of values of n doesn't exceed 200.

Output

For each test case output on the first line an integer c (0 ≤ c ≤ 2) — the number of inverted cells.

In i-th of the following c lines, print the coordinates of the i-th cell you inverted. You may not invert the same cell twice.  Note that you cannot invert cells (1, 1) and (n, n).

Example

Input


3
4
S010
0001
1000
111F
3
S10
101
01F
5
S0101
00000
01111
11111
0001F


Output


1
3 4
2
1 2
2 1
0

Note

For the first test case, after inverting the cell, we get the following grid:
    
    
      
    S010  
    0001  
    1001  
    111F  
    

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.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);
			BPuttingBricksInTheWall solver = new BPuttingBricksInTheWall();
			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 BPuttingBricksInTheWall {
		public BPuttingBricksInTheWall() {
		}

		public void solve(int kase, InputReader in, Output pw) {
			int n = in.nextInt();
			int[][] arr = in.nextIntChar(n, o -> o-'0');
			int a = arr[0][1], b = arr[1][0], c = arr[n-2][n-1], d = arr[n-1][n-2];
			int cnt = (a^1)+(b^1)+c+d;
			ArrayList<String> al = new ArrayList<>(2);
			if(cnt>2) {
				if(a!=0) {
					al.add("1 2");
				}
				if(b!=0) {
					al.add("2 1");
				}
				if(c!=1) {
					al.add((n-1)+" "+(n));
				}
				if(d!=1) {
					al.add((n)+" "+(n-1));
				}
			}else {
				if(a!=1) {
					al.add("1 2");
				}
				if(b!=1) {
					al.add("2 1");
				}
				if(c!=0) {
					al.add((n-1)+" "+(n));
				}
				if(d!=0) {
					al.add((n)+" "+(n-1));
				}
			}
			pw.println(al.size());
			for(var v: al) {
				pw.println(v);
			}
		}

	}

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

		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();
			}
		}

	}

	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;
		}

	}
}