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 PYTHON3):

Solution:
# Contest: https://codeforces.com/contest/1481
# Problem: https://codeforces.com/contest/1481/problem/D
# Trials: 
# Solved: after contest
# Duration:
# Attempt: 

from collections import Counter
import itertools

t = int(input())

for tc in range(1, t + 1):
  n, m = map(int, input().split())
  g = []
  acs = []
  bcs = []
  # The big idea: no need to set msg to NO or YES, as res will tell the story.d
  #msg = "NO"
  res = []
  if m == 1:
    #msg = "YES"
    res = [1, 2]
  for i in range(n):
    #if msg == "YES":
    if res:
      _ = input()
    else:
      r = input().replace(" ", "")
      
      if i > 0:
        for j in range(i):
          if r[j] == g[j][i]:
            #msg = "YES"
            if m % 2 == 0:
              res = [i + 1, j + 1] * (m // 2)
              res.append(i + 1)
            else:
              res = [i + 1, j + 1] * ((m + 1) // 2)
            break
      
      #if msg == "NO":
      if not res:
        ac = r.count("a")
        acs.append(ac)
        bcs.append(n - 1 - ac)
        g.append(r)
  
  # Trivial: if ij is the same as ji.
  
  #if msg == "NO":
  if not res:
    # I know that if ij is a then ji is b and vica versa.
    # 50% a's and 50% b's
    if m % 2 == 1:
      #msg = "YES"
      if m % 2 == 0:
        res = [1, 2] * (m // 2)
        res.append(1)
      else: 
        res = [1, 2] * ((m + 1) // 2)
    elif m == 2:
      for i in range(n):
        for j in range(n):
          # MISTAKE 1: did not prepare that i cannot be equal to j!
          if i != j:
            c = g[i][j]
            k = g[j].find(c)
            #print(i, j, c, k, g[j])
            if k != -1:
              #msg = "YES"
              res = [i + 1, j + 1, k + 1]
              break

      # MISTAKE 7: the below was indented one more, ending it after the first row.
      # MISTAKE 8: it should break if it is non-empty, and indented after one to enable double break of from a nested loop!
        if res:
          break
    else:
      # If there is no identical letters in loop of size 3, 
      # there is no identical letters in loop of higher size.
      # IDEA: do not go from 1st row, but from highest count of char in row.
      for rs in itertools.permutations(range(n), 3):
        c0 = g[rs[0]][rs[1]]
        c1 = g[rs[1]][rs[2]]
        if c0 == c1:
          c2 = g[rs[2]][rs[0]]
          if c1 == c2:
            #msg = "YES"
            #  m   
            #  4  1*3+2
            #  6  2*3+1
            #  8  2*3+3
            # 10  3*3+2
            # 12  4*3+1
            # 
            # MISTAKE 2: wrote res before, instead of rs!
            # MISTAKE 2: l = [res[0] + 1, res[1] + 1, res[2] + 1]
            l = [rs[0] + 1, rs[1] + 1, rs[2] + 1]
            res = l * ((m + 1) // 3)
            res.extend(l[:(m + 1) % 3])
            break
      
      #print("Loop of 3", res)
      # QUESTION: should I merge the above with the below?
      # NO wishful thinking: solutions where m > 2 and a and b must be mixed.
      # MISTAKE 3: not thinking of solutions where it can a...ab...b.
      # The solution is a m+1 length list of numbers from 1 to n+1 (0 to n)
      # where no consecutive numbers can be the same.
      # This is the full solution space.
      # The middle.
      for x, y, z in itertools.permutations(range(n), 3):
        # Can it be declared outside of the loop?
        fs = [] # first half
        ss = [] # second half
        f = g[x][y]
        s = g[y][z]
        #print(x+1, y+1, z+1, f, s)
        # MISTAKE 9: If they are not the same, it shouldn't escape the loop!
        #if f != s:
        #  break
        if f == s:
          fs.append(x)
          ss.append(z)
          
          find = False
          l = 2
          r = 0
          c = 0
          #i = 0
          while l < m and l >= 2:
            #print(l, x, z, r, c, fs, ss)
            #i += 1
            #if i == 10:
            #  assert False
              
            if r == x:
              r += 1
            elif c == z:
              c += 1
            elif r >= n:
              r = 0
              c += 1
            elif c >= n:
              # FROM HERE, it's not the below.
              # TODO: better naming, better pop / append policy.
              x = fs.pop() + 1
              z = ss.pop()
              r = 0
              c = 0
              # MISTAKE 4: modified m instead of l (also below). What is input and fixed, and what is a tracker, to keep afloat in the memory with no telling names takes practice.
              l -= 2
            elif x >= n:
              x = 0
              z += 1
              r = 0
              c = 0
            elif z >= n:
              x = fs.pop() + 1
              z = ss.pop()
              r = 0
              c = 0
              l -= 2
            else:
              f = g[r][x]
              s = g[z][c]
              if f == s:
                fs.append(r)
                ss.append(c)
                x = r
                z = c
                r = 0
                c = 0
                l += 2
              else:
                r += 1
                
          if l == m:
            # Instead of using reversed(), use a stack and a queue for 1st and 2nd halves fs and ss.
            # MISTAKE 5: used only reversed(), but that gives back an iterator.
            res = map(lambda x: x + 1, list(reversed(fs)) + [y] + ss)
            break
  
  # Hacking to get the failing test case 
  # b/c I couldn't reproduce myself with my script.
  #if tc == 74 and n == 3:
  #  print("YES", n, m, ",".join(g), sep=";")
  #elif res:
  if res:
    print("YES")
    print(" ".join(map(str, res)))
  else:
    print("NO")