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:
"""
#If FastIO not needed, use this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
 
import os
import sys
from io import BytesIO, IOBase
import heapq as h 
from bisect import bisect_left, bisect_right
import time
 
from types import GeneratorType
BUFSIZE = 8192
 
class FastIO(IOBase):
    newlines = 0
 
    def __init__(self, file):
        import os
        self.os = os
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
 
    def read(self):
        while True:
            b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            self.os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
 
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string
 
#start_time = time.time()
 
def getInts():
    return [int(s) for s in input().split()]
 
def getInt():
    return int(input())
 
def getStrs():
    return [s for s in input().split()]
 
def getStr():
    return input()
 
def listStr():
    return list(input())
 
def getMat(n):
    return [getInts() for _ in range(n)]
 
def isInt(s):
    return '0' <= s[0] <= '9'
 
MOD = 10**9 + 7 
 
"""
Definitely possible if N > 2?
To make an odd palindrome:
    can go back and forward between literally ANY pair
To make an even one:
    
    if N > 2 and M % 4 = 0: find a vertex where you can arrive and leave in different ways
    ABBAABBAABBAABBAABBA
    if M % 4 = 2: find a red red sequence or a black black sequence
    AABBAABBAA
"""
 
def solve():
    N, M = getInts()
    grid = [listStr() for j in range(N)]
    if M % 2 == 1:
        ans = []
        for i in range(M+1):
            if i % 2 == 0: ans.append(1)
            else: ans.append(2)
        print("YES")
        print(*ans)
        return
    a_out = [-1]*N
    a_in = [-1]*N
    b_out = [-1]*N
    b_in = [-1]*N
    for i in range(N):
        for j in range(N):
            if i == j: continue
            if grid[i][j] == grid[j][i]:
                ans = []
                for k in range(M+1):
                    if k % 2 == 0: ans.append(i+1)
                    else: ans.append(j+1)
                print("YES")
                print(*ans)
                return
            if grid[i][j] == 'a':
                a_out[i] = j
                a_in[j] = i
            else:
                b_out[i] = j
                b_in[j] = i
    if N == 2:
        print("NO")
        return
    for i in range(N):
        if a_out[i] >= 0 and a_in[i] >= 0:
            x, y, z = i+1, a_out[i]+1, a_in[i]+1
            ans = []
            if M % 4 == 0:
                for j in range(M+1):
                    if j % 2 == 0:
                        ans.append(x)
                    elif j % 4 == 1:
                        ans.append(y)
                    else:
                        ans.append(z)
            else:
                for j in range(M+1):
                    if j % 2 == 1:
                        ans.append(x)
                    elif j % 4 == 0:
                        ans.append(z)
                    else:
                        ans.append(y)
        elif b_out[i] >= 0 and b_in[i] >= 0:
            x, y, z = i+1, b_out[i]+1, b_in[i]+1
            ans = []
            if M % 4 == 0:
                for j in range(M+1):
                    if j % 2 == 0:
                        ans.append(x)
                    elif j % 4 == 1:
                        ans.append(y)
                    else:
                        ans.append(z)
            else:
                for j in range(M+1):
                    if j % 2 == 1:
                        ans.append(x)
                    elif j % 4 == 0:
                        ans.append(z)
                    else:
                        ans.append(y)
    print("YES")
    print(*ans)
    return
    
for _ in range(getInt()):
    #print(solve())
    solve()
 
 
#print(time.time()-start_time)