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:
import sys
import re
import random
import math
import copy
from heapq import heappush, heappop, heapify
from functools import cmp_to_key
from bisect import bisect_left, bisect_right
from collections import defaultdict, deque, Counter
# sys.setrecursionlimit(1000000)

# input aliases
input = sys.stdin.readline
getS = lambda: input().strip()
getN = lambda: int(input())
getList = lambda: list(map(int, input().split()))
getZList = lambda: [int(x) - 1 for x in input().split()]

INF = float("inf")
MOD = 10**9 + 7
divide = lambda x: pow(x, MOD-2, MOD)


def nck(n, k, kaijyo):
    return (npk(n, k, kaijyo) * divide(kaijyo[k])) % MOD

def npk(n, k, kaijyo):
    if k == 0 or k == n:
        return n % MOD
    return (kaijyo[n] * divide(kaijyo[n-k])) % MOD

def fact_and_inv(SIZE):
    inv = [0] * SIZE  # inv[j] = j^{-1} mod MOD
    fac = [0] * SIZE  # fac[j] = j! mod MOD
    finv = [0] * SIZE  # finv[j] = (j!)^{-1} mod MOD
    inv[1] = 1
    fac[0] = fac[1] = 1
    finv[0] = finv[1] = 1
    for i in range(2, SIZE):
        inv[i] = MOD - (MOD // i) * inv[MOD % i] % MOD
        fac[i] = fac[i - 1] * i % MOD
        finv[i] = finv[i - 1] * inv[i] % MOD

    return fac, finv

def renritsu(A, Y):
    # example 2x + y = 3, x + 3y = 4
    # A = [[2,1], [1,3]])
    # Y = [[3],[4]] または [3,4]
    A = np.matrix(A)
    Y = np.matrix(Y)
    Y = np.reshape(Y, (-1, 1))
    X = np.linalg.solve(A, Y)

    # [1.0, 1.0]
    return X.flatten().tolist()[0]

class TwoDimGrid:
    # 2次元座標 -> 1次元
    def __init__(self, h, w, wall="#"):
        self.h = h
        self.w = w
        self.size = (h+2) * (w+2)
        self.wall = wall
        self.get_grid()
        # self.init_cost()

    def get_grid(self):
        grid = [self.wall * (self.w + 2)]
        for i in range(self.h):
            grid.append(self.wall + getS() + self.wall)

        grid.append(self.wall * (self.w + 2))
        self.grid = grid
    def init_cost(self):
        self.cost = [INF] * self.size

    def pos(self, x, y):
        # 壁も含めて0-indexed 元々の座標だけ考えると1-indexed
        return y * (self.w + 2) + x
    def getgrid(self, x, y):
        return self.grid[y][x]
    def get(self, x, y):
        return self.cost[self.pos(x, y)]
    def set(self, x, y, v):
        self.cost[self.pos(x, y)] = v
        return
    def show(self):
        for i in range(self.h+2):
            print(self.cost[(self.w + 2) * i:(self.w + 2) * (i+1)])
    def showsome(self, tgt):
        for t in tgt:
            print(t)
        return
    def showsomejoin(self, tgt):
        for t in tgt:
            print("".join(t))
        return

    def search(self):
        grid = self.grid
        move = [(0, 1), (0, -1), (1, 0), (-1, 0)]
        move_eight = [(0, 1), (0, -1), (1, 0), (-1, 0), (1, 1), (1, -1), (-1, 1), (-1, -1)]
        # for i in range(1, self.h+1):
        #     for j in range(1, self.w+1):
        #         cx, cy = j, i
        #         for dx, dy in move:
        #             nx, ny = dx + cx, dy + cy


class BIT():
    # A1 ... AnのBIT(1-indexed)
    def __init__(self, n):
        self.n = n
        self.BIT = [0] * (n + 2)

    # A1 ~ Aiまでの和 O(logN)
    def query(self, idx):
        res_sum = 0
        while idx > 0:
            res_sum += self.BIT[idx]
            idx -= idx & (-idx)
        return res_sum

    # Ai += x O(logN)
    def update(self, idx, x):
        # print(idx)
        while idx <= self.n:
            # print(idx)
            self.BIT[idx] += x
            idx += idx & (-idx)
        return

    def show(self):
        print(self.BIT)


def solve():
    n, m = getList()

    graph = [getS() for i 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

    if (n == 2):
        if graph[0][1] != graph[1][0]:
            print("NO")
            return
        else:
            ans = []
            for i in range(m + 1):
                if (i % 2 == 0):
                    ans.append(1)
                else:
                    ans.append(2)
            print("YES")
            print(*ans)
            return

    anses = []
    for i in range(n):
        for j in range(n):
            if i == j:
                continue
            if graph[i][j] == graph[j][i]:
                anses = [i, j]
                break

    if anses:
        ans = []
        for i in range(m + 1):
            ans.append(anses[i % 2] + 1)
        print("YES")
        print(*ans)
        return

    if graph[0][1] == graph[1][2]:
        if graph[1][2] == graph[2][0]:
            ans = []
            for i in range(m + 1):
                ans.append((i % 3) + 1)
            print("YES")
            print(*ans)
            return

        else:
            p4 = [2,3,2,1]
            p6 = [1,2,3,2]

    else:
        if graph[0][2] == graph[0][1]:
            p4 = [3, 2, 3, 1]
            p6 = [1, 3, 2, 3]
        else:
            p4 = [1, 2, 1, 3]
            p6 = [3, 1, 2, 1]


    if m % 4 == 0:
        ans = []
        for i in range(m + 1):
            ans.append(p4[i%4])
        print("YES")
        print(*ans)
        return

    else:
        ans = []
        for i in range(m + 1):
            ans.append(p6[i%4])
        print("YES")
        print(*ans)
        return

def main():
    n = getN()
    for _ in range(n):
        solve()

    return
if __name__ == "__main__":
    main()
    # solve()