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.*;


public class Main implements Runnable {
    int n, m, k;
    static boolean use_n_tests = true;


    void solve(FastScanner in, PrintWriter out, int testNumber) {
        n = in.nextInt();
        m = in.nextInt();
        char[][] s = new char[n][];
        for (int i = 0; i < n; i++) {
            s[i] = in.next().toCharArray();
        }
        int sr = -1, ds = -1;
        int sr1 = -1, tr = -1, en = -1;
        int[][] in1 = new int[n][2];
        int[][] out1 = new int[n][2];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (i == j) {
                    continue;
                }
                if (s[i][j] == s[j][i]) {
                    sr = i;
                    ds = j;
                }
                in1[i][s[j][i] - 'a'] = j + 1;
                out1[i][s[i][j] - 'a'] = j + 1;
            }
        }
        for (int i = 0; i < n; i++) {
            if (in1[i][0] != 0 && out1[i][0] != 0) {
                sr1 = in1[i][0] - 1;
                tr = i;
                en = out1[i][0] - 1;
                break;
            }
            if (in1[i][1] != 0 && out1[i][1] != 0) {
                sr1 = in1[i][1] - 1;
                tr = i;
                en = out1[i][1] - 1;
                break;
            }
        }

        if (sr != -1) {
            out.println("YES");
            print(m, sr, ds);
            return;
        }

        if (m % 2 == 1) {
            out.println("YES");
            print(m, 0, 1);
        } else {
            if (sr1 != -1) {
                out.println("YES");
                int m2 = m / 2;
                if (m2 % 2 == 1) {
                    for (int i = 0; i < m2; i++) {
                        if (i % 2 == 0) {
                            out.printf("%d %d %d ", sr1 + 1, tr + 1, en + 1);
                        } else {
                            out.printf("%d ", tr + 1);
                        }
                    }
                } else {
                    for (int i = 0; i < m2; i++) {
                        if (i % 2 == 0) {
                            out.printf("%d %d %d ", tr + 1, en + 1, tr + 1);
                        } else {
                            out.printf("%d ", sr1 + 1);
                        }
                    }
                    out.printf("%d\n", tr + 1);
                }
            } else {
                out.println("NO");
            }
        }
    }

    void print(int m, int s, int d) {
        s++;
        d++;
        out.print(s + " ");
        for (int i = 0; i < m; i++) {
            if (i % 2 == 0) {
                out.print(d + " ");
            } else {
                out.print(s + " ");
            }
        }
        out.println();
    }

    // ****************************** template code ***********
    static int stack_size = 1 << 27;

    void pre_solve() {
    }

    class Coeff {
        long mod;
        long[][] C;
        long[] fact;
        boolean cycleWay = false;

        Coeff(int n, long mod) {
            this.mod = mod;
            fact = new long[n + 1];
            fact[0] = 1;
            for (int i = 1; i <= n; i++) {
                fact[i] = i;
                fact[i] %= mod;
                fact[i] *= fact[i - 1];
                fact[i] %= mod;
            }
        }

        Coeff(int n, int m, long mod) {
            // n > m
            cycleWay = true;
            this.mod = mod;
            C = new long[n + 1][m + 1];
            for (int i = 0; i <= n; i++) {
                for (int j = 0; j <= Math.min(i, m); j++) {
                    if (j == 0 || j == i) {
                        C[i][j] = 1;
                    } else {
                        C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
                        C[i][j] %= mod;
                    }
                }
            }
        }

        public long C(int n, int m) {
            if (cycleWay) {
                return C[n][m];
            }
            return fC(n, m);
        }

        private long fC(int n, int m) {
            return (fact[n] * inv(fact[n - m] * fact[m] % mod)) % mod;
        }

        private long inv(long r) {
            if (r == 1)
                return 1;
            return ((mod - mod / r) * inv(mod % r)) % mod;
        }
    }

    class Pair {
        int first;
        int second;

        public int getFirst() {
            return first;
        }

        public int getSecond() {
            return second;
        }
    }

    class MultisetTree<T> {
        int size = 0;

        TreeMap<T, Integer> mp = new TreeMap<>();

        void add(T x) {
            mp.merge(x, 1, Integer::sum);
            size++;
        }

        void remove(T x) {
            if (mp.containsKey(x)) {
                mp.merge(x, -1, Integer::sum);
                if (mp.get(x) == 0) {
                    mp.remove(x);
                }
                size--;
            }
        }

        boolean contains(T x) {
            return mp.containsKey(x);
        }

        T greatest() {
            return mp.lastKey();
        }

        T smallest() {
            return mp.firstKey();
        }

        int size() {
            return size;
        }

        int diffSize() {
            return mp.size();
        }

        void clear() {
            mp.clear();
            size = 0;
        }
    }

    class Multiset<T> {
        int size = 0;

        Map<T, Integer> mp = new HashMap<>();

        void add(T x) {
            mp.merge(x, 1, Integer::sum);
            size++;
        }

        boolean contains(T x) {
            return mp.containsKey(x);
        }

        void remove(T x) {
            if (mp.containsKey(x)) {
                mp.merge(x, -1, Integer::sum);
                if (mp.get(x) == 0) {
                    mp.remove(x);
                }
                size--;
            }
        }

        int size() {
            return size;
        }

        int diffSize() {
            return mp.size();
        }
    }

    static class Range {
        int l, r;
        int id;

        public int getL() {
            return l;
        }

        public int getR() {
            return r;
        }

        public Range(int l, int r, int id) {
            this.l = l;
            this.r = r;
            this.id = id;
        }
    }

    static class Array {
        static Range[] readRanges(int n, FastScanner in) {
            Range[] result = new Range[n];
            for (int i = 0; i < n; i++) {
                result[i] = new Range(in.nextInt(), in.nextInt(), i);
            }
            return result;
        }

        static boolean isSorted(Integer[] a) {
            for (int i = 0; i < a.length - 1; i++) {
                if (a[i] > a[i + 1]) {
                    return false;
                }
            }
            return true;
        }

        static public <T> long sum(T[] a) {
            long sum = 0;
            for (T x : a) {
                if (x instanceof Integer) {
                    sum += (Integer) x;
                } else {
                    sum += (Long) x;
                }
            }
            return sum;
        }

        static public int min(Integer[] a) {
            int mn = Integer.MAX_VALUE;
            for (int x : a) {
                mn = Math.min(mn, x);
            }
            return mn;
        }

        static public int min(int[] a) {
            int mn = Integer.MAX_VALUE;
            for (int x : a) {
                mn = Math.min(mn, x);
            }
            return mn;
        }

        static public int max(Integer[] a) {
            int mx = Integer.MIN_VALUE;
            for (int x : a) {
                mx = Math.max(mx, x);
            }
            return mx;
        }

        static public int max(int[] a) {
            int mx = Integer.MIN_VALUE;
            for (int x : a) {
                mx = Math.max(mx, x);
            }
            return mx;
        }

        static public int[] readint(int n, FastScanner in) {
            int[] out = new int[n];
            for (int i = 0; i < out.length; i++) {
                out[i] = in.nextInt();
            }
            return out;
        }
    }

    class Graph {
        List<List<Integer>> graph;

        Graph(int n) {
            create(n);
        }

        void create(int n) {
            List<List<Integer>> graph = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                graph.add(new ArrayList<>());
            }
            this.graph = graph;
        }

        void readBi(int m, FastScanner in) {
            for (int i = 0; i < m; i++) {
                int v = in.nextInt() - 1;
                int u = in.nextInt() - 1;
                graph.get(v).add(u);
                graph.get(u).add(v);
            }
        }
    }

    class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(InputStream io) {
            br = new BufferedReader(new InputStreamReader(io));
        }

        public String line() {
            String result = "";
            try {
                result = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return result;
        }

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

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

        public int[] nextArray(int n) {
            int[] res = new int[n];
            for (int i = 0; i < n; i++) {
                res[i] = in.nextInt();
            }
            return res;
        }

        public long[] nextArrayL(int n) {
            long[] res = new long[n];
            for (int i = 0; i < n; i++) {
                res[i] = in.nextLong();
            }
            return res;
        }

        public Long[] nextArrayL2(int n) {
            Long[] res = new Long[n];
            for (int i = 0; i < n; i++) {
                res[i] = in.nextLong();
            }
            return res;
        }


        public Integer[] nextArray2(int n) {
            Integer[] res = new Integer[n];
            for (int i = 0; i < n; i++) {
                res[i] = in.nextInt();
            }
            return res;
        }

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

    }

    void run_t_tests() {
        pre_solve();
        int t = in.nextInt();
        int i = 0;
        while (t-- > 0) {
            solve(in, out, i++);
        }
    }

    void run_one() {
        solve(in, out, -1);
    }

    @Override
    public void run() {
        in = new FastScanner(System.in);
        out = new PrintWriter(System.out);
        if (use_n_tests) {
            run_t_tests();
        } else {
            run_one();
        }
        out.close();
    }

    static FastScanner in;
    static PrintWriter out;

    public static void main(String[] args) throws InterruptedException {
        Thread thread = new Thread(null, new Main(), "", stack_size);
        thread.start();
        thread.join();
    }
}