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.io.*;
import java.util.*;
import static java.lang.Math.*;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class TheTreasureOfTheSegments {
    public static void main(String[] args) throws IOException {
        // InputStream inputStream = new FileInputStream("input.txt");
        // OutputStream outputStream = new FileOutputStream("output.txt");
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskA solver = new TaskA();
        solver.solve(in.nextInt(), in, out);
        out.close();
    }
    static class TaskA {
        long mod = (long)(1e9) + 7;
        long fact[];
        int depth[];
        int parentTable[][];
        int degree[];
        ArrayList<Integer> leaves;
        int max = Integer.MIN_VALUE;
        int min = Integer.MAX_VALUE;
        int diam = 0;

        int res = 0;

        public void solve(int testNumber, InputReader in, PrintWriter out) throws IOException {
            while(testNumber-->0){
                int n = in.nextInt();
                int m = in.nextInt();
                char g[][] = new char[n][];
                for (int i = 0; i < n; i++) {
                    g[i] = in.next().toCharArray();
                }
         
                if (n == 2) {
                    if (m % 2 == 0 && g[0][1] != g[1][0])
                        out.println("NO");
                    else {
                        out.println("YES");
                        for (int k = 0; k <= m; k++) {
                            out.print((2 - k % 2) + " ");
                        }
                        out.println();
                    }
                    continue;
                }
                
                boolean found = false;
                X:
                for (int i = 0; i < n; i++) {
                    for (int j = i + 1; j < n; j++) {
                        if (g[i][j] == g[j][i]) {
                            out.println("YES");
                            for (int k = 0; k <= m; k++) {
                                if (k % 2 == 0)
                                    out.print((i + 1) + " ");
                                else
                                    out.print((j + 1) + " ");
                            }
                            out.println();
                            // return;
                            found = true;
                            break X;
                        }
                    }
                }
                if(found)
                    continue;
                int v1 = -1, v2 = -1, v3 = -1;
                for (int i = 0; i < n; i++) {
                    v2 = -1;
                    v3 = -1;
                    for (int j = 0; j < n; j++) {
                        if (g[i][j] == 'a')
                            v2 = j;
                        else if (g[i][j] == 'b')
                            v3 = j;
                    }
                    if (v2 != -1 && v3 != -1) {
                        v1 = i;
                        break;
                    }
                }
         
                out.println("YES");
                if (m % 4 == 0) {
                    for (int k = 0; k < m / 4; k++) {
                        out.print((v1 + 1) + " " + (v2 + 1) + " " + (v1 + 1) + " " + (v3 + 1) + " ");
                    }
                    out.println(v1 + 1);
                } else if (m % 4 == 2) {
                    out.print((v2 + 1) + " ");
                    for (int k = 0; k < m / 4; k++) {
                        out.print((v1 + 1) + " " + (v2 + 1) + " " + (v1 + 1) + " " + (v3 + 1) + " ");
                    }
                    out.println((v1 + 1) + " " + (v3 + 1));
                } else {
                    for (int k = 0; k <= m / 2; k++) {
                        out.print((v1 + 1) + " " + (v2 + 1) + " ");
                    }
                    out.println();
                }
            }
        }

        public int distance(ArrayList<ArrayList<Integer>> a , int u , int v){
            return depth[u]+depth[v] - 2*depth[lca(a , u , v)];
        }

        public int lca(ArrayList<ArrayList<Integer>> a , int u , int v){
            if(depth[v]<depth[u]){
                int x = u;
                u = v;
                v = x;
            }
            int diff = depth[v] - depth[u];
            for(int i=0;i<parentTable[v].length;i++){
                // checking whether the ith bit is set in the diff variable
                if(((diff>>i)&1) == 1)
                    v = parentTable[v][i];
            }
            if(v == u)
                return v;
            for(int i=parentTable[v].length-1;i>=0;i--){
                if(parentTable[u][i] != parentTable[v][i]){
                    v = parentTable[v][i];
                    u = parentTable[u][i];
                }
            }
            return parentTable[u][0];
        }
        

        public int[][] multiply(int a[][] , int b[][]){
            int c[][] = new int[a.length][b[0].length];
            for(int i=0;i<a.length;i++){
                for(int j=0;j<b[0].length;j++){
                    for(int k=0;k<b.length;k++)
                        c[i][j] += a[i][k]*b[k][j];
                }
            }
            return c;
        }

        public int[][] multiply(int a[][] , int b[][] , int mod){
            int c[][] = new int[a.length][b[0].length];
            for(int i=0;i<a.length;i++){
                for(int j=0;j<b[0].length;j++){
                    for(int k=0;k<b.length;k++){
                        c[i][j] += a[i][k]*b[k][j];
                        c[i][j]%=mod;
                    }
                }
            }
            return c;
        }

        public int[][] pow(int a[][] , long b){
            int res[][] = new int[a.length][a[0].length];
            for(int i=0;i<a.length;i++)
                res[i][i] = 1;
            while(b>0){
                if((b&1) == 1)
                    res = multiply(res , a , 10);
                a = multiply(a , a , 10);
                b>>=1;
            }
            return res;
        }

        

        // for the min max problems
        public void build(int lookup[][] , int arr[], int n) { 
            for (int i = 0; i < n; i++) 
                lookup[i][0] = arr[i]; 
            for (int j = 1; (1 << j) <= n; j++) { 
                for (int i = 0; (i + (1 << j) - 1) < n; i++) { 
                    if (lookup[i][j - 1] > lookup[i + (1 << (j - 1))][j - 1]) 
                        lookup[i][j] = lookup[i][j - 1]; 
                    else
                        lookup[i][j] = lookup[i + (1 << (j - 1))][j - 1]; 
                } 
            } 
        }
        public int query(int lookup[][] , int L, int R) { 
            int j = (int)(Math.log(R - L + 1)/Math.log(2)); 
            if (lookup[L][j] >= lookup[R - (1 << j) + 1][j]) 
                return lookup[L][j]; 
            else
                return lookup[R - (1 << j) + 1][j]; 
        }

        // for printing purposes
        public void print1d(int a[] , PrintWriter out){
            for(int i=0;i<a.length;i++)
                out.print(a[i] + " ");
            out.println();
        }
        public void print1d(boolean a[] , PrintWriter out){
            for(int i=0;i<a.length;i++)
                out.print(a[i] + " ");
            out.println();
        }
        public void print1d(long a[] , PrintWriter out){
            for(int i=0;i<a.length;i++)
                out.print(a[i] + " ");
            out.println();
        }
        public void print2d(int a[][] , PrintWriter out){
            for(int i=0;i<a.length;i++){
                for(int j=0;j<a[i].length;j++)
                    out.print(a[i][j] + " ");
                out.println();
            }
            // out.println();
        }
        public void print2d(long a[][] , PrintWriter out){
            for(int i=0;i<a.length;i++){
                for(int j=0;j<a[i].length;j++)
                    out.print(a[i][j] + " ");
                out.println();
            }
            // out.println();
        }


        public void sieve(int a[]){
            a[0] = a[1] = 1;
            int i;
            for(i=2;i*i<=a.length;i++){
                if(a[i] != 0)
                    continue;
                a[i] = i;
                for(int k = (i)*(i);k<a.length;k+=i){
                    if(a[k] != 0)
                        continue;
                    a[k] = i;
                }
            }
        }

        public long nCrPFermet(int n , int r , long p){
            if(r==0)
                return 1l;
            // long fact[] = new long[n+1];
            // fact[0] = 1;
            // for(int i=1;i<=n;i++)
            //     fact[i] = (i*fact[i-1])%p;
            long modInverseR = pow(fact[r] , p-2 , p);
            long modInverseNR = pow(fact[n-r] , p-2 , p);
            long w = (((fact[n]*modInverseR)%p)*modInverseNR)%p;
            return w;
        }
        public long pow(long a, long b, long m) {
            a %= m;
            long res = 1;
            while (b > 0) {
                long x = b&1;
                if (x == 1)
                    res = res * a % m;
                a = a * a % m;
                b >>= 1;
            }
            return res;
        }
        public long pow(long a, long b) {
            long res = 1;
            while (b > 0) {
                long x = b&1;
                if (x == 1)
                    res = res * a;
                a = a * a;
                b >>= 1;
            }
            return res;
        }
        public void sortedArrayToBST(TreeSet<Integer> a , int start, int end) { 
            if (start > end) {
                return;
            }
            int mid = (start + end) / 2;
            a.add(mid);
            sortedArrayToBST(a, start, mid - 1);
            sortedArrayToBST(a, mid + 1, end); 
        }

        public int lowerLastBound(ArrayList<Integer> a , int x){
            int l = 0;
            int r = a.size()-1;
            if(a.get(l)>=x)
                return -1;
            if(a.get(r)<x)
                return r;
            int mid = -1;
            while(l<=r){
                mid = (l+r)/2;
                if(a.get(mid) == x && a.get(mid-1)<x)
                    return mid-1;
                else if(a.get(mid)>=x)
                    r = mid-1;
                else if(a.get(mid)<x && a.get(mid+1)>=x)
                    return mid;
                else if(a.get(mid)<x && a.get(mid+1)<x)
                    l = mid+1;
            }
            return mid;
        }
        public int upperFirstBound(ArrayList<Integer> a , int x){
            int l = 0;
            int r = a.size()-1;
            if(a.get(l)>x)
                return l;
            if(a.get(r)<=x)
                return r+1;
            int mid = -1;
            while(l<=r){
                mid = (l+r)/2;
                if(a.get(mid) == x && a.get(mid+1)>x)
                    return mid+1;
                else if(a.get(mid)<=x)
                    l = mid+1;
                else if(a.get(mid)>x && a.get(mid-1)<=x)
                    return mid;
                else if(a.get(mid)>x && a.get(mid-1)>x)
                    r = mid-1;
            }
            return mid;
        }
        public int lowerLastBound(long a[] , long x){
            int l = 0;
            int r = a.length-1;
            if(a[l]>=x)
                return -1;
            if(a[r]<x)
                return r;
            int mid = -1;
            while(l<=r){
                mid = (l+r)/2;
                if(a[mid] == x && a[mid-1]<x)
                    return mid-1;
                else if(a[mid]>=x)
                    r = mid-1;
                else if(a[mid]<x && a[mid+1]>=x)
                    return mid;
                else if(a[mid]<x && a[mid+1]<x)
                    l = mid+1;
            }
            return mid;
        }
        public int upperFirstBound(long a[] , long x){
            int l = 0;
            int r = a.length-1;
            if(a[l]>x)
                return l;
            if(a[r]<=x)
                return r+1;
            int mid = -1;
            while(l<=r){
                mid = (l+r)/2;
                if(a[mid] == x && a[mid+1]>x)
                    return mid+1;
                else if(a[mid]<=x)
                    l = mid+1;
                else if(a[mid]>x && a[mid-1]<=x)
                    return mid;
                else if(a[mid]>x && a[mid-1]>x)
                    r = mid-1;
            }
            return mid;
        }
        public long log(float number , int base){
            return (long) Math.ceil((Math.log(number) / Math.log(base)) + 1e-9);
        }
        public long gcd(long a , long b){
            if(a<b){
                long c = b;
                b = a;
                a = c;
            }
            while(b!=0){
                long c = a;
                a = b;
                b = c%a;
            }
            return a;
        }
        public long[] gcdEx(long p, long q) {
            if (q == 0)
                return new long[] { p, 1, 0 };
            long[] vals = gcdEx(q, p % q);
            long d = vals[0];
            long a = vals[2];
            long b = vals[1] - (p / q) * vals[2];
            // 0->gcd 1->xValue 2->yValue
            return new long[] { d, a, b };
        }
        public void sievePhi(int a[]){
            a[0] = 0;
            a[1] = 1;
            for(int i=2;i<a.length;i++)
                a[i] = i-1;
            for(int i=2;i<a.length;i++)
                for(int j = 2*i;j<a.length;j+=i)
                    a[j] -= a[i];
        }
        public void lcmSum(long a[]){
            int sievePhi[] = new int[(int)1e6 + 1];
            sievePhi(sievePhi);
            a[0] = 0;
            for(int i=1;i<a.length;i++)
                for(int j = i;j<a.length;j+=i)
                    a[j] += (long)i*sievePhi[i];
        }
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

    }
}