Problem: Your friend Salem is Warawreh's brother and only loves math and geometry problems. He has solved plenty of such problems, but according to Warawreh, in order to graduate from university he has to solve more graph problems. Since Salem is not good with graphs he asked your help with the following problem.

<image>

You are given a complete directed graph with n vertices without self-loops. In other words, you have n vertices and each pair of vertices u and v (u ≠ v) has both directed edges (u, v) and (v, u).

Every directed edge of the graph is labeled with a single character: either 'a' or 'b' (edges (u, v) and (v, u) may have different labels).

You are also given an integer m > 0. You should find a path of length m such that the string obtained by writing out edges' labels when going along the path is a palindrome. The length of the path is the number of edges in it.

You can visit the same vertex and the same directed edge any number of times.

Input

The first line contains a single integer t (1 ≤ t ≤ 500) — the number of test cases.

The first line of each test case contains two integers n and m (2 ≤ n ≤ 1000; 1 ≤ m ≤ 10^{5}) — the number of vertices in the graph and desirable length of the palindrome.

Each of the next n lines contains n characters. The j-th character of the i-th line describes the character on the edge that is going from node i to node j.

Every character is either 'a' or 'b' if i ≠ j, or '*' if i = j, since the graph doesn't contain self-loops.

It's guaranteed that the sum of n over test cases doesn't exceed 1000 and the sum of m doesn't exceed 10^5.

Output

For each test case, if it is possible to find such path, print "YES" and the path itself as a sequence of m + 1 integers: indices of vertices in the path in the appropriate order. If there are several valid paths, print any of them.

Otherwise, (if there is no answer) print "NO".

Example

Input


5
3 1
*ba
b*b
ab*
3 3
*ba
b*b
ab*
3 4
*ba
b*b
ab*
4 6
*aaa
b*ba
ab*a
bba*
2 6
*a
b*


Output


YES
1 2
YES
2 1 3 2
YES
1 3 1 3 1
YES
1 2 1 3 4 1 4
NO

Note

The graph from the first three test cases is shown below:

<image>

In the first test case, the answer sequence is [1,2] which means that the path is:

$$$1 \xrightarrow{b} 2$$$

So the string that is obtained by the given path is b.

In the second test case, the answer sequence is [2,1,3,2] which means that the path is:

$$$2 \xrightarrow{b} 1 \xrightarrow{a} 3 \xrightarrow{b} 2$$$

So the string that is obtained by the given path is bab.

In the third test case, the answer sequence is [1,3,1,3,1] which means that the path is:

$$$1 \xrightarrow{a} 3 \xrightarrow{a} 1 \xrightarrow{a} 3 \xrightarrow{a} 1$$$

So the string that is obtained by the given path is aaaa.

The string obtained in the fourth test case is abaaba.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
import java.math.*;

/**
 *
 * @Har_Har_Mahadev
 */

public class D {

	private static long INF = 2000000000000000000L, M = 1000000007, MM = 998244353;
	private static int N = 0;

	public static void process() throws IOException {

		int n = sc.nextInt(),m = sc.nextInt();
		int arr[][] = new int[n+1][n+1];
		for(int i=0; i<n;i++) {
			String str = sc.next();
			for(int j=0; j<n; j++) {
				if(i == j)continue;
				arr[i][j] = str.charAt(j) - 'a'+1;
			}
		}
		if(m%2 == 1) {
			System.out.println("YES");
			for(int i=0; i<=m; i++) {
				if(i%2 == 1)System.out.print(1+" ");
				else System.out.print(2+" ");
			}
			System.out.println();
			return;
		}
		for(int i=0; i<n; i++) {
			for(int j=0; j<n; j++) {
				if(i == j)continue;
				if(arr[i][j] == arr[j][i]) {
					System.out.println("YES");
					for(int ii=0; ii<=m; ii++) {
						if(ii%2 == 1)System.out.print((i+1)+" ");
						else System.out.print((j+1)+" ");
					}
					System.out.println();
					return;
				}
			}
		}
		
		for(int i=0; i<n; i++) {
			int a = 0,b = 0;
			for(int j=0; j<n; j++) {
				if(arr[i][j] == 1)a = j+1;
				if(arr[i][j] == 2)b = j+1;
			}
			if(a > 0 && b > 0) {
				System.out.println("YES");
				if(m%4==0)
                {
                    for(int j=0;j<=m;j++)
                    {
                        if(j%2==0)
                            System.out.print((i+1)+" ");
                        else
                        {
                            if(j<m/2)
                            	System.out.print(a+" ");
                            else System.out.print(b+" ");
                            
                        }
                    }
                }
                else
                {
                    for(int j=0;j<=m;j++)
                    {
                        if(j%2==1)
                            System.out.print((i+1)+" ");
                        else
                        {
                            if(j<m/2)
                                System.out.print(a+" ");
                            else System.out.print(b+" ");
                        }
                    }
                }
                return;
			}
			
		}
		System.out.println("NO");

	}

	//=============================================================================
	//--------------------------The End---------------------------------
	//=============================================================================

	static FastScanner sc;
	static PrintWriter out;

	public static void main(String[] args) throws IOException {
		boolean oj = true;
		if (oj) {
			sc = new FastScanner();
			out = new PrintWriter(System.out);
		} else {
			sc = new FastScanner(100);
			out = new PrintWriter("output.txt");
		}
		int t = 1;
		t = sc.nextInt();
		while (t-- > 0) {
			process();
		}
		out.flush();
		out.close();
	}

	static class Pair implements Comparable<Pair> {
		int x, y;

		Pair(int x, int y) {
			this.x = x;
			this.y = y;
		}

		@Override
		public int compareTo(Pair o) {
			return Integer.compare(this.x, o.x);
		}

		//		 @Override
		//		    public boolean equals(Object o) {
		//		        if (this == o) return true;
		//		        if (!(o instanceof Pair)) return false;
		//		        Pair key = (Pair) o;
		//		        return x == key.x && y == key.y;
		//		    }
		//		 
		//		    @Override
		//		    public int hashCode() {
		//		        int result = x;
		//		        result = 31 * result + y;
		//		        return result;
		//		    }
	}

	/////////////////////////////////////////////////////////////////////////////////////////////////////////

	static void println(Object o) {
		out.println(o);
	}

	static void println() {
		out.println();
	}

	static void print(Object o) {
		out.print(o);
	}

	static void pflush(Object o) {
		out.println(o);
		out.flush();
	}

	static int ceil(int x, int y) {
		return (x % y == 0 ? x / y : (x / y + 1));
	}

	static long ceil(long x, long y) {
		return (x % y == 0 ? x / y : (x / y + 1));
	}

	static int max(int x, int y) {
		return Math.max(x, y);
	}

	static int min(int x, int y) {
		return Math.min(x, y);
	}

	static int abs(int x) {
		return Math.abs(x);
	}

	static long abs(long x) {
		return Math.abs(x);
	}

	static int log2(int N) {
		int result = (int) (Math.log(N) / Math.log(2));
		return result;
	}

	static long max(long x, long y) {
		return Math.max(x, y);
	}

	static long min(long x, long y) {
		return Math.min(x, y);
	}

	public static int gcd(int a, int b) {
		BigInteger b1 = BigInteger.valueOf(a);
		BigInteger b2 = BigInteger.valueOf(b);
		BigInteger gcd = b1.gcd(b2);
		return gcd.intValue();
	}

	public static long gcd(long a, long b) {
		BigInteger b1 = BigInteger.valueOf(a);
		BigInteger b2 = BigInteger.valueOf(b);
		BigInteger gcd = b1.gcd(b2);
		return gcd.longValue();
	}

	public static long lca(long a, long b) {
		return (a * b) / gcd(a, b);
	}

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

	/////////////////////////////////////////////////////////////////////////////////////////////////////////

	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		FastScanner() throws FileNotFoundException {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		FastScanner(int a) throws FileNotFoundException {
			br = new BufferedReader(new FileReader("input.txt"));
		}

		String next() throws IOException {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		String nextLine() throws IOException {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}

		int[] readArray(int n) throws IOException {
			int[] A = new int[n];
			for (int i = 0; i != n; i++) {
				A[i] = sc.nextInt();
			}
			return A;
		}

		long[] readArrayLong(int n) throws IOException {
			long[] A = new long[n];
			for (int i = 0; i != n; i++) {
				A[i] = sc.nextLong();
			}
			return A;
		}
	}

	static void ruffleSort(int[] a) {
		Random get = new Random();
		for (int i = 0; i < a.length; i++) {
			int r = get.nextInt(a.length);
			int temp = a[i];
			a[i] = a[r];
			a[r] = temp;
		}
		Arrays.sort(a);
	}

	static void ruffleSort(long[] a) {
		Random get = new Random();
		for (int i = 0; i < a.length; i++) {
			int r = get.nextInt(a.length);
			long temp = a[i];
			a[i] = a[r];
			a[r] = temp;
		}
		Arrays.sort(a);
	}
}