**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are the constraints on all the variables. You can make hacks only if both versions of the problem are solved.

Tsovak and Narek are playing a game. They have an array $a$ and a matrix $b$ of integers with $n$ rows and $m$ columns, numbered from $1$. The cell in the $i$-th row and the $j$-th column is $(i, j)$.

They are looking for the elements of $a$ in turns; Tsovak starts first. Each time a player looks for a cell in the matrix containing the current element of $a$ (Tsovak looks for the first, then Narek looks for the second, etc.). Let's say a player has chosen the cell $(r, c)$. The next player has to choose his cell in the submatrix starting at $(r + 1, c + 1)$ and ending in $(n, m)$ (the submatrix can be empty if $r=n$ or $c=m$). If a player cannot find such a cell (or the remaining submatrix is empty) or the array ends (the previous player has chosen the last element), then he loses.

Your task is to determine the winner if the players play optimally.

Input: The first line of the input contains $t$ ($1 \le t \le 300$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 300$) – the size of the array and the sizes of the matrix.

The second line contains $l$ integers $a_1, a_2, a_3, \ldots a_l$ ($1 \le a_i \le \min(7, n \cdot m)$) – the elements of the array $a$.

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le \min(7, n \cdot m)$) – representing the $i$-th row of the matrix.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^5$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $300$.

Output: You should output $t$ lines, the $i$-th of them containing a character representing the answer of the $i$-th test case: "T" if Tsovak wins or "N", otherwise (without quotes).

Examples:
input:
3
2 2 3
1 2
1 3 5
4 5 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 5
5 5
5 5
output:
N
T
N

Note: In the first example, Tsovak starts by looking for $1$. There is only one occurrence of $1$ at $(1,1)$, so he chooses it. Then Narek needs to look for $2$ in the submatrix of $(2, 2)$, which consists of just the last two elements: $5$ and $2$. He chooses $2$, and then Tsovak loses since the array has ended.

In the second example, Tsovak needs to choose $1$. There is a $1$ at the cell $(n,m)$, so he chooses that cell. Then, since the submatrix of $(n + 1, m + 1)$ is empty, Narek cannot find $2$, so he loses.

Buggy Code:
#include <bits/stdc++.h>

using namespace std;

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set; // find_by_order, order_of_key

#define int int64_t
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define vi vector<int>
#define vii vector<vi>

namespace io
{
    template <typename T, typename F>
    istream &operator>>(istream &cin, pair<T, F> &pr)
    {
        cin >> pr.first >> pr.second;
        return cin;
    }

    template <typename T, typename F>
    ostream &operator<<(ostream &cout, pair<T, F> &pr)
    {
        cout << pr.first << ' ' << pr.second;
        return cout;
    }

    template <typename T>
    istream &operator>>(istream &cin, vector<T> &vec)
    {
        for (T &i : vec)
            cin >> i;
        return cin;
    }

    template <typename T>
    ostream &operator<<(ostream &cout, vector<T> vec)
    {
        for (T i : vec)
            cout << i << ' ';
        return cout;
    }
}
using namespace io;

int binpow(int x, int y, int p)
{
    x %= p;
    if (y == 0)
        return 1;
    int res = 1;
    while (y)
    {
        if (y & 1)
            res *= x, res %= p;
        x = x * x % p;
        y /= 2;
    }
    return res;
}

int M = 998244353;
int inv(int a)
{
    return binpow(a, M - 2, M);
}

vector<int> z_function(string s)
{
    int n = s.size();
    vector<int> z(n);
    int l = 0, r = 0;
    for (int i = 1; i < n; i++)
    {
        if (i < r)
        {
            z[i] = min(r - i, z[i - l]);
        }
        while (i + z[i] < n && s[z[i]] == s[i + z[i]])
        {
            z[i]++;
        }
        if (i + z[i] > r)
        {
            l = i;
            r = i + z[i];
        }
    }
    return z;
}

void solve()
{
    int l, n, m;
    cin >> l >> n >> m;
    vi a(l);
    vii v(n, vi(m));
    cin >> a >> v;
    int x = 0, y = 0;
    for (int i = 0; i < l; i++)
    {
        bool f = 0;
        int nx = 0, ny = 0;
        for (int j = x; j < n; j++)
        {
            for (int k = y; k < m; k++)
            {
                if (v[j][k] == a[i])
                {
                    nx = max(nx, j);
                    ny = max(ny, k);
                    f = 1;
                }
            }
        }
        if (!f)
            return cout << (i % 2 ? "T" : "N"), void();
        x = nx;
        y = ny;
    }
    return cout << (l % 2 ? "T" : "N"), void();
}

signed main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
        cout << '\n';
    }
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
2 2 2
1 2
1 2
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are the constraints on all the variables. You can make hacks only if both versions of the problem are solved.

Tsovak and Narek are playing a game. They have an array $a$ and a matrix $b$ of integers with $n$ rows and $m$ columns, numbered from $1$. The cell in the $i$-th row and the $j$-th column is $(i, j)$.

They are looking for the elements of $a$ in turns; Tsovak starts first. Each time a player looks for a cell in the matrix containing the current element of $a$ (Tsovak looks for the first, then Narek looks for the second, etc.). Let's say a player has chosen the cell $(r, c)$. The next player has to choose his cell in the submatrix starting at $(r + 1, c + 1)$ and ending in $(n, m)$ (the submatrix can be empty if $r=n$ or $c=m$). If a player cannot find such a cell (or the remaining submatrix is empty) or the array ends (the previous player has chosen the last element), then he loses.

Your task is to determine the winner if the players play optimally.

Input: The first line of the input contains $t$ ($1 \le t \le 300$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 300$) – the size of the array and the sizes of the matrix.

The second line contains $l$ integers $a_1, a_2, a_3, \ldots a_l$ ($1 \le a_i \le \min(7, n \cdot m)$) – the elements of the array $a$.

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le \min(7, n \cdot m)$) – representing the $i$-th row of the matrix.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^5$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $300$.

Output: You should output $t$ lines, the $i$-th of them containing a character representing the answer of the $i$-th test case: "T" if Tsovak wins or "N", otherwise (without quotes).

Examples:
input:
3
2 2 3
1 2
1 3 5
4 5 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 5
5 5
5 5
output:
N
T
N

Note: In the first example, Tsovak starts by looking for $1$. There is only one occurrence of $1$ at $(1,1)$, so he chooses it. Then Narek needs to look for $2$ in the submatrix of $(2, 2)$, which consists of just the last two elements: $5$ and $2$. He chooses $2$, and then Tsovak loses since the array has ended.

In the second example, Tsovak needs to choose $1$. There is a $1$ at the cell $(n,m)$, so he chooses that cell. Then, since the submatrix of $(n + 1, m + 1)$ is empty, Narek cannot find $2$, so he loses.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;

#define int long long
#define all(v)                ((v).begin()), ((v).end())
#define allr(v)                ((v).rbegin()), ((v).rend())
#define sz(v)                ((int)((v).size()))
#define clr(v, d)            memset(v, d, sizeof(v))
#define rep(i, v)        for(int i=0;i<sz(v);++i)
#define lp(i, n)        for(int i=0;i<(int)(n);++i)
#define lpi(i, j, n)    for(int i=(j);i<(int)(n);++i)
#define lpd(i, j, n)    for(int i=(j);i>=(int)(n);--i)
#define ceils(x, y) (x + y - 1) / y

//for pairs
#define pii pair<int, int>
#define vii vector<pii>

const double EPS = (1e-7);
int dcmp(double x, double y) {    return fabs(x-y) <= EPS ? 0 : x < y ? -1 : 1;    }
int fix(int a,int n) {
    return (a % n + n) % n;
}

#define pb                    push_back
#define MP                    make_pair
#define P(x)                cout<<#x<<" = { "<<x<<" }\n"
typedef long double         ld;
typedef vector<int>       vi;
typedef vector<double>    vd;
typedef vector< vi >      vvi;
typedef vector< vd >      vvd;
typedef vector<string>    vs;

#define sumv(v) accumulate(all(v), 0ll)
#define maxv(v) *max_element(all(v))
#define minv(v) *min_element(all(v))

// Handling vector, vector of pairs
template <typename T1, typename T2> // cin >> pair<T1, T2>
istream& operator>>(istream& istream, pair<T1, T2>& p){return (istream >> p.first >> p.second);}
template <typename T1, typename T2> // cout << pair<T1, T2>
ostream& operator<<(ostream& ostream, const pair<T1, T2>& p){return (ostream << p.first << " " << p.second);}
template<typename T> // cin >> vector<T>
istream& operator>>(istream &in, vector<T> &v){for(auto &it: v) cin >> it; return in;}
template<typename T> //cout << vector<T>
ostream& operator<<(ostream &out, const vector<T> &c){for (auto &it: c) cout << it << " "; return out;}

int binpow(int a, int b) {int res = 1;while (b > 0) {if (b & 1)res = res * a;a = a * a;b >>= 1;}return res;}

int gcd (int a, int b) { while (b) {a %= b;swap(a, b);} return a;}
int lcm (int a, int b) {return a / gcd(a, b) * b;}
int int_sqrt (long long x) {int ans = 0;for (int k = 1LL << 30; k != 0; k /= 2) {if ((ans + k) * (ans + k) <= x) {ans += k;}}return ans;}
bool isOverflow(int x, int y){return (x > LLONG_MAX / y or y > LLONG_MAX / x);}

const int mod = 1e9 + 7;

int inv(int i) {if (i == 1) return 1; return (mod - ((mod / i) * inv(mod % i)) % mod) % mod;}
int mod_mul(int a, int b) {a = a % mod; b = b % mod; return (((a * b) % mod) + mod) % mod;}
int mod_add(int a, int b) {a = a % mod; b = b % mod; return (((a + b) % mod) + mod) % mod;}

const int inf=1e18;

int solve(){

    int l,n,m;
    cin>>l>>n>>m;

    vi vec(l);
    cin>>vec;

    vvi grid(n,vi(m));

    lp(i,n)
    lp(j,m)cin>>grid[i][j];

    function<bool(int,int,int)>fun=[&](int x,int y,int idx){

      if(idx>=l)
      return false;

      if(x>=n)
      return false;
      if(y>=m)
      return false;
      if(grid[x][y]==vec[idx])
      {
          bool ch1=!fun(x+1,y+1,idx+1);
          bool ch2=(fun(x+1,y,idx))||(fun(x,y+1,idx));

          return !(ch1&&ch2);

      }
      return !!fun(x+1,y+1,idx+1);

    };

    int ans=fun(0,0,0);
    if(ans)
    cout<<"T";
    else
    cout<<"N";

   return 0;
}

signed main(){
  ios_base::sync_with_stdio(false);
    cin.tie(NULL);cout.tie(NULL);

int t=1;
cin>>t;
while(t--)
{
    solve();
    cout<<"\n";
}

    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
2 2 2
1 2
1 2
3 4
2 2 2
1 2
1 1
2 2
2 3 3
1 2 3
1 2 3
4 5 6
7 8 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are the constraints on all the variables. You can make hacks only if both versions of the problem are solved.

Tsovak and Narek are playing a game. They have an array $a$ and a matrix $b$ of integers with $n$ rows and $m$ columns, numbered from $1$. The cell in the $i$-th row and the $j$-th column is $(i, j)$.

They are looking for the elements of $a$ in turns; Tsovak starts first. Each time a player looks for a cell in the matrix containing the current element of $a$ (Tsovak looks for the first, then Narek looks for the second, etc.). Let's say a player has chosen the cell $(r, c)$. The next player has to choose his cell in the submatrix starting at $(r + 1, c + 1)$ and ending in $(n, m)$ (the submatrix can be empty if $r=n$ or $c=m$). If a player cannot find such a cell (or the remaining submatrix is empty) or the array ends (the previous player has chosen the last element), then he loses.

Your task is to determine the winner if the players play optimally.

Input: The first line of the input contains $t$ ($1 \le t \le 300$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 300$) – the size of the array and the sizes of the matrix.

The second line contains $l$ integers $a_1, a_2, a_3, \ldots a_l$ ($1 \le a_i \le \min(7, n \cdot m)$) – the elements of the array $a$.

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le \min(7, n \cdot m)$) – representing the $i$-th row of the matrix.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^5$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $300$.

Output: You should output $t$ lines, the $i$-th of them containing a character representing the answer of the $i$-th test case: "T" if Tsovak wins or "N", otherwise (without quotes).

Examples:
input:
3
2 2 3
1 2
1 3 5
4 5 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 5
5 5
5 5
output:
N
T
N

Note: In the first example, Tsovak starts by looking for $1$. There is only one occurrence of $1$ at $(1,1)$, so he chooses it. Then Narek needs to look for $2$ in the submatrix of $(2, 2)$, which consists of just the last two elements: $5$ and $2$. He chooses $2$, and then Tsovak loses since the array has ended.

In the second example, Tsovak needs to choose $1$. There is a $1$ at the cell $(n,m)$, so he chooses that cell. Then, since the submatrix of $(n + 1, m + 1)$ is empty, Narek cannot find $2$, so he loses.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;

#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#define for_(i, s, e) for (ll i = s; i < (ll)e; i++)
#define for__(i, s, e) for (ll i = s; i > (ll)e; i--)
#define mod(x) ((x) % MOD)
#define mid(x) midpoint(x)
#define popcount(x) __builtin_popcount(x)
#define popcountll(x) __builtin_popcountll(x)
#define bg begin()
#define ed end()
#define sz(a) int(a.size())
#define all(x) (x).bg, (x).ed
#define allr(x) x.rbegin(), x.rend()
#define uniq(v) (v).erase(unique(all(v)), (v).end())
#define nl "\n"
#define _ << ' ' <<
#define pb emplace_back
#define MP make_pair
#define UB upper_bound
#define LB lower_bound
#define ff first
#define ss second

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<vpii> vvpii;
typedef vector<vpll> vvpll;
const ll MOD = 1e9 + 7; // 998244353;

template <typename T1, typename T2>
inline istream &operator>>(istream &in, pair<T1, T2> &a)
{
  in >> a.ff >> a.ss;
  return in;
}
template <typename T1, typename T2>
inline ostream &operator<<(ostream &out, pair<T1, T2> a)
{
  out << a.ff << " " << a.ss;
  return out;
}
template <typename T>
istream &operator>>(istream &in, vector<T> &v)
{
  for_(i, 0, sz(v)) cin >> v[i];
  return in;
}
template <typename T>
ostream &operator<<(ostream &out, vector<T> &v)
{
  for_(i, 0, sz(v)) cout << v[i] << " ";
  return out;
}
template <typename T1, typename T2>
pair<T1, T2> operator+(pair<T1, T2> &l, pair<T1, T2> &r) { return {l.ff + r.ff, l.ss + r.ss}; }

void solve()
{
  int l,n,m;
  cin>>l>>n>>m;
  int dp[n][m][8];
  memset(dp,0,sizeof(dp));
  vector<int>a(l);
  for(int i=0;i<l;i++){
    cin>>a[i];
  }
  vector<vector<int>>b(n, vector<int>(m));
  cin>>b;
  for(int i=1;i<=7;i++){
    dp[n-1][m-1][i]=0;
  }
  dp[n-1][m-1][b[n-1][m-1]]=1;
  for(int i=n-1;i>=0;i--){
    for(int j=m-1;j>=0;j--){
      if(i==n-1 and j==m-1){
        continue;
      }
      for(int val=1;val<=7;val++){
        if(i+1<n){
          dp[i][j][val]=max(dp[i][j][val], dp[i+1][j][val]);
        }
        if(j+1<m){
          dp[i][j][val]=max(dp[i][j][val], dp[i][j+1][val]);
        }
      }
      dp[i][j][b[i][j]]=1;
    }
  }

  vector<pair<int,int>>cur;
  for(int i=0;i<n;i++){
    for(int j=0;j<m;j++){
      cur.push_back({i,j});
    }
  }
  for(int idx=0;idx<l;idx++){
    sort(cur.begin(),cur.end());
    int temp[n][m];
    memset(temp, 0, sizeof(temp));
    for(auto el:cur){
      if(temp[el.first][el.second]){
        continue;
      }
      for(int x=el.first;x<n;x++){
        for(int y=el.first;y<m;y++){
          if(temp[x][y]){
            break;
          }
          temp[x][y]=1;
        }
      }
    }
    vector<pair<int,int>>nxt;

    if(idx%2==0){
      for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
          if(temp[i][j] and a[idx]==b[i][j]){
            nxt.push_back({i,j});
          }
        }
      }
      if(nxt.size()==0){
        cout<<"N"<<nl;
        return;
      }
      cur.clear();
      cur = nxt;
    }else{
      for(auto el:cur){
        if(dp[el.first][el.second][a[idx]]==0){
          cout<<"T"<<nl;return;
        }
      }
      for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
          if(temp[i][j] and a[idx]==b[i][j]){
            nxt.push_back({i,j});
          }
        }
      }
      if(nxt.size()==0){
        cout<<"T"<<nl;
        return;
      }
      cur.clear();
      cur = nxt;
    }
  }
  if(l%2){
    cout<<"T"<<nl;
  }else{
    cout<<"N"<<nl;
  }
}

int main()
{
  ios_base::sync_with_stdio(false);
  cin.tie(0);

  int tt = 1;
  cin >> tt;
  while (tt--)
    solve();

  return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
2 2 2
1 2
1 2
2 1
2 3 3
1 2 3
1 2 3
4 5 6
7 8 9
1 3 2
1 2 3
4 5 6
7 8 9
3 3 3
1 2 3
1 2 3
4 5 6
7 8 9
1 2 3
4 5 6
7 8 9
1 2 3
4 5 6
7 8 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are the constraints on all the variables. You can make hacks only if both versions of the problem are solved.

Tsovak and Narek are playing a game. They have an array $a$ and a matrix $b$ of integers with $n$ rows and $m$ columns, numbered from $1$. The cell in the $i$-th row and the $j$-th column is $(i, j)$.

They are looking for the elements of $a$ in turns; Tsovak starts first. Each time a player looks for a cell in the matrix containing the current element of $a$ (Tsovak looks for the first, then Narek looks for the second, etc.). Let's say a player has chosen the cell $(r, c)$. The next player has to choose his cell in the submatrix starting at $(r + 1, c + 1)$ and ending in $(n, m)$ (the submatrix can be empty if $r=n$ or $c=m$). If a player cannot find such a cell (or the remaining submatrix is empty) or the array ends (the previous player has chosen the last element), then he loses.

Your task is to determine the winner if the players play optimally.

Input: The first line of the input contains $t$ ($1 \le t \le 300$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 300$) – the size of the array and the sizes of the matrix.

The second line contains $l$ integers $a_1, a_2, a_3, \ldots a_l$ ($1 \le a_i \le \min(7, n \cdot m)$) – the elements of the array $a$.

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le \min(7, n \cdot m)$) – representing the $i$-th row of the matrix.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^5$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $300$.

Output: You should output $t$ lines, the $i$-th of them containing a character representing the answer of the $i$-th test case: "T" if Tsovak wins or "N", otherwise (without quotes).

Examples:
input:
3
2 2 3
1 2
1 3 5
4 5 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 5
5 5
5 5
output:
N
T
N

Note: In the first example, Tsovak starts by looking for $1$. There is only one occurrence of $1$ at $(1,1)$, so he chooses it. Then Narek needs to look for $2$ in the submatrix of $(2, 2)$, which consists of just the last two elements: $5$ and $2$. He chooses $2$, and then Tsovak loses since the array has ended.

In the second example, Tsovak needs to choose $1$. There is a $1$ at the cell $(n,m)$, so he chooses that cell. Then, since the submatrix of $(n + 1, m + 1)$ is empty, Narek cannot find $2$, so he loses.

Buggy Code:
import sys

input = lambda: sys.stdin.readline().strip()
maput = lambda: map(int, input().split())
printf = print
print = sys.stdout.write

t = int(input())

for _ in range(t):
    l,n,m = maput()
    ls = maput()
    tt = {1: [], 2:[], 3:[], 4:[],5:[],6:[],7:[]}
    for i in range(n):
        a = list(maput())
        for j in range(m):
            tt[a[j]].append((i, j))

    for i in tt:
        tt[i].sort(key = lambda x: (n - x[0] - 1) * (m - x[1] - 1))

    used = [False] * 7
    mn = -1
    mm = -1
    nwin = True
    for i in ls:
        if used[i-1] == False and tt[i]:
            used[i-1] = True
            ans = tt[i][0]
            if ans[0] > mn and ans[1] > mm:
                mn = ans[0]
                mm = ans[1]
                nwin = not nwin
            else:
                break
        else:
            break
    print('N\n' if nwin else 'T\n')

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
1 1 1
1
1
1 1 1
1
1
2 2 2
1 2
1 2
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are the constraints on all the variables. You can make hacks only if both versions of the problem are solved.

Tsovak and Narek are playing a game. They have an array $a$ and a matrix $b$ of integers with $n$ rows and $m$ columns, numbered from $1$. The cell in the $i$-th row and the $j$-th column is $(i, j)$.

They are looking for the elements of $a$ in turns; Tsovak starts first. Each time a player looks for a cell in the matrix containing the current element of $a$ (Tsovak looks for the first, then Narek looks for the second, etc.). Let's say a player has chosen the cell $(r, c)$. The next player has to choose his cell in the submatrix starting at $(r + 1, c + 1)$ and ending in $(n, m)$ (the submatrix can be empty if $r=n$ or $c=m$). If a player cannot find such a cell (or the remaining submatrix is empty) or the array ends (the previous player has chosen the last element), then he loses.

Your task is to determine the winner if the players play optimally.

Input: The first line of the input contains $t$ ($1 \le t \le 300$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 300$) – the size of the array and the sizes of the matrix.

The second line contains $l$ integers $a_1, a_2, a_3, \ldots a_l$ ($1 \le a_i \le \min(7, n \cdot m)$) – the elements of the array $a$.

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le \min(7, n \cdot m)$) – representing the $i$-th row of the matrix.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^5$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $300$.

Output: You should output $t$ lines, the $i$-th of them containing a character representing the answer of the $i$-th test case: "T" if Tsovak wins or "N", otherwise (without quotes).

Examples:
input:
3
2 2 3
1 2
1 3 5
4 5 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 5
5 5
5 5
output:
N
T
N

Note: In the first example, Tsovak starts by looking for $1$. There is only one occurrence of $1$ at $(1,1)$, so he chooses it. Then Narek needs to look for $2$ in the submatrix of $(2, 2)$, which consists of just the last two elements: $5$ and $2$. He chooses $2$, and then Tsovak loses since the array has ended.

In the second example, Tsovak needs to choose $1$. There is a $1$ at the cell $(n,m)$, so he chooses that cell. Then, since the submatrix of $(n + 1, m + 1)$ is empty, Narek cannot find $2$, so he loses.

Buggy Code:
for _ in range(int(input())):
    l, n, m = list(map(int, input().split()))
    arr = list(map(int, input().split()))
    matrix = []
    for _ in range(n):
        matrix.append(list(map(int, input().split())))

    prev_i = -1
    prev_j = -1

    cur_choice = 0
    for a in arr:
        if prev_i >= n - 1 and prev_j >= m - 1:
            break
        for i in range(n - 1, prev_i, -1):
            fl = False
            for j in range(m - 1, prev_j, -1):
                if matrix[i][j] == a:
                    fl = True
                    prev_i = i + 1
                    prev_j = j + 1
                    break
            if fl:
                break
        cur_choice += 1
        cur_choice %= 2
    if cur_choice == 1:
        print('T')
    else:
        print('N')

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
2 2 2
1 2
1 1
2 2
2 2 2
1 2
1 2
2 1
2 2 2
2 1
2 2
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are the constraints on all the variables. You can make hacks only if both versions of the problem are solved.

Tsovak and Narek are playing a game. They have an array $a$ and a matrix $b$ of integers with $n$ rows and $m$ columns, numbered from $1$. The cell in the $i$-th row and the $j$-th column is $(i, j)$.

They are looking for the elements of $a$ in turns; Tsovak starts first. Each time a player looks for a cell in the matrix containing the current element of $a$ (Tsovak looks for the first, then Narek looks for the second, etc.). Let's say a player has chosen the cell $(r, c)$. The next player has to choose his cell in the submatrix starting at $(r + 1, c + 1)$ and ending in $(n, m)$ (the submatrix can be empty if $r=n$ or $c=m$). If a player cannot find such a cell (or the remaining submatrix is empty) or the array ends (the previous player has chosen the last element), then he loses.

Your task is to determine the winner if the players play optimally.

Input: The first line of the input contains $t$ ($1 \le t \le 300$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 300$) – the size of the array and the sizes of the matrix.

The second line contains $l$ integers $a_1, a_2, a_3, \ldots a_l$ ($1 \le a_i \le \min(7, n \cdot m)$) – the elements of the array $a$.

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le \min(7, n \cdot m)$) – representing the $i$-th row of the matrix.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^5$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $300$.

Output: You should output $t$ lines, the $i$-th of them containing a character representing the answer of the $i$-th test case: "T" if Tsovak wins or "N", otherwise (without quotes).

Examples:
input:
3
2 2 3
1 2
1 3 5
4 5 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 5
5 5
5 5
output:
N
T
N

Note: In the first example, Tsovak starts by looking for $1$. There is only one occurrence of $1$ at $(1,1)$, so he chooses it. Then Narek needs to look for $2$ in the submatrix of $(2, 2)$, which consists of just the last two elements: $5$ and $2$. He chooses $2$, and then Tsovak loses since the array has ended.

In the second example, Tsovak needs to choose $1$. There is a $1$ at the cell $(n,m)$, so he chooses that cell. Then, since the submatrix of $(n + 1, m + 1)$ is empty, Narek cannot find $2$, so he loses.

Buggy Code:
def find_winner(array, matrix):
  """
  Determines the winner of the game between Tsovak and Narek.

  Args:
    array: The array of integers.
    matrix: The matrix of integers.

  Returns:
    'T' if Tsovak wins, 'N' otherwise.
  """

  row = 1
  col = 1
  current_element = 0

  while current_element < len(array):
    # Find the element in the matrix
    while row <= len(matrix) and col <= len(matrix[0]) and matrix[row - 1][col - 1] != array[current_element]:
      row += 1
      col += 1

    # Check if the element was found
    if row > len(matrix) or col > len(matrix[0]):
      return 'N'  # Narek wins

    # Move to the next submatrix
    row += 1
    col += 1
    current_element += 1

  # If the array ends before Narek runs out of elements, Tsovak wins
  return 'T'

def main():
  """
  Main function to handle multiple test cases.
  """

  t = int(input())

  for _ in range(t):
    l, n, m = map(int, input().split())
    array = list(map(int, input().split()))
    matrix = []
    for _ in range(n):
      matrix.append(list(map(int, input().split())))
    winner = find_winner(array, matrix)
    print(winner)

if __name__ == '__main__':
  main()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
2 2 2
1 2
1 2
2 1
2 2 2
1 2
1 1
2 2
2 2 2
1 2
2 2
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are the constraints on all the variables. You can make hacks only if both versions of the problem are solved.

Tsovak and Narek are playing a game. They have an array $a$ and a matrix $b$ of integers with $n$ rows and $m$ columns, numbered from $1$. The cell in the $i$-th row and the $j$-th column is $(i, j)$.

They are looking for the elements of $a$ in turns; Tsovak starts first. Each time a player looks for a cell in the matrix containing the current element of $a$ (Tsovak looks for the first, then Narek looks for the second, etc.). Let's say a player has chosen the cell $(r, c)$. The next player has to choose his cell in the submatrix starting at $(r + 1, c + 1)$ and ending in $(n, m)$ (the submatrix can be empty if $r=n$ or $c=m$). If a player cannot find such a cell (or the remaining submatrix is empty) or the array ends (the previous player has chosen the last element), then he loses.

Your task is to determine the winner if the players play optimally.

Input: The first line of the input contains $t$ ($1 \le t \le 300$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 300$) – the size of the array and the sizes of the matrix.

The second line contains $l$ integers $a_1, a_2, a_3, \ldots a_l$ ($1 \le a_i \le \min(7, n \cdot m)$) – the elements of the array $a$.

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le \min(7, n \cdot m)$) – representing the $i$-th row of the matrix.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^5$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $300$.

Output: You should output $t$ lines, the $i$-th of them containing a character representing the answer of the $i$-th test case: "T" if Tsovak wins or "N", otherwise (without quotes).

Examples:
input:
3
2 2 3
1 2
1 3 5
4 5 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 5
5 5
5 5
output:
N
T
N

Note: In the first example, Tsovak starts by looking for $1$. There is only one occurrence of $1$ at $(1,1)$, so he chooses it. Then Narek needs to look for $2$ in the submatrix of $(2, 2)$, which consists of just the last two elements: $5$ and $2$. He chooses $2$, and then Tsovak loses since the array has ended.

In the second example, Tsovak needs to choose $1$. There is a $1$ at the cell $(n,m)$, so he chooses that cell. Then, since the submatrix of $(n + 1, m + 1)$ is empty, Narek cannot find $2$, so he loses.

Buggy Code:
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

public class Main {

    //------------------------------------------main--------------------------------------------------//
    public static void main(String[] args) throws java.lang.Exception {
        OutputStream outputStream = System.out;
        PrintWriter out = new PrintWriter(outputStream);
        FastReader sc = new FastReader();
        Task s = new Task(sc, out);
        s.solve();
        out.close();
        // your code goes here"
    }
}

class FastReader {

    BufferedReader br;
    StringTokenizer st;

    public FastReader() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }

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

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

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

    double nextDouble() {
        return Double.parseDouble(next());
    }

    String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }

    public int[] readIntArray(int n) throws IOException {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = nextInt();
        }
        return arr;
    }

    public long[] readLongArray(int n) throws IOException {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++) {
            arr[i] = nextLong();
        }
        return arr;
    }
}

class Task {

    PrintWriter out;
    FastReader sc;

    Task(FastReader sc, PrintWriter out) {
        this.sc = sc;
        this.out = out;
    }
    //----------------------------------------------inputfunction------------------------------------------//

    String st() {
        return sc.nextLine();
    }

    int i() {
        return sc.nextInt();
    }

    long l() {
        return sc.nextLong();
    }

    double d() {
        return sc.nextDouble();
    }

    //---------------------------------------------absolute----------------------------------------------//
    public long abs(long a) {
        return a < 0 ? -a : a;
    }

    public int abs(int a) {
        return a < 0 ? -a : a;
    }

    //-------------------------------------------max--------------------------------------------------//
    public int max(int a, int b) {
        return a >= b ? a : b;
    }

    public long max(long a, long b) {
        return a >= b ? a : b;
    }

    public int max(int a, int b, int c) {
        return Math.max(a, Math.max(b, c));
    }

    public long max(long a, long b, long c) {
        return Math.max(a, Math.max(b, c));
    }

    //------------------------------------------min---------------------------------------------------//
    public int min(int a, int b) {
        return a <= b ? a : b;
    }

    public long min(long a, long b) {
        return a <= b ? a : b;
    }

    public int min(int a, int b, int c) {
        return Math.min(a, Math.min(b, c));
    }

    public long min(long a, long b, long c) {
        return Math.min(a, Math.min(b, c));
    }

    //----------------------------------------sortingfunction---------------------------------------------//
    private void sort(int[] arr) {
        int n = arr.length;
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void sort(char[] arr) {
        int n = arr.length;
        List<Character> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(char[] arr) {
        int n = arr.length;
        List<Character> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(int[] arr) {
        int n = arr.length;
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void sort(long[] arr) {
        int n = arr.length;
        List<Long> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(long[] arr) {
        int n = arr.length;
        List<Long> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }
    //---------------------------------------power----------------------------------------------------//

    public long pow(long a, int b) {
        long ans = 1L;
        while (b > 0) {
            if ((b & 1) > 0) {
                ans *= a;
            }
            a *= a;
            b >>= 1;
        }
        return b;
    }
    //----------------------------------------binary_exponentiation-----------------------------------//

    public long exp(long a, long b, long mod) {
        long ans = 1L;
        while (b > 0) {
            if ((b & 1) > 0) {
                ans = (ans * a) % mod;
            }
            a = (a * a) % mod;
            b >>= 1;
        }
        return ans;
    }

    //---------------------------------------helpingmethods-------------------------------------------//
    //---------------------------------------main-----------------------------------------------------//
    int t, n;
    Boolean[][][] dp;
    int m , l;
    int[] val;
    int[][] a;
    List<List<int[]>> list = new ArrayList<>();
    boolean help(int i, int j, int index){
        if(i>=n||j>=m||index>=l) return false;
        if(dp[i][j][index]!= null) return dp[i][j][index];
        boolean win  = false;
        for(int[] e: list.get(val[index])){
            int x= e[0];
            int y = e[1];
            if(a[x][y]==val[index]){
                    win|= !help(x+1, y+1, index+1);
            }
        }
        return dp[i][j][index]=win;
    }
    public void solve() throws IOException {
        t = sc.nextInt();
        while (t-- > 0) {
            l = i();
            n = i();
            m = i();
            list = new ArrayList<>(7);
            for(int i = 0;i<=7;i++) list.add(new ArrayList<>());
            val = sc.readIntArray(l);
            a = new int[n][m];
            dp = new Boolean[n][m][l];
            for(int i = 0;i<n;i++) for(int j = 0;j<m;j++) a[i][j] = i();
            for(int i = 0;i<n;i++) for(int j =0;j<m;j++) list.get(a[i][j]).add(new int[]{i,j});
            out.print(help(0, 0, 0)?'T':'N');
            out.print("\n");
        }
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
2 2 2
1 2
1 2
3 4
2 2 2
1 1
1 2
3 4
2 3 3
1 2
1 2 3
4 5 6
7 8 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are the constraints on all the variables. You can make hacks only if both versions of the problem are solved.

Tsovak and Narek are playing a game. They have an array $a$ and a matrix $b$ of integers with $n$ rows and $m$ columns, numbered from $1$. The cell in the $i$-th row and the $j$-th column is $(i, j)$.

They are looking for the elements of $a$ in turns; Tsovak starts first. Each time a player looks for a cell in the matrix containing the current element of $a$ (Tsovak looks for the first, then Narek looks for the second, etc.). Let's say a player has chosen the cell $(r, c)$. The next player has to choose his cell in the submatrix starting at $(r + 1, c + 1)$ and ending in $(n, m)$ (the submatrix can be empty if $r=n$ or $c=m$). If a player cannot find such a cell (or the remaining submatrix is empty) or the array ends (the previous player has chosen the last element), then he loses.

Your task is to determine the winner if the players play optimally.

Input: The first line of the input contains $t$ ($1 \le t \le 300$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 300$) – the size of the array and the sizes of the matrix.

The second line contains $l$ integers $a_1, a_2, a_3, \ldots a_l$ ($1 \le a_i \le \min(7, n \cdot m)$) – the elements of the array $a$.

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le \min(7, n \cdot m)$) – representing the $i$-th row of the matrix.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^5$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $300$.

Output: You should output $t$ lines, the $i$-th of them containing a character representing the answer of the $i$-th test case: "T" if Tsovak wins or "N", otherwise (without quotes).

Examples:
input:
3
2 2 3
1 2
1 3 5
4 5 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 5
5 5
5 5
output:
N
T
N

Note: In the first example, Tsovak starts by looking for $1$. There is only one occurrence of $1$ at $(1,1)$, so he chooses it. Then Narek needs to look for $2$ in the submatrix of $(2, 2)$, which consists of just the last two elements: $5$ and $2$. He chooses $2$, and then Tsovak loses since the array has ended.

In the second example, Tsovak needs to choose $1$. There is a $1$ at the cell $(n,m)$, so he chooses that cell. Then, since the submatrix of $(n + 1, m + 1)$ is empty, Narek cannot find $2$, so he loses.

Buggy Code:
import java.util.*;
import java.lang.*;
import java.io.*;

public class Main {
    public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    public static Scanner sc = new Scanner(System.in);
    public static PrintWriter out = new PrintWriter(System.out);

    public static void printArr(int[] arr) {
        for (int i = 0; i < arr.length; i++) {
            out.print(arr[i] + " ");
        }
        out.println();
    } // change array type as needed

    public static long gcd(long a, long b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    public static long lcm(long a, long b) {
        return (a / gcd(a, b)) * b;
    }

    public static void readArr(int[] arr) throws IOException {
        int n = arr.length;
        StringTokenizer st = new StringTokenizer(br.readLine());
        for (int i = 0; i < n; i++) {
            arr[i] = Integer.parseInt(st.nextToken());
        }
    }

    public static void readArr(long[] arr) throws IOException {
        int n = arr.length;
        StringTokenizer st = new StringTokenizer(br.readLine());
        for (int i = 0; i < n; i++) {
            arr[i] = Long.parseLong(st.nextToken());
        }
    }

    public static boolean isPrime(int n) {
        if (n == 1)
            return false;
        for (int i = 2; i <= Math.sqrt(n); i++) {
            if (n % i == 0) {
                return false;
            }
        }
        return true;
    }

    public static void sort(long[] arr) {
        int n = arr.length;
        Random rnd = new Random();
        for (int i = 0; i < n; i++) {
            long tmp = arr[i];
            int randomPos = i + rnd.nextInt(n - i);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
        Arrays.sort(arr);
    }

    public static void sort(int[] arr) {
        int n = arr.length;
        Random rnd = new Random();
        for (int i = 0; i < n; i++) {
            int tmp = arr[i];
            int randomPos = i + rnd.nextInt(n - i);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
        Arrays.sort(arr);
    }

    public static void solve() throws IOException {
        StringTokenizer st = new StringTokenizer(br.readLine());
        int l = Integer.parseInt(st.nextToken());
        int n = Integer.parseInt(st.nextToken());
        int m = Integer.parseInt(st.nextToken());
        int[] arr = new int[l];
        readArr(arr);
        int[][] grid = new int[n][m];
        for(int[] x : grid) readArr(x);

        HashMap<Integer, ArrayList<int[]>> map = new HashMap<>();
        for(int i = 1; i <= 7; i++) map.put(i, new ArrayList<>());
        for(int i = 1; i <= 7; i++) {
            int y = -1;
            for(int r = n-1; r >= 0; r--) {
                for(int c = y+1; c < m; c++) {
                    if(grid[r][c] == i) {
                        map.get(i).add(new int[]{r, c});
                        y = c;
                        break;
                    }
                }
            }
        }
        // for(int i = 1; i <= 7; i++) {
        //     out.print(i);
        //     for(int[] x : map.get(i)) out.print(Arrays.toString(x));
        //     out.println();
        // }

        boolean[][] dp = new boolean[n][m];
        for(int[] x : map.get(arr[l-1])) {
            dp[x[0]][x[1]] = true;
        }
        for(int i = l-2; i >= 0; i--) {
            for(int[] x : map.get(arr[i])) {
                boolean seen = false;
                for(int[] y : map.get(arr[i+1])) {
                    if(x[0] < y[0] && x[1] < y[1]) {
                        seen = true;
                        if(!dp[y[0]][y[1]]) {
                            dp[x[0]][x[1]] = true;
                        }
                        break;
                    }
                }
                if(!seen) dp[x[0]][x[1]] = true;
            }
        }
        // for(boolean[] x : dp) {
        //     out.println(Arrays.toString(x));
        // }
        boolean win = false;
        for(int[] x : map.get(arr[0])) {
            win = win || dp[x[0]][x[1]];
        }
        if(win) out.println("T");
        else out.println("N");
    }

    public static void main(String[] args) throws IOException {
        int t = Integer.parseInt(br.readLine());
        // int t=1;
        while (t-->0) {
            solve();
        }
        out.close();
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases where the algorithm might fail. The key points to consider are:

1. **Empty Submatrix**: Ensure that the submatrix for the next player can sometimes be empty.
2. **Multiple Occurrences**: Ensure that there are multiple occurrences of the same number in the matrix to test the logic of choosing the correct cell.
3. **Optimal Play**: Ensure that the players play optimally, which means they will always choose the best available cell to win.

Given these considerations, let's create a test case that challenges the algorithm:

```plaintext
3
3 3 3
1 2 3
1 2 3
4 5 6
7 8 9
2 3 3
1 2
1 2 3
4 5 6
7 8 9
2 3 2
1 2
1 2
3 4
5 6
```

### Explanation:
1. **Test Case 1**:
   - Array: `[1, 2, 3]`
   - Matrix:
     ```
     1 2 3
     4 5 6
     7 8 9
     ```
   - **Expected Output**: `T`
   - **Reason**: Tsovak can choose (1,1), Narek can choose (2,2), and Tsovak can choose (3,3). Narek loses because the next submatrix is empty.

2. **Test Case 2**:
   - Array: `[1, 2]`
   - Matrix:
     ```
     1 2 3
     4 5 6
     7 8 9
     ```
   - **Expected Output**: `T`
   - **Reason**: Tsovak can choose (1,1), and Narek cannot find a 2 in the submatrix starting at (2,2).

3. **Test Case 3**:
   - Array: `[1, 2]`
   - Matrix:
     ```
     1 2
     3 4
     5 6
     ```
   - **Expected Output**: `N`
   - **Reason**: Tsovak can choose (1,1), but Narek can choose (2,2), and Tsovak cannot find a 2 in the submatrix starting at (3,3) (which is empty).

This test input should help expose the bug in the provided buggy code by covering edge cases and ensuring the algorithm handles them correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are the constraints on all the variables. You can make hacks only if both versions of the problem are solved.

Tsovak and Narek are playing a game. They have an array $a$ and a matrix $b$ of integers with $n$ rows and $m$ columns, numbered from $1$. The cell in the $i$-th row and the $j$-th column is $(i, j)$.

They are looking for the elements of $a$ in turns; Tsovak starts first. Each time a player looks for a cell in the matrix containing the current element of $a$ (Tsovak looks for the first, then Narek looks for the second, etc.). Let's say a player has chosen the cell $(r, c)$. The next player has to choose his cell in the submatrix starting at $(r + 1, c + 1)$ and ending in $(n, m)$ (the submatrix can be empty if $r=n$ or $c=m$). If a player cannot find such a cell (or the remaining submatrix is empty) or the array ends (the previous player has chosen the last element), then he loses.

Your task is to determine the winner if the players play optimally.

Input: The first line of the input contains $t$ ($1 \le t \le 300$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 300$) – the size of the array and the sizes of the matrix.

The second line contains $l$ integers $a_1, a_2, a_3, \ldots a_l$ ($1 \le a_i \le \min(7, n \cdot m)$) – the elements of the array $a$.

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le \min(7, n \cdot m)$) – representing the $i$-th row of the matrix.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^5$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $300$.

Output: You should output $t$ lines, the $i$-th of them containing a character representing the answer of the $i$-th test case: "T" if Tsovak wins or "N", otherwise (without quotes).

Examples:
input:
3
2 2 3
1 2
1 3 5
4 5 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 5
5 5
5 5
output:
N
T
N

Note: In the first example, Tsovak starts by looking for $1$. There is only one occurrence of $1$ at $(1,1)$, so he chooses it. Then Narek needs to look for $2$ in the submatrix of $(2, 2)$, which consists of just the last two elements: $5$ and $2$. He chooses $2$, and then Tsovak loses since the array has ended.

In the second example, Tsovak needs to choose $1$. There is a $1$ at the cell $(n,m)$, so he chooses that cell. Then, since the submatrix of $(n + 1, m + 1)$ is empty, Narek cannot find $2$, so he loses.

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/trie_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
#ifndef ONLINE_JUDGE
  #include "debug.h"
#else
  #define dbg(...) 42
  template<typename T>ostream&operator<<(ostream&os,vector<T>&vec){for(signed i=0;i+1<vec.size();++i){os<<vec[i]<<" ";}if(vec.size()>0)os<<vec.back();return os;}
#endif

#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()

template<typename T>istream&operator>>(istream&is,vector<T>&vec){for(T&element:vec){is>>element;}return is;}
template<typename T>void chmin(T&x,T y){x=min(x,y);}
template<typename T>void chmax(T&x,T y){x=max(x,y);}

void test_case(signed ___) {
  int l, n, m;
  cin >> l >> n >> m;
  vector<vector<int>> mat(n, vector<int>(m));
  vector<int> ar(l);
  cin >> ar >> mat;
  for (int& i : ar) --i;
  for (auto& x : mat) for (auto& y : x) --y;
  vector<vector<vector<bool>>> winning(n, vector<vector<bool>>(m, vector<bool>(l+1, false)));
  vector<vector<vector<bool>>> cnt_loss(n, vector<vector<bool>>(m, vector<bool>(7, false)));
  vector<vector<vector<int>>> cnt_num(n, vector<vector<int>>(m, vector<int>(7, 0)));

  for (int num = 0; num < 7; ++num) {
    for (int i = n - 1; i >= 0; --i) {
      for (int j = m - 1; j >= 0; --j) {
        cnt_num[i][j][num] = int(mat[i][j] == num);
        if (j < m-1) cnt_num[i][j][num] += cnt_num[i][j+1][num];
      }
    }
    for (int i = n - 2; i >= 0; --i) {
      for (int j = m - 1; j >= 0; --j) {
        cnt_num[i][j][num] += cnt_num[i+1][j][num];
      }
    }
  }

  for (int move = l-1; move >= 0; --move) {
    for (int num = 0; num < 7; ++num) {
      for (int i = n - 1; i >= 0; --i) {
        for (int j = m - 1; j >= 0; --j) {
          cnt_loss[i][j][num] = int(mat[i][j] == num && !winning[i][j][move+1]);
          if (j < m-1) cnt_loss[i][j][num] = cnt_loss[i][j][num] | cnt_loss[i][j+1][num];
        }
      }
      for (int i = n - 2; i >= 0; --i) {
        for (int j = m - 1; j >= 0; --j) {
          cnt_loss[i][j][num] = cnt_loss[i][j][num] | cnt_loss[i+1][j][num];
        }
      }
    }
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < m; ++j) {
        if (mat[i][j] != ar[move]) continue;
        if (move == l-1 || i == n-1 || j == m-1 || (cnt_loss[i+1][j+1][ar[move+1]] != false) || cnt_num[i+1][j+1][ar[move+1]] == 0) {
          winning[i][j][move] = true;
        }
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (winning[i][j][0]) {
        cout << "T\n";
        return;
      }
    }
  }
  cout << "N\n";
}

signed main(void) {
  cin.tie(0)->sync_with_stdio(false);

  signed tt = 1;
  cin >> tt;
  while (tt--)
    test_case(69420);
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
1 1 1
1
1
1 1 1
1
1
2 2 2
1 2
1 2
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are the constraints on all the variables. You can make hacks only if both versions of the problem are solved.

Tsovak and Narek are playing a game. They have an array $a$ and a matrix $b$ of integers with $n$ rows and $m$ columns, numbered from $1$. The cell in the $i$-th row and the $j$-th column is $(i, j)$.

They are looking for the elements of $a$ in turns; Tsovak starts first. Each time a player looks for a cell in the matrix containing the current element of $a$ (Tsovak looks for the first, then Narek looks for the second, etc.). Let's say a player has chosen the cell $(r, c)$. The next player has to choose his cell in the submatrix starting at $(r + 1, c + 1)$ and ending in $(n, m)$ (the submatrix can be empty if $r=n$ or $c=m$). If a player cannot find such a cell (or the remaining submatrix is empty) or the array ends (the previous player has chosen the last element), then he loses.

Your task is to determine the winner if the players play optimally.

Input: The first line of the input contains $t$ ($1 \le t \le 300$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 300$) – the size of the array and the sizes of the matrix.

The second line contains $l$ integers $a_1, a_2, a_3, \ldots a_l$ ($1 \le a_i \le \min(7, n \cdot m)$) – the elements of the array $a$.

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le \min(7, n \cdot m)$) – representing the $i$-th row of the matrix.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^5$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $300$.

Output: You should output $t$ lines, the $i$-th of them containing a character representing the answer of the $i$-th test case: "T" if Tsovak wins or "N", otherwise (without quotes).

Examples:
input:
3
2 2 3
1 2
1 3 5
4 5 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 5
5 5
5 5
output:
N
T
N

Note: In the first example, Tsovak starts by looking for $1$. There is only one occurrence of $1$ at $(1,1)$, so he chooses it. Then Narek needs to look for $2$ in the submatrix of $(2, 2)$, which consists of just the last two elements: $5$ and $2$. He chooses $2$, and then Tsovak loses since the array has ended.

In the second example, Tsovak needs to choose $1$. There is a $1$ at the cell $(n,m)$, so he chooses that cell. Then, since the submatrix of $(n + 1, m + 1)$ is empty, Narek cannot find $2$, so he loses.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
int main() {
	int t;
	cin>>t;
	while(t--) {
		int n,m,i,j,k=1,c,a,b;
		cin>>n>>a>>b;
		vector<int>q(n);
		vector<int>x(10,-1);
		vector<int>y(10,-1);
		int p[305][305];
		for(i=0;i<n;i++){
			cin>>q[i];
		}
		for(i=0;i<a;i++){
			for(j=0;j<b;j++){
				cin>>p[i][j];
				x[p[i][j]]=i;
				y[p[i][j]]=j;
			}
		}
		int xx=0,yy=0;
		for(i=0;i<n;i++){
			if(x[q[i]]>=xx&&y[q[i]]>=yy){
				xx=x[q[i]]+1;
				yy=y[q[i]]+1;
			}else{
				k=0;
				if(i%2==0){
					cout<<"N"<<endl;
				}else cout<<"T"<<endl;
				break;
			}
		}
		if(k==1){
			if(n%2==0){
				cout<<"N"<<endl;
			}else cout<<"T"<<endl;
		}
	}
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify potential edge cases or scenarios where the code might fail.

### Analysis of the Buggy Code
1. **Initialization and Input Reading:**
   - The code reads the number of test cases, the dimensions of the matrix, and the array elements.
   - It initializes vectors and a 2D array to store the matrix.

2. **Matrix and Array Processing:**
   - The code populates the matrix and updates the positions of the elements in the vectors `x` and `y`.

3. **Game Logic:**
   - The code iterates through the array elements and checks if the current element can be found in the submatrix starting from `(xx, yy)`.
   - If the element is found, it updates the starting point for the next player.
   - If the element is not found, it determines the winner based on the turn and breaks the loop.
   - If all elements are found, it determines the winner based on the parity of the array length.

### Potential Bugs
1. **Incorrect Submatrix Checking:**
   - The code uses `x[p[i][j]]` and `y[p[i][j]]` to store the positions of elements, but it does not correctly handle cases where multiple elements in the matrix have the same value.
   - The code assumes that the first occurrence of the element is the correct one, which might not always be the case.

2. **Edge Cases:**
   - Small matrices and arrays with repeated elements.
   - Matrices with elements in the last row or column.
   - Matrices with elements that are not in the array.

### Challenging Test Input
To expose these potential bugs, we can create a test case with a small matrix and array that includes repeated elements and edge cases.

```plaintext
3
2 2 2
1 1
1 1
1 1
2 2 2
1 2
1 2
2 1
2 2 2
2 1
2 2
1 1
```

### Explanation
1. **First Test Case:**
   - Array: `[1, 1]`
   - Matrix:
     ```
     1 1
     1 1
     ```
   - Tsovak can choose any `(1, 1)`, and Narek can also choose any `(1, 1)`. Since the submatrix is empty after Narek's turn, Tsovak loses.

2. **Second Test Case:**
   - Array: `[1, 2]`
   - Matrix:
     ```
     1 2
     2 1
     ```
   - Tsovak can choose `(1, 1)` and Narek can choose `(2, 2)`. The submatrix is empty, so Tsovak wins.

3. **Third Test Case:**
   - Array: `[2, 1]`
   - Matrix:
     ```
     2 2
     1 1
     ```
   - Tsovak can choose `(1, 1)` and Narek can choose `(2, 1)`. The submatrix is empty, so Tsovak loses.

This test input should help expose the bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are the constraints on all the variables. You can make hacks only if both versions of the problem are solved.

Tsovak and Narek are playing a game. They have an array $a$ and a matrix $b$ of integers with $n$ rows and $m$ columns, numbered from $1$. The cell in the $i$-th row and the $j$-th column is $(i, j)$.

They are looking for the elements of $a$ in turns; Tsovak starts first. Each time a player looks for a cell in the matrix containing the current element of $a$ (Tsovak looks for the first, then Narek looks for the second, etc.). Let's say a player has chosen the cell $(r, c)$. The next player has to choose his cell in the submatrix starting at $(r + 1, c + 1)$ and ending in $(n, m)$ (the submatrix can be empty if $r=n$ or $c=m$). If a player cannot find such a cell (or the remaining submatrix is empty) or the array ends (the previous player has chosen the last element), then he loses.

Your task is to determine the winner if the players play optimally.

Input: The first line of the input contains $t$ ($1 \le t \le 300$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 300$) – the size of the array and the sizes of the matrix.

The second line contains $l$ integers $a_1, a_2, a_3, \ldots a_l$ ($1 \le a_i \le \min(7, n \cdot m)$) – the elements of the array $a$.

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le \min(7, n \cdot m)$) – representing the $i$-th row of the matrix.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^5$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $300$.

Output: You should output $t$ lines, the $i$-th of them containing a character representing the answer of the $i$-th test case: "T" if Tsovak wins or "N", otherwise (without quotes).

Examples:
input:
3
2 2 3
1 2
1 3 5
4 5 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 5
5 5
5 5
output:
N
T
N

Note: In the first example, Tsovak starts by looking for $1$. There is only one occurrence of $1$ at $(1,1)$, so he chooses it. Then Narek needs to look for $2$ in the submatrix of $(2, 2)$, which consists of just the last two elements: $5$ and $2$. He chooses $2$, and then Tsovak loses since the array has ended.

In the second example, Tsovak needs to choose $1$. There is a $1$ at the cell $(n,m)$, so he chooses that cell. Then, since the submatrix of $(n + 1, m + 1)$ is empty, Narek cannot find $2$, so he loses.

Buggy Code:
def solve_game(l, n, m, a, b):
    def can_win(row, col, idx):
        if idx == l:
            return False

        for i in range(row, n):
            for j in range(col if i == row else 0, m):
                if b[i][j] == a[idx]:
                    if not can_win(i + 1, j + 1, idx + 1):
                        return True
        return False

    return can_win(0, 0, 0)

t = int(input())
for _ in range(t):
    l, n, m = map(int, input().split())
    a = list(map(int, input().split()))
    b = [list(map(int, input().split())) for _ in range(n)]

    winner = "T" if solve_game(l, n, m, a, b) else "N"
    print(winner)

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
1 1 1
1
1
2 2 2
1 2
1 2
2 2 2
1 2
2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are the constraints on all the variables. You can make hacks only if both versions of the problem are solved.

Tsovak and Narek are playing a game. They have an array $a$ and a matrix $b$ of integers with $n$ rows and $m$ columns, numbered from $1$. The cell in the $i$-th row and the $j$-th column is $(i, j)$.

They are looking for the elements of $a$ in turns; Tsovak starts first. Each time a player looks for a cell in the matrix containing the current element of $a$ (Tsovak looks for the first, then Narek looks for the second, etc.). Let's say a player has chosen the cell $(r, c)$. The next player has to choose his cell in the submatrix starting at $(r + 1, c + 1)$ and ending in $(n, m)$ (the submatrix can be empty if $r=n$ or $c=m$). If a player cannot find such a cell (or the remaining submatrix is empty) or the array ends (the previous player has chosen the last element), then he loses.

Your task is to determine the winner if the players play optimally.

Input: The first line of the input contains $t$ ($1 \le t \le 300$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 300$) – the size of the array and the sizes of the matrix.

The second line contains $l$ integers $a_1, a_2, a_3, \ldots a_l$ ($1 \le a_i \le \min(7, n \cdot m)$) – the elements of the array $a$.

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le \min(7, n \cdot m)$) – representing the $i$-th row of the matrix.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^5$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $300$.

Output: You should output $t$ lines, the $i$-th of them containing a character representing the answer of the $i$-th test case: "T" if Tsovak wins or "N", otherwise (without quotes).

Examples:
input:
3
2 2 3
1 2
1 3 5
4 5 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 5
5 5
5 5
output:
N
T
N

Note: In the first example, Tsovak starts by looking for $1$. There is only one occurrence of $1$ at $(1,1)$, so he chooses it. Then Narek needs to look for $2$ in the submatrix of $(2, 2)$, which consists of just the last two elements: $5$ and $2$. He chooses $2$, and then Tsovak loses since the array has ended.

In the second example, Tsovak needs to choose $1$. There is a $1$ at the cell $(n,m)$, so he chooses that cell. Then, since the submatrix of $(n + 1, m + 1)$ is empty, Narek cannot find $2$, so he loses.

Buggy Code:
import java.io.*;

public class Main {
		static PrintWriter out=new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		static StreamTokenizer in=new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
		static int[][] arr=new int[8][2];
		static int[] arrL=new int[300+5];
		public static void main(String[] args) throws IOException {
				in.nextToken();int t=(int)in.nval;
				while(t-->0){
						for(int i=0;i<arr.length;i++){
								arr[i][0]=0;
								arr[i][1]=0;
						}
						in.nextToken();int l=(int)in.nval;
						in.nextToken();int n=(int)in.nval;
						in.nextToken();int m=(int)in.nval;
						for(int i=1;i<=l;i++){
								in.nextToken();arrL[i]=(int)in.nval;
						}
						for(int i=1;i<=n;i++){
								for(int j=1;j<=m;j++){
										in.nextToken();int x=(int)in.nval;
										arr[x][0]=i;
										arr[x][1]=j;
								}
						}
						int count=0,curX=1,curY=1;
						for(int i=1;i<=l;i++){
								if(arr[arrL[i]][0]==0||curX>n||curY>m){
										break;
								}else if(arr[arrL[i]][0]<curX||(arr[arrL[i]][0]==curX&&arr[arrL[i]][1]<curY)){
										break;
								}else{
										count++;
										curX=arr[arrL[i]][0]+1;
										curY=arr[arrL[i]][1]+1;
								}
						}
						out.println(count%2==0?"N":"T");
				}
				out.close();
		}
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
2 2 2
1 2
1 2
2 1
2 2 2
1 2
2 1
1 2
3 3 3
1 2 3
1 2 3
4 5 6
7 8 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are the constraints on all the variables. You can make hacks only if both versions of the problem are solved.

Tsovak and Narek are playing a game. They have an array $a$ and a matrix $b$ of integers with $n$ rows and $m$ columns, numbered from $1$. The cell in the $i$-th row and the $j$-th column is $(i, j)$.

They are looking for the elements of $a$ in turns; Tsovak starts first. Each time a player looks for a cell in the matrix containing the current element of $a$ (Tsovak looks for the first, then Narek looks for the second, etc.). Let's say a player has chosen the cell $(r, c)$. The next player has to choose his cell in the submatrix starting at $(r + 1, c + 1)$ and ending in $(n, m)$ (the submatrix can be empty if $r=n$ or $c=m$). If a player cannot find such a cell (or the remaining submatrix is empty) or the array ends (the previous player has chosen the last element), then he loses.

Your task is to determine the winner if the players play optimally.

Input: The first line of the input contains $t$ ($1 \le t \le 300$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 300$) – the size of the array and the sizes of the matrix.

The second line contains $l$ integers $a_1, a_2, a_3, \ldots a_l$ ($1 \le a_i \le \min(7, n \cdot m)$) – the elements of the array $a$.

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le \min(7, n \cdot m)$) – representing the $i$-th row of the matrix.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^5$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $300$.

Output: You should output $t$ lines, the $i$-th of them containing a character representing the answer of the $i$-th test case: "T" if Tsovak wins or "N", otherwise (without quotes).

Examples:
input:
3
2 2 3
1 2
1 3 5
4 5 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 5
5 5
5 5
output:
N
T
N

Note: In the first example, Tsovak starts by looking for $1$. There is only one occurrence of $1$ at $(1,1)$, so he chooses it. Then Narek needs to look for $2$ in the submatrix of $(2, 2)$, which consists of just the last two elements: $5$ and $2$. He chooses $2$, and then Tsovak loses since the array has ended.

In the second example, Tsovak needs to choose $1$. There is a $1$ at the cell $(n,m)$, so he chooses that cell. Then, since the submatrix of $(n + 1, m + 1)$ is empty, Narek cannot find $2$, so he loses.

Buggy Code:
import java.util.*;
import java.lang.*;
import java.io.*;

public class Main {
    public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    public static Scanner sc = new Scanner(System.in);
    public static PrintWriter out = new PrintWriter(System.out);

    public static void printArr(int[] arr) {
        for (int i = 0; i < arr.length; i++) {
            out.print(arr[i] + " ");
        }
        out.println();
    } // change array type as needed

    public static long gcd(long a, long b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    public static long lcm(long a, long b) {
        return (a / gcd(a, b)) * b;
    }

    public static void readArr(int[] arr) throws IOException {
        int n = arr.length;
        StringTokenizer st = new StringTokenizer(br.readLine());
        for (int i = 0; i < n; i++) {
            arr[i] = Integer.parseInt(st.nextToken());
        }
    }

    public static void readArr(long[] arr) throws IOException {
        int n = arr.length;
        StringTokenizer st = new StringTokenizer(br.readLine());
        for (int i = 0; i < n; i++) {
            arr[i] = Long.parseLong(st.nextToken());
        }
    }

    public static boolean isPrime(int n) {
        if (n == 1)
            return false;
        for (int i = 2; i <= Math.sqrt(n); i++) {
            if (n % i == 0) {
                return false;
            }
        }
        return true;
    }

    public static void sort(long[] arr) {
        int n = arr.length;
        Random rnd = new Random();
        for (int i = 0; i < n; i++) {
            long tmp = arr[i];
            int randomPos = i + rnd.nextInt(n - i);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
        Arrays.sort(arr);
    }

    public static void sort(int[] arr) {
        int n = arr.length;
        Random rnd = new Random();
        for (int i = 0; i < n; i++) {
            int tmp = arr[i];
            int randomPos = i + rnd.nextInt(n - i);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
        Arrays.sort(arr);
    }

    public static void solve() throws IOException {
        StringTokenizer st = new StringTokenizer(br.readLine());
        int l = Integer.parseInt(st.nextToken());
        int n = Integer.parseInt(st.nextToken());
        int m = Integer.parseInt(st.nextToken());
        int[] arr = new int[l];
        readArr(arr);
        int[][] grid = new int[n][m];
        for(int[] x : grid) readArr(x);

        HashMap<Integer, ArrayList<int[]>> map = new HashMap<>();
        for(int i = 1; i <= 7; i++) map.put(i, new ArrayList<>());
        // for(int i = 1; i <= 7; i++) {
        //     int y = -1;
        //     for(int r = n-1; r >= 0; r--) {
        //         for(int c = m-1; c > y; c--) {
        //             if(grid[r][c] == i) {
        //                 map.get(i).add(new int[]{r, c});
        //                 y = c;
        //                 break;
        //             }
        //         }
        //     }
        // }
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                map.get(grid[i][j]).add(new int[]{i, j});
            }
        }
        // for(int i = 1; i <= 7; i++) {
        //     out.print(i);
        //     for(int[] x : map.get(i)) out.print(Arrays.toString(x));
        //     out.println();
        // }

        boolean[][] dp = new boolean[n][m];
        for(int[] x : map.get(arr[l-1])) {
            dp[x[0]][x[1]] = true;
        }
        for(int i = l-2; i >= 0; i--) {
            for(int[] x : map.get(arr[i])) {
                boolean seen = false;
                dp[x[0]][x[1]] = false;
                for(int[] y : map.get(arr[i+1])) {
                    if(x[0] < y[0] && x[1] < y[1]) {
                        seen = true;
                        if(!dp[y[0]][y[1]]) {
                            dp[x[0]][x[1]] = true;
                        }
                        break;
                    }
                }
                if(!seen) dp[x[0]][x[1]] = true;
            }
        }
        // for(boolean[] x : dp) {
        //     out.println(Arrays.toString(x));
        // }
        boolean win = false;
        for(int[] x : map.get(arr[0])) {
            win = win || dp[x[0]][x[1]];
        }
        if(win) out.println("T");
        else out.println("N");
    }

    public static void main(String[] args) throws IOException {
        int t = Integer.parseInt(br.readLine());
        // int t=1;
        while (t-->0) {
            solve();
        }
        out.close();
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
2 2 2
1 2
1 2
2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are the constraints on all the variables. You can make hacks only if both versions of the problem are solved.

Tsovak and Narek are playing a game. They have an array $a$ and a matrix $b$ of integers with $n$ rows and $m$ columns, numbered from $1$. The cell in the $i$-th row and the $j$-th column is $(i, j)$.

They are looking for the elements of $a$ in turns; Tsovak starts first. Each time a player looks for a cell in the matrix containing the current element of $a$ (Tsovak looks for the first, then Narek looks for the second, etc.). Let's say a player has chosen the cell $(r, c)$. The next player has to choose his cell in the submatrix starting at $(r + 1, c + 1)$ and ending in $(n, m)$ (the submatrix can be empty if $r=n$ or $c=m$). If a player cannot find such a cell (or the remaining submatrix is empty) or the array ends (the previous player has chosen the last element), then he loses.

Your task is to determine the winner if the players play optimally.

Input: The first line of the input contains $t$ ($1 \le t \le 300$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 300$) – the size of the array and the sizes of the matrix.

The second line contains $l$ integers $a_1, a_2, a_3, \ldots a_l$ ($1 \le a_i \le \min(7, n \cdot m)$) – the elements of the array $a$.

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le \min(7, n \cdot m)$) – representing the $i$-th row of the matrix.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^5$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $300$.

Output: You should output $t$ lines, the $i$-th of them containing a character representing the answer of the $i$-th test case: "T" if Tsovak wins or "N", otherwise (without quotes).

Examples:
input:
3
2 2 3
1 2
1 3 5
4 5 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 5
5 5
5 5
output:
N
T
N

Note: In the first example, Tsovak starts by looking for $1$. There is only one occurrence of $1$ at $(1,1)$, so he chooses it. Then Narek needs to look for $2$ in the submatrix of $(2, 2)$, which consists of just the last two elements: $5$ and $2$. He chooses $2$, and then Tsovak loses since the array has ended.

In the second example, Tsovak needs to choose $1$. There is a $1$ at the cell $(n,m)$, so he chooses that cell. Then, since the submatrix of $(n + 1, m + 1)$ is empty, Narek cannot find $2$, so he loses.

Buggy Code:
def II():
    return(int(input()))
def LMI():
    return(list(map(int,input().split())))
def I():
    return(input())
def MII():
    return(map(int,input().split()))
import sys
input=sys.stdin.readline
# import io,os
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# from collections import Counter
# int(math.log(len(L)))
# import math
# from collections import defaultdict
# mod=10**9+7
# from collections import deque
import  math

# import bisect
# from functools import cache
# @cache
for _ in range(II()):
    l,n,m=MII()
    A=LMI()
    X=[]
    for _ in range(n):
        B=LMI()
        X.append(B)
    dp=[[[[0]*4 for _ in range(l)] for _ in range(m)] for _ in range(n)]
    for i in range(n-1,-1,-1):
        for j in range(m-1,-1,-1):
            for s in range(l):
                if X[i][j]==A[s]:
                    if i!=n-1 and j!=m-1:
                        if s!=l-1:
                            dp[i][j][s][0]|=not(dp[i+1][j+1][s+1][3])
                        else:
                            dp[i][j][s][0]=1
                    else:
                        dp[i][j][s][0]=1
                dp[i][j][s][1]=dp[i][j][s][0]
                dp[i][j][s][2]=dp[i][j][s][0]
                dp[i][j][s][3]=dp[i][j][s][0]
                if j!=m-1:
                    dp[i][j][s][1]|=dp[i][j+1][s][1]
                if i!=n-1:
                    dp[i][j][s][2]|=dp[i+1][j][s][2]
                if i!=n-1 and j!=m-1:
                    dp[i][j][s][3]|=dp[i+1][j+1][s][3]
                dp[i][j][s][3]|=dp[i][j][s][1]|dp[i][j][s][2]
    if dp[0][0][0][3]:
        print("T")
    else:
        print("N")

# if __name__=="__main__":

#     for _ in range(II()):
#         t()
    # t()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. Here are the steps to consider:

1. **Small but Complex Test Case**: Since the input constraints allow for up to 300 for \( l \), \( n \), and \( m \), we can use smaller values that still capture the complexity. For example, \( l = 3 \), \( n = 3 \), and \( m = 3 \).

2. **Matrix and Array Configuration**: We need to ensure that the elements in the array \( a \) are present in the matrix \( b \) in such a way that the game logic can be tested thoroughly. We should place the elements in a way that forces the players to make strategic decisions.

3. **Optimal Play**: The players play optimally, so the test case should be designed to ensure that the logic of the game is correctly implemented. We need to consider scenarios where the submatrix constraints and the order of elements in the array \( a \) can lead to a specific outcome.

Given these considerations, let's create a test case:

```plaintext
1
3 3 3
1 2 3
1 2 3
4 5 6
7 8 9
```

### Explanation:
- **Test Case 1**:
  - \( l = 3 \)
  - \( n = 3 \)
  - \( m = 3 \)
  - Array \( a \) is \([1, 2, 3]\)
  - Matrix \( b \) is:
    ```
    1 2 3
    4 5 6
    7 8 9
    ```

### Analysis:
- **Tsovak's First Move**: Tsovak can choose the element 1 at position (1, 1).
- **Narek's Move**: Narek needs to choose the element 2 from the submatrix starting at (2, 2). The submatrix is:
  ```
  5 6
  8 9
  ```
  Narek can choose the element 2 at position (2, 2).
- **Tsovak's Second Move**: Tsovak needs to choose the element 3 from the submatrix starting at (3, 3). The submatrix is:
  ```
  9
  ```
  Tsovak can choose the element 3 at position (3, 3).

Since Tsovak can make all the necessary moves, Tsovak wins.

### Expected Output:
```plaintext
T
```

This test case is small enough to be complete and valid while still being challenging enough to expose potential bugs in the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are the constraints on all the variables. You can make hacks only if both versions of the problem are solved.

Tsovak and Narek are playing a game. They have an array $a$ and a matrix $b$ of integers with $n$ rows and $m$ columns, numbered from $1$. The cell in the $i$-th row and the $j$-th column is $(i, j)$.

They are looking for the elements of $a$ in turns; Tsovak starts first. Each time a player looks for a cell in the matrix containing the current element of $a$ (Tsovak looks for the first, then Narek looks for the second, etc.). Let's say a player has chosen the cell $(r, c)$. The next player has to choose his cell in the submatrix starting at $(r + 1, c + 1)$ and ending in $(n, m)$ (the submatrix can be empty if $r=n$ or $c=m$). If a player cannot find such a cell (or the remaining submatrix is empty) or the array ends (the previous player has chosen the last element), then he loses.

Your task is to determine the winner if the players play optimally.

Input: The first line of the input contains $t$ ($1 \le t \le 300$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 300$) – the size of the array and the sizes of the matrix.

The second line contains $l$ integers $a_1, a_2, a_3, \ldots a_l$ ($1 \le a_i \le \min(7, n \cdot m)$) – the elements of the array $a$.

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le \min(7, n \cdot m)$) – representing the $i$-th row of the matrix.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^5$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $300$.

Output: You should output $t$ lines, the $i$-th of them containing a character representing the answer of the $i$-th test case: "T" if Tsovak wins or "N", otherwise (without quotes).

Examples:
input:
3
2 2 3
1 2
1 3 5
4 5 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 5
5 5
5 5
output:
N
T
N

Note: In the first example, Tsovak starts by looking for $1$. There is only one occurrence of $1$ at $(1,1)$, so he chooses it. Then Narek needs to look for $2$ in the submatrix of $(2, 2)$, which consists of just the last two elements: $5$ and $2$. He chooses $2$, and then Tsovak loses since the array has ended.

In the second example, Tsovak needs to choose $1$. There is a $1$ at the cell $(n,m)$, so he chooses that cell. Then, since the submatrix of $(n + 1, m + 1)$ is empty, Narek cannot find $2$, so he loses.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;

#define all(v) v.begin(),v.end()
#define sz(v) ((ll)v.size())
#define lp(p) p << 1
#define rp(p) p << 1 | 1

using ll = long long;

void solve(){
    int l,n,m;
    cin >> l >> n >> m;
    vector<int> a(l+5);
    for(int i = 1; i <= l; i ++ )cin >> a[i];
    vector<vector<int>> g(n+5,vector<int>(m+5));
	for(int i = 1; i <= n; i ++ ){
		for(int j = 1; j <= m; j ++ ){
			cin >> g[i][j];
		}
	}
	vector<vector<int>> SG(n+5,vector<int>(m+5,-1));
	auto dfs = [&](auto && self,int x,int y,int now)->int{
		if(x > n || y > m)return 0;
		if(now > l)return 0;
		if(SG[x][y] != -1)return SG[x][y];
		set<int> st;
		for(int i = x; i <= n; i ++ ){
			for(int j = y; j <= m; j ++ ){
				if(g[i][j] == now){
					st.insert(self(self,i+1,j+1,now+1));
				}
			}
		}
		int res = 0;
		for(auto x:st){
			if(x != res)break;
			res ++;
		}
		return SG[x][y] = res;
	};
	cout << (dfs(dfs,1,1,1) ? 'T' : 'N');
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    //freopen("input.txt","r",stdin);
    //freopen("bad.txt","w",stdout);

    //cout << fixed << setprecision(10);

    //auto st = clock();

    int T = 1;
    cin >> T;
    for(int i = 1; i <= T; i ++ ){
        solve();
        if(i!=T)cout << '\n';
    }

    //cerr << '\n' << clock() - st << "ms\n";

    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
2 2 2
1 2
1 2
3 4
2 2 2
1 2
1 1
2 2
2 2 2
1 1
1 1
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The differences between the two versions are the constraints on all the variables. You can make hacks only if both versions of the problem are solved.

Tsovak and Narek are playing a game. They have an array $a$ and a matrix $b$ of integers with $n$ rows and $m$ columns, numbered from $1$. The cell in the $i$-th row and the $j$-th column is $(i, j)$.

They are looking for the elements of $a$ in turns; Tsovak starts first. Each time a player looks for a cell in the matrix containing the current element of $a$ (Tsovak looks for the first, then Narek looks for the second, etc.). Let's say a player has chosen the cell $(r, c)$. The next player has to choose his cell in the submatrix starting at $(r + 1, c + 1)$ and ending in $(n, m)$ (the submatrix can be empty if $r=n$ or $c=m$). If a player cannot find such a cell (or the remaining submatrix is empty) or the array ends (the previous player has chosen the last element), then he loses.

Your task is to determine the winner if the players play optimally.

Input: The first line of the input contains $t$ ($1 \le t \le 300$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 300$) – the size of the array and the sizes of the matrix.

The second line contains $l$ integers $a_1, a_2, a_3, \ldots a_l$ ($1 \le a_i \le \min(7, n \cdot m)$) – the elements of the array $a$.

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le \min(7, n \cdot m)$) – representing the $i$-th row of the matrix.

It is guaranteed that the sum of $n \cdot m$ over all test cases does not exceed $10^5$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $300$.

Output: You should output $t$ lines, the $i$-th of them containing a character representing the answer of the $i$-th test case: "T" if Tsovak wins or "N", otherwise (without quotes).

Examples:
input:
3
2 2 3
1 2
1 3 5
4 5 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 5
5 5
5 5
output:
N
T
N

Note: In the first example, Tsovak starts by looking for $1$. There is only one occurrence of $1$ at $(1,1)$, so he chooses it. Then Narek needs to look for $2$ in the submatrix of $(2, 2)$, which consists of just the last two elements: $5$ and $2$. He chooses $2$, and then Tsovak loses since the array has ended.

In the second example, Tsovak needs to choose $1$. There is a $1$ at the cell $(n,m)$, so he chooses that cell. Then, since the submatrix of $(n + 1, m + 1)$ is empty, Narek cannot find $2$, so he loses.

Buggy Code:
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 3e2 + 5;
const ll MOD = 1e9 + 7, MAX = 1e18;
void fastio()
{
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}

ll n, l, m = 0, k;
ll arr[N][N], a[N];
set<pair<int, int>> gg[10];
bool dp[N][N][N];
int vis[N][N][N], id = 0;
bool solve(int i, int j, int k)
{
  // if (i == l + 1 && j == n + 1 && k == m + 1)
  //   return 1;
  if (i == l + 1 || j == n + 1 || k == m + 1)
    return 1;
  bool &ret = dp[i][j][k];
  if (vis[i][j][k] == id)
    return ret;
  vis[i][j][k] = id;
  ret = 0;
  for (auto [x, y] : gg[a[i]])
  {
    if (x < j || y < k)
      continue;
    ret |= solve(i + 1, x + 1, y + 1);
    if (ret)
      break;
  }
  return !ret;
}
void fn()
{
  cin >> l >> n >> m;
  id++;
  for (int i = 1; i <= l; i++)
  {
    cin >> a[i];
    gg[a[i]].clear();
  }
  for (int i = 1; i <= n; i++)
  {
    for (int j = 1; j <= m; j++)
    {
      cin >> arr[i][j];
      gg[arr[i][j]].insert({i, j});
    }
  }
  ll ans = solve(1, 1, 1);
  if (!ans)
    cout << "T\n";
  else
    cout << "N\n";
}

int main()
{
  fastio();
  int t = 1;
  cin >> t;
  while (t--)
    fn();
  return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
3 3 3
1 2 3
1 2 3
4 5 6
7 8 9
```