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 CPP):

Solution:
//~ while (clock()<=69*CLOCKS_PER_SEC)
//~ #pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("O3")
#pragma GCC optimize("-Ofast","-funroll-all-loops","-ffast-math")
//~ #pragma GCC target ("avx2")
//~ #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//~ 火影三板斧
// #pragma GCC target ("avx2,fma")
// #pragma GCC optimize ("O3")
// #pragma GCC optimize ("unroll-loops")
#include <algorithm>
#include <cassert>
#include <bits/stdc++.h>
#include <functional>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
using namespace std;
using LL = long long;
#define int LL
#define eps 1e-8
#define fi first
#define se second
#define endl '\12'
#define eb emplace_back
#define SZ(a) int32_t(a.size())
#define ALL(x) (x).begin(),(x).end()
#define trav(a,x) for (auto& a: x)
#define LOG(FMT...) fprintf(stderr, FMT)
#define close std::ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr),cout<<fixed<<setprecision(10)
#define FOR(i, x, y) for (LL i = (x), _##i = (y); i < _##i; ++i)
#define FORD(i, x, y) for (LL i = (x), _##i = (y); i > _##i; --i)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define CASET int ___T; cin>>___T; for(int __CS=1;__CS<=___T;__CS++)
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef pair<int,int> pii;
constexpr int mod = 1e9+7; // 998244353
// mt19937 dlsrand(random_device{}());
// mt19937 mrand(std::chrono::system_clock::now().time_since_epoch().count()); 
// int rnd(int x) { return mrand() % x;}
// ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
ll ex_gcd(ll a, ll b, ll& x, ll& y){if(!b){x=1;y=0;return a;}ll ret=ex_gcd(b,a%b,y,x);y-=a/b*x;return ret;}
LL mul(LL u, LL v, LL p) { return (u * v - LL((long double) u * v / p) * p + p) % p; }
LL bin(LL x, LL n, LL MOD) {LL ret = MOD != 1;for (x %= MOD; n; n >>= 1, x = x * x % MOD)if (n & 1) ret = ret * x % MOD;return ret;}
int norm(int x) { return x >= mod ? (x - mod) : x; }
inline LL get_inv(LL x, LL p) { return bin(x, p - 2, p); }
inline ll get_inv(ll a) { ll x, y; ex_gcd(a, mod, x, y); return norm(x + mod);}
template<class T>inline void umin(T &x, T y) {x = x > y ? y : x;}
template<class T>inline void umax(T &x, T y) {x = x < y ? y : x;}
template<class T>inline void dec(T &x, T y) {x -= y; if(x < 0) x += mod;}
template<class T>inline void add(T &x, T y) {x += y; if(x >= mod) x -= mod;}
template<typename T> void read(T &x){ cin >> x; }
template<typename T, typename... Args> void read(T &first, Args& ... args) { read(first); read(args...); }
template<class T>int ctz(T a){return sizeof(T)>2?__builtin_ctzll(a):__builtin_ctz(a);}
template<class T>inline T gcd(T _a,T _b){T a=abs(_a),b=abs(_b);int shift=ctz(a|b);for(b>>=ctz(b),a>>=ctz(a);a;a-=b,a>>=ctz(a))if(a<b)swap(a,b);return b<<shift;}
const double PI = acos(-1.0);
constexpr int INF = 0x3f3f3f3f;
constexpr ll linf = 0x3f3f3f3f3f3f3f3f;
constexpr ull base=2333, P_1=19260817, P_2=999998639;
constexpr int maxn = 1e6+10; // remember to calculate. if tle, check maxn first.
// 及时当勉励
char s[1110][1110];
void up(){
    int n , m;
    cin >> n >> m;
    for(int i = 0; i< n; i++){
        cin >> s[i];
    }
    if(m&1){
        cout <<"YES" << endl;
        for(int i = 0; i <= m; i++){
            cout << (i&1?1:2) << " \n"[i == m];
        }
        return ;
    }
    for(int i = 0; i < n; i++){
        for(int j = i+1; j < n; j++){
            if(s[i][j]==s[j][i]){
                cout << "YES\n";
                for(int k = 0; k <= m; k++){
                    cout << (k&1 ? i+1:j+1) << " \n"[k == m];
                }
                return ;
            }
        }
    }
    for(int i = 0; i < n; i++){
        int x = -1, y = -1;
        for(int j = 0; j < n; j++){
            if(s[i][j] == 'a') x = j;
            if(s[i][j] == 'b') y = j;
        }
        if(x!=-1 and y !=-1){
            cout << "YES\n";
            for(int k = m / 2; k >= 1; k --) cout << (k & 1 ? x + 1 : i + 1) << " ";
            cout << i + 1 << " ";
            for(int k = 1; k <= m / 2; k ++) cout << (k & 1 ? y + 1 : i + 1) << " ";
            return;
        }
    }
    cout << "NO\n";
}
int32_t main()
{
    close;
    CASET{
        up();
    }
    return 0;
}
/*
Though leaves are many , the root is one.
Through all the lying days of my youth
I swayed my leaves and flowers in the sun.
Now I may wither into the truth.
                    - William Butler Yeats
*/