**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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.

Note: since the input is large, you may need to optimize input/output for this problem.

For example, in C++, it is enough to use the following lines at the start of the main() function:

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

Input: The first line of the input contains $t$ ($1 \le t \le 1500$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 1500$) – the size of the array and the sizes of the matrix.

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

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le n \cdot m$) – 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 $3 \cdot 10^6$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $1500$.

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 6
4 6 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 6
7 8
8 8
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: $6$ 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:
#pragma GCC optimize("O3")
#include <bits/stdc++.h>
#pragma GCC target("avx,avx2,sse,sse4,popcnt")
#if __has_include(<tr2/dynamic_bitset>)
#include <tr2/dynamic_bitset>
#endif
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;

using ll = long long;
using ull = unsigned long long;
using ldbl = long double;
//using lll = __int128_t;
template<typename T> using two = pair<T, T>;
#define all(X...) (X).begin(), (X).end()
#define all_r(X...) (X).rbegin(), (X).rend()
using _ntests_t = unsigned;
#define ntests _ntests_t _nt; cin >> _nt; [[maybe_unused]] NEXT_TEST: while (_nt --)
#define next_test goto NEXT_TEST
#define line "\n"

constexpr int mod  [[maybe_unused]] = 998244353;
constexpr int mod1 [[maybe_unused]] = 1000000007;
#define mod mod1
template<class T, class Cmp> using pq_cmp = priority_queue<T, vector<T>, Cmp>;
template<typename T> T sq(T x) { return x * x; }
template<typename T> T id(T x) { return x; }
template<typename T> void upd(T &x, T y) { x = max(x, y); }
template<typename T> void updm(T &x, T y) { x = min(x, y); }

vector<int> a;
vector<int> last;
array<array<int, 1501>, 1501> dp0, dp1, b;

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

	ntests {
		int l, n, m;
		cin >> l >> n >> m;
		a.resize(n);
		last.assign(n * m + 10, -1);
		for (auto &i : a) cin >> i;
		for (int i = 0; i < n; i ++) {
			if (last[a[i]] == -1) {
				last[a[i]] = i;
			} else {
				l = i;
				break;
			}
		}
		for (int i = 0; i < n; i ++) {
			for (int j = 0; j < m; j ++) {
				cin >> b[i][j];
			}
		}
		for (int i = 0; i <= n; i ++) {
			for (int j = 0; j <= m; j ++) {
				dp0[i][j] = dp1[i][j] = INT_MAX / 2;
			}
		}
		for (int i = n - 1; i >= 0; i --) {
			for (int j = m - 1; j >= 0; j --) {
				dp0[i][j] = min(dp0[i + 1][j], dp0[i][j + 1]);
				dp1[i][j] = min(dp1[i + 1][j], dp1[i][j + 1]);
				if (last[b[i][j]] == -1) {
					continue;
				}
				if (last[b[i][j]] % 2 == 1 && last[b[i][j]] + 3 <= dp1[i + 1][j + 1]) {
					updm(dp0[i][j], last[b[i][j]]);
				}
				if (last[b[i][j]] % 2 == 0 && last[b[i][j]] + 3 <= dp0[i + 1][j + 1]) {
					updm(dp1[i][j], last[b[i][j]]);
				}
			}
		}
		if (dp1[0][0] == 0) {
			cout << "T\n";
		} else {
			cout << "N\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 3
4 2
2 2 2
1 2
1 1
2 2
2 2 2
2 1
3 1
4 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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.

Note: since the input is large, you may need to optimize input/output for this problem.

For example, in C++, it is enough to use the following lines at the start of the main() function:

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

Input: The first line of the input contains $t$ ($1 \le t \le 1500$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 1500$) – the size of the array and the sizes of the matrix.

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

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le n \cdot m$) – 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 $3 \cdot 10^6$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $1500$.

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 6
4 6 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 6
7 8
8 8
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: $6$ 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:
/**
 *    author:  Shubham Kr. Verma
 *    created: 15.09.2024 15:23:10
**/
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
// const int N=1e9+7;
#define endl '\n'
#define ff first
#define ss second
#define pb push_back
// #define int long long
#define ll long long
#define INF 1e18+7
#define vi vector<int>
#define pi pair<int,int>
#define sz(v) v.size()
#define all(v)begin(v),end(v)
#define cint(y...) int y;IN(y);
#define vvi vector<vector<int>>
#define cstr(y...) string y;IN(y);
#define setBits(x) __builtin_popcountll(x)
#define print(v)for(auto x:v)cout<<x<<' ';cout<<endl;
#define cvect(v,n)vector<int>v(n);for(auto&x:v)cin>>x;
#define minpq priority_queue<int, vector<int>, greater<int>>
#define mrg(v1,v2,v3) merge(all(v1),all(v2),back_inserter(v3));
#define cvect1(v,n)vector<int>v(n+1);for(int i=1;i<=n;i++)cin>>v[i];
#define getunique(v) {sort(all(v));v.erase(unique(all(v)),v.end());}
void IN(auto&x){cin>>x;}void IN(auto &x,auto&... y){IN(x);IN(y...);}
template<class T> using oset = tree<T, null_type, less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template<class T> using omultiset = tree<T, null_type, less_equal<T>,rb_tree_tag,tree_order_statistics_node_update>;

#ifndef ONLINE_JUDGE
#define dbg true
#include<bits/debugger.h>
#else
#define dbg false
#define debug(x...)
#define debugptr(x,y)
#endif

const int L =1502;

void solve(){
    cint(l,n,m)
    cvect1(v,l)
    vvi mat(n+1,vi(m+1));
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            cin>>mat[i][j];
        }
    }

    map<int,bitset<L>>v1;
    for(int i=1;i<=l;i++){
        int x = v[i];
        v1[x] |= (1<<i);
    }

    vector<bitset<L>> next(m+1), curr(m+1);
    for(int i=n;i>=1;i--){
        for(int j=m;j>=1;j--){
            if(i!=n){
                curr[j] = next[j];
            }
            if(j!=m){
                curr[j] |= curr[j+1];
            }

            bitset<L> bit = v1[mat[i][j]];
            if(i==n || j == m){
                curr[j] |= bit;
            }else{
                bitset<L> x = next[j+1] >> 1;
                curr[j] |= (bit & (~x));
                if(bit[l] == 1){
                    curr[j][l] = 1;
                }else{
                    curr[j][l] = 0;
                }
            }

            if(curr[j][1] == 1){
                cout << "T" << endl;
                return;
            }
        }
        swap(next,curr);
        curr.assign(m+1,0);
    }
    cout << "N" << endl;
}

signed main(){
    ios_base::sync_with_stdio(false);
    // cout<<setprecision(10)<<fixed;
    // #ifndef ONLINE_JUDGE
    //     freopen("input.txt","r",stdin);
    //     freopen("output.txt","w",stdout);
    // #endif
    cin.tie(NULL);
    int t=1,T=0;
    cin>>t;
    while(t--){
        cerr<<endl<<"[TESTCASE : "<<++T <<"]"<<endl;
        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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases or scenarios that are likely to cause the code to fail. Here are some considerations:

1. **Boundary Conditions**: Test with the maximum and minimum values for \( l \), \( n \), and \( m \).
2. **Special Cases**: Test with specific patterns in the array \( a \) and matrix \( b \) that might cause issues in the bitwise operations or the logic of the game.
3. **Empty Submatrix**: Test cases where the submatrix becomes empty, which can cause issues in the bitwise operations.

Let's construct a test case that focuses on these aspects:

- **Test Case**: \( l = 2 \), \( n = 2 \), \( m = 2 \)
- **Array \( a \)**: [1, 2]
- **Matrix \( b \)**:
  - Row 1: [1, 3]
  - Row 2: [4, 2]

This test case is designed to:
- Ensure that the first element of \( a \) (1) is found at the top-left corner of the matrix.
- Ensure that the second element of \( a \) (2) is found at the bottom-right corner of the matrix.
- Create a scenario where the submatrix for the second player is empty, which should cause the second player to lose.

Here is the test input:

```plaintext
1
2 2 2
1 2
1 3
4 2
```

This test case should expose the bug in the bitwise operations and the logic of the game, particularly in handling the empty submatrix scenario.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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.

Note: since the input is large, you may need to optimize input/output for this problem.

For example, in C++, it is enough to use the following lines at the start of the main() function:

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

Input: The first line of the input contains $t$ ($1 \le t \le 1500$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 1500$) – the size of the array and the sizes of the matrix.

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

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le n \cdot m$) – 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 $3 \cdot 10^6$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $1500$.

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 6
4 6 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 6
7 8
8 8
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: $6$ 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.*;
import java.util.*;

public class D {

	static IOHandler sc = new IOHandler();
	static PrintWriter out = new PrintWriter(System.out);

	public static void main(String[] args) throws IOException {

        new Thread(null, new Solution(), "solution", 1 << 26 ).start();
    }

	private static class Solution implements Runnable {

		public void run() {
			// TODO Auto-generated method stub
			int testCases = sc.nextInt();

			for (int i = 1; i <= testCases; ++i) {
				solve(i);
			}

			out.flush();
		}

		private void advice() {
			// Read question and Test cases thoroughly
			// Check for overflow , Avoid Integer ==
			// Handle invalid cases
			// Think before solving
			// Search for concepts
			// Double check Constraints
			// expected alg like seg tree, trie, binSearch, sqrt decom ...

			// Sample Intuitions
			// Hashing, Inout Red, binary partition

			// Source of Error
			// Int overflow, Not returning after special case
			// Not taking mod properly
			// sorting input groups after reading certain values
			// Not using the right vairable eg idx in method for i in loop
			// Accounting for special case in general logic. eg special in dp
			// for BIT. Do not use tree maps with update query
			// Assuming Constraints on an unbound variable
			// Not properly calculating time complexity
		}

		private void solve(int t) {
			int l = sc.nextInt();

			int n = sc.nextInt();
			int m = sc.nextInt();

			int [] vals = sc.readArray(l);
			int val;

			int [][] arr = new int [n][];

			for (int i = 0; i < n; ++i) {
				arr[i] = sc.readArray(m);
			}

			Map<Integer, List<int []>> map = new HashMap<>();

			List<int []> list;

			for (int i = n - 1; i >= 0; --i) {
				for (int j = m - 1; j >= 0; --j) {
					if (!map.containsKey(arr[i][j]))
						map.put(arr[i][j], new ArrayList<>());

					list = map.get(arr[i][j]);

					if (!list.isEmpty() && list.get(list.size() - 1)[1] >= j)
						continue;

					list.add(new int [] {i, j});
				}
			}

			List<int []> def = new ArrayList<>();

			List<int []> prev = new ArrayList<>();
			List<int []> cur = new ArrayList<>();

			int [] cArr;

			for (int a = l - 1; a >= 0; --a) {
				val = vals[a];
				cur = new ArrayList<>();

				int minIdx = -1;

				list = map.getOrDefault(val, def);
				int pIdx = 0;

				for (int i = 0; i < list.size(); ++i) {

					// System.out.println(Arrays.toString(list.get(i)));

					while (pIdx < prev.size() && prev.get(pIdx)[0] > list.get(i)[0]) {

						cArr = prev.get(pIdx++);
						minIdx = Math.max(cArr[1], minIdx);
					}

					if (list.get(i)[1] >= minIdx) {
						cur.add(list.get(i));
					}

					// System.out.println(a + " " + i + " " + minIdx);
				}

				prev = cur;
			}

			out.println(!cur.isEmpty() ? "T" : "N");
		}

	}

	private static class IOHandler {
      BufferedReader br;
      StringTokenizer st;

      public IOHandler() {
         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());
      }

      int [] readArray(int n) {
    	  int [] res = new int [n];

    	  for (int i = 0; i < n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [] readArray2(int n) {
    	  int [] res = new int [n + 1];

    	  for (int i = 1; i <= n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [][] readGraph(int n, int m, int c) {
    	  int [][] graph = new int [n + 1][];

    	  int [] count = new int [n + 1];
    	  int [][] arr = new int [m][1 + c];

    	  int a, b;

    	  for (int i = 0; i < m; ++i) {
    		  a = sc.nextInt();
    		  b = sc.nextInt();

    		  arr[i][0] = a;
    		  arr[i][1] = b;

    		  ++count[a];
    		  ++count[b];

    		  for (int j = 2; j <= c; ++j) {
    			  arr[i][j] = sc.nextInt();
    		  }
    	  }

    	  for (int i = 0; i <= n; ++i) {
    		  graph[i] = new int [count[i] * c];
    		  count[i] = 0;
    	  }
    	  int swap;

    	  for (int [] tArr : arr) {
    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }

    		  swap = tArr[0];
    		  tArr[0] = tArr[1];
    		  tArr[1] = swap;

    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }
    	  }

    	  return graph;
      }

      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;
      }

   }

}

**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, let's consider the following aspects:

1. **Constraints**: The input should respect the constraints \(1 \le l, n, m \le 1500\).
2. **Edge Cases**: We should consider edge cases where the matrix and array configurations can lead to unexpected behavior.
3. **Optimal Play**: Both players play optimally, so the test case should challenge the logic that determines the winner.

### Step-by-Step Reasoning:

1. **Small-Scale Test Case**:
   - Use small values for \(l\), \(n\), and \(m\) to ensure the test case is manageable and easy to verify.
   - Choose values that can expose potential issues in the logic, such as:
     - \(l = 2\): This is the smallest non-trivial length for the array.
     - \(n = 2\): This is the smallest non-trivial size for the matrix.
     - \(m = 2\): This is the smallest non-trivial size for the matrix.

2. **Matrix and Array Configuration**:
   - Create a matrix and array that can lead to a situation where the logic might fail.
   - For example, place the elements in the matrix in such a way that the submatrix constraints are challenging to satisfy.

### Challenging Test Case:

```plaintext
1
2 2 2
1 2
1 2
3 4
```

### Explanation:

- **Test Case**:
  - \(l = 2\)
  - \(n = 2\)
  - \(m = 2\)
  - Array \(a = [1, 2]\)
  - Matrix \(b\):
    - \[
    \begin{bmatrix}
    1 & 2 \\
    3 & 4 \\
    \end{bmatrix}
    \]

- **Game Play**:
  - Tsovak starts by looking for \(1\). The only occurrence of \(1\) is at \((1,1)\).
  - Narek then needs to look for \(2\) in the submatrix starting at \((2,2)\), which is just the element \(4\). Since \(4\) is not \(2\), Narek cannot find \(2\), and Tsovak wins.

This test case is designed to be small and simple, yet it can expose the logic of the game where the submatrix constraints are crucial. The correct output for this test case should be "T" (Tsovak wins).


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Subtangle Game (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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.

Note: since the input is large, you may need to optimize input/output for this problem.

For example, in C++, it is enough to use the following lines at the start of the main() function:

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

Input: The first line of the input contains $t$ ($1 \le t \le 1500$) – the number of test cases.

The first line of each test case contains three integers $l$, $n$, and $m$ ($1 \le l, n, m \le 1500$) – the size of the array and the sizes of the matrix.

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

The $i$-th of the last $n$ lines contains $m$ integers $b_{i,1}, b_{i,2}, b_{i,3}, \ldots b_{i,m}$ ($1 \le b_{i,j} \le n \cdot m$) – 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 $3 \cdot 10^6$.

It is guaranteed that the sum of $l$ over all test cases does not exceed $1500$.

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 6
4 6 2
2 2 4
1 2
1 1 3 2
4 2 5 1
2 4 2
1 2
3 4
5 6
7 8
8 8
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: $6$ 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:
#I know exactly where you are!!
import math
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)]

    ind = [0] * (n * m + 1)  # Create the ind array with size large enough

    # Process array a and populate ind[]
    for i in range(1, l + 1):
        if ind[a[i - 1]]:  # If a[i] has already appeared
            l = i - 1
            break
        ind[a[i - 1]] = i

    # Initialize dp arrays with a large value
    dp0 = [[int(1e9)] * (m + 2) for _ in range(n + 2)]
    dp1 = [[int(1e9)] * (m + 2) for _ in range(n + 2)]

    # Fill dp arrays using dynamic programming
    for i in range(n - 1, -1, -1):
        for j in range(m - 1, -1, -1):
            dp0[i][j] = min(dp0[i + 1][j], dp0[i][j + 1])
            dp1[i][j] = min(dp1[i + 1][j], dp1[i][j + 1])

            if not ind[b[i][j]]:
                continue

            if ind[b[i][j]] % 2 == 0 and ind[b[i][j]] + 3 <= dp1[i + 1][j + 1]:
                dp0[i][j] = min(dp0[i][j], ind[b[i][j]])

            if ind[b[i][j]] % 2 == 1 and ind[b[i][j]] + 3 <= dp0[i + 1][j + 1]:
                dp1[i][j] = min(dp1[i][j], ind[b[i][j]])

    if dp1[0][0] == 1:
        print("T")
    else:
        print("N")

    # Reset ind[] array for the next test case
    for i in range(1, l + 1):
        ind[a[i - 1]] = 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 2 2
1 2 3
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: Cube
time_limit_ms: 3000
memory_limit_mb: 1024
Description: You are given an $$$n\times n\times n$$$ big three-dimensional cube that contains $$$n^3$$$ numbers. You have to choose $$$n$$$ of those numbers so that their sum is as small as possible. It is, however, forbidden to choose two numbers that are located in the same plane. That is, if we identify the positions in the cube by three Cartesian coordinates, then choosing two numbers from positions $$$(x,y,z)$$$ and $$$(x',y',z')$$$ is forbidden if $$$x=x'$$$, $$$y=y'$$$, or $$$z=z'$$$.

Input: The input consists of the number $$$n$$$ followed by $$$n^3$$$ numbers in the cube. The numbers are presented as $$$n$$$ two-dimensional matrices, one for each layer of the cube. More precisely, there will be $$$n^2$$$ lines follow, each having $$$n$$$ numbers. For each $$$x, y, z$$$ ($$$1\le x, y, z\le n$$$), the number at the position $$$(x, y, z)$$$ is listed as the $$$z$$$-th number in the $$$((x-1)\times n+y)$$$-th line.

- $$$2 \leq n \leq 12$$$
- All numbers in the cube are integers between $$$0$$$ and $$$2\times 10^7$$$.

Output: The output consists of a single number. It is the minimum sum of $$$n$$$ numbers chosen from the cube according to the above rules.

Examples:
input:
3
1 2 3
4 5 6
7 8 9
1 1 1
2 2 2
3 3 3
4 3 0
2 1 4
9 8 9
output:
5

Note:

Buggy Code:
//
// Created by Tang Yi on 12/9/2024.
//
// 使用动态规划和二进制（位运算）法
#include <iostream>
#include <vector>
#include <climits>
using namespace std;

int main() {
    int n;
    cin >> n;

    // 读入三维数组
    vector<vector<vector<int>>> cube(n, vector<vector<int>>(n, vector<int>(n)));
    for (int x = 0; x < n; ++x) {
        for (int y = 0; y < n; ++y) {
            for (int z = 0; z < n; ++z) {
                cin >> cube[x][y][z];
            }
        }
    }

    // 使用 long long 来避免溢出
    vector<long long> dp(1 << n, LLONG_MAX);  // 使用 long long 类型，初始化为最大值
    dp[0] = 0;  // 初始状态没有选择任何数，和为0

    // 遍历每一层
    for (int layer = 0; layer < n; ++layer) {
        vector<long long> next_dp(1 << n, LLONG_MAX);  // 保存当前层的 dp 状态

        // 遍历当前选择的数的状态
        for (int mask = 0; mask < (1 << n); ++mask) {
            // 如果 mask 状态已经选择了 n 个数，则继续
            if (__builtin_popcount(mask) >= n) continue;

            // 更新下一步状态，遍历每个点
            for (int i = 0; i < n; ++i) {
                if ((mask >> i) & 1) continue;  // i 已经被选择过了

                // 遍历 cube[layer][i] 中的每个元素
                for (int j = 0; j < n; ++j) {
                    // 新状态 mask 选择了 i
                    int new_mask = mask | (1 << i);
                    next_dp[new_mask] = min(next_dp[new_mask], dp[mask] + cube[layer][i][j]);
                }
            }
        }

        // 更新 dp
        dp = next_dp;
    }

    // 结果是 dp 中已经选择了 n 个数的最小值
    long long result = LLONG_MAX;
    for (int mask = 0; mask < (1 << n); ++mask) {
        if (__builtin_popcount(mask) == n) {
            result = min(result, dp[mask]);
        }
    }

    cout << result << endl;
    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
2
1 2
3 4
5 6
7 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cube
time_limit_ms: 3000
memory_limit_mb: 1024
Description: You are given an $$$n\times n\times n$$$ big three-dimensional cube that contains $$$n^3$$$ numbers. You have to choose $$$n$$$ of those numbers so that their sum is as small as possible. It is, however, forbidden to choose two numbers that are located in the same plane. That is, if we identify the positions in the cube by three Cartesian coordinates, then choosing two numbers from positions $$$(x,y,z)$$$ and $$$(x',y',z')$$$ is forbidden if $$$x=x'$$$, $$$y=y'$$$, or $$$z=z'$$$.

Input: The input consists of the number $$$n$$$ followed by $$$n^3$$$ numbers in the cube. The numbers are presented as $$$n$$$ two-dimensional matrices, one for each layer of the cube. More precisely, there will be $$$n^2$$$ lines follow, each having $$$n$$$ numbers. For each $$$x, y, z$$$ ($$$1\le x, y, z\le n$$$), the number at the position $$$(x, y, z)$$$ is listed as the $$$z$$$-th number in the $$$((x-1)\times n+y)$$$-th line.

- $$$2 \leq n \leq 12$$$
- All numbers in the cube are integers between $$$0$$$ and $$$2\times 10^7$$$.

Output: The output consists of a single number. It is the minimum sum of $$$n$$$ numbers chosen from the cube according to the above rules.

Examples:
input:
3
1 2 3
4 5 6
7 8 9
1 1 1
2 2 2
3 3 3
4 3 0
2 1 4
9 8 9
output:
5

Note:

Buggy Code:
#include<iostream>
#include<algorithm>
#include<map>
#include<cmath>
#include<cstring>
#include<climits>
#include<cstdio>
#include<string>
#include<numeric>
#include<set>
#include<vector>
#include<queue>
#include<unordered_map>
#include<unordered_set>
#include<iomanip>
#include<array>
#include<functional>
using namespace std;
#define LL long long
#define uLL unsigned long long
#define LD long double
//#define LL int
#define INF 0x3f3f3f3f
#define IOS ios::sync_with_stdio(0);std::cin.tie(0);std::cout.tie(0);
const double PI = 3.1415926535897932384626433832795;
const LL N = 2e5 + 10;
const LL mod = 998244353;
void YES() { cout << "YES\n"; }
void NO() { cout << "NO\n"; }
LL gcd(LL x, LL y) { return y ? gcd(y, x % y) : x; }
LL lcm(LL x, LL y) { return x / gcd(x, y) * y; }
LL lowbit(LL x) { return x & (-x); }
struct node {
    LL x, y;
    bool operator<(const node& a)const
    {
        if (x != a.x) return x < a.x;
        return y < a.y;
    }
};
LL n, m, k, T, ans = 0;
bool cmp(LL x, LL y) { return x > y; }
LL quickpow(LL base, LL up) {
    if (base == 0 && up == 0)
        return 0;
    LL num = 1;
    while (up > 0)
    {
        if (up & 1)
            num = num * base;
        base = base * base;
        up >>= 1;
    }
    return num;
}

int popcount(unsigned int x) {
    int count = 0;
    while (x) {
        count += x & 1;
        x >>= 1;
    }
    return count;
}

const int MAX_N = 12;  // 最大n = 12
int cube[MAX_N][MAX_N][MAX_N];  // 三维立方体数组

int dp[1 << MAX_N][1 << MAX_N];  // 动态规划状态压缩

void init_dp(int n) {
    // 初始化 dp 数组为无穷大，表示没有解的状态
    for (int i = 0; i < (1 << n); ++i) {
        for (int j = 0; j < (1 << n); ++j) {
            dp[i][j] = INT_MAX;  // 初始值设为无穷大
        }
    }
}

int solve(int n) {
    init_dp(n);

    // 初始状态，表示没有选择任何行、列
    dp[0][0] = 0;

    // 动态规划：逐步选择行、列
    for (int z = 0; z < n; z++)
    {
        for (int mask_x = 0; mask_x < (1 << n); ++mask_x) {
            for (int mask_y = 0; mask_y < (1 << n); ++mask_y) {
                if (dp[mask_x][mask_y] == INT_MAX) {
                    continue;
                }

                int current_sum = dp[mask_x][mask_y];  // 当前最小和

                // 遍历所有可能的行、列组合，尝试选择新的元素

                for (int x = 0; x < n; ++x) {
                    if ((mask_x & (1 << x)) == 0) {  // 如果行 x 没有被选择
                        for (int y = 0; y < n; ++y) {
                            if ((mask_y & (1 << y)) == 0) {  // 如果列 y 没有被选择
                                int new_sum = current_sum + cube[x][y][z];  // 选定 x 行和 y 列
                                dp[mask_x | (1 << x)][mask_y | (1 << y)] =
                                    min(dp[mask_x | (1 << x)][mask_y | (1 << y)], new_sum);
                            }
                        }
                    }
                }

            }
        }
    }

    int result = INT_MAX;
    for (int mask_x = 0; mask_x < (1 << n); ++mask_x) {
        for (int mask_y = 0; mask_y < (1 << n); ++mask_y) {
            if (popcount(mask_x) == n && popcount(mask_y) == n) {
                result = min(result, dp[mask_x][mask_y]);
            }
        }
    }

    return result;
}

int main()
{
    IOS;
    T = 1;
    //cin >> T;
    while (T--)
    {
        cin >> n;

        // 输入立方体的数据
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                for (int k = 0; k < n; ++k) {
                    cin >> cube[i][j][k];
                }
            }
        }

        // 求解并输出结果
        cout << solve(n) << endl;
    }
    return 0;
}

/* ------------我是模板-------------- */
/*
//  状压dp
//  互不侵犯
LL cnt = 0, sit[2010], gs[2010];
LL f[20][1010][110] = { 0 };
void dfs(LL he, LL sum, LL pos)
{
    if (pos >= n)
    {
        cnt++;
        sit[cnt] = he;
        gs[cnt] = sum;
        return;
    }
    dfs(he, sum, pos + 1);//不用第pos个
    dfs(he + (1 << pos), sum + 1, pos + 2);//用第pos个
}
int main()
{
    IOS;
    T = 1;
    while (T--)
    {
        cin >> n >> m;
        dfs(0, 0, 0);
        for (LL i = 1; i <= cnt; i++)
            f[1][i][gs[i]] = 1;
        for (LL i = 2; i <= n; i++)
            for (LL j = 1; j <= cnt; j++)// 第i行
                for (LL k = 1; k <= cnt; k++)// 第i-1行
                {
                    if (sit[j] & sit[k])
                        continue;
                    if ((sit[j] << 1) & sit[k])
                        continue;
                    if (sit[j] & (sit[k] << 1))
                        continue;
                    for (LL l = m; l >= gs[j]; l--)
                    {
                        f[i][j][l] += f[i - 1][k][l - gs[j]];
                    }
                }
        ans = 0;
        for (LL i = 1; i <= cnt; i++)
            ans += f[n][i][m];
        cout << ans << "\n";
    }
    return 0;
}
//  炮兵阵地
LL mp[110][15];
LL sta[210], f[110][210][210];
LL ck(LL a)
{
    LL M = 0;
    for (LL i = 0; i < a; i++)
        if ((i & (i >> 2)) == 0 && (i & (i << 2)) == 0 && (i & (i >> 1)) == 0 && (i & (i << 1)) == 0)
            sta[++M] = i;
    return M;
}
LL ckk(LL i, LL x)
{
    LL sum = 0;
    for (LL j = m; j >= 1; j--)
    {
        if (x & 1)
            sum += mp[i][j];
        x >>= 1;
    }
    return sum;
}
int main()
{
    //IOS;
    T = 1;
    //cin >> T;
    while (T--)
    {
        cin >> n >> m;
        LL M = ck(1 << m);
        for (LL i = 1; i <= n; i++)
            for (LL j = 1; j <= m; j++)
            {
                char c;
                cin >> c;
                if (c == 'P')
                    mp[i][j] = 1;
                else
                    mp[i][j] = 0;
            }
        for (LL i = 1; i <= n; i++)
            for (LL j = 1; j <= M; j++)// i行
                for (LL k = 1; k <= M; k++)// i-1行
                {
                    if (i == 1)
                    {
                        f[i][j][k] = ckk(i, sta[j]);
                        ans = max(ans, f[i][j][k]);
                        continue;
                    }
                    if (sta[j] & sta[k])
                        continue;
                    LL tmp = 0;
                    for (LL p = 1; p <= M; p++)// i-2行
                    {
                        if (sta[p] & sta[k])
                            continue;
                        if (sta[p] & sta[j])
                            continue;
                        tmp = max(tmp, f[i - 1][k][p]);
                    }
                    f[i][j][k] = tmp + ckk(i, sta[j]);
                    ans = max(ans, f[i][j][k]);
                }
        cout << ans << "\n";
    }
    return 0;
}
*/
/*
// Corn Fields G
LL a[13][13], f[13][1 << 13];
LL F[1 << 13], g[1 << 13];
int main()
{
    IOS;
    T = 1;
    //cin >> T;
    while (T--)
    {
        cin >> n >> m;
        for (LL i = 1; i <= n; i++)
            for (LL j = 1; j <= m; j++)
                cin >> a[i][j], F[i] = (F[i] << 1) + a[i][j];
        for (LL i = 0; i < (1 << m); i++)
            if (!(i & (i >> 1)) && !(i & (i << 1)))
            {
                g[i] = 1;
                if ((i & F[1]) == i)
                    f[1][i] = 1;
            }
        for (LL i = 2; i <= n; i++)
            for (LL j = 0; j < (1 << m); j++)
                if ((j & F[i - 1]) == j && g[j])
                    for (LL k = 0; k < (1 << m); k++)
                        if ((k & F[i]) == k && !(j & k) && g[k])
                            f[i][k] = (f[i][k] + f[i - 1][j]) % mod;
        for (LL i = 0; i < (1 << m); i++)
            ans = (ans + f[n][i]) % mod;
        cout << ans << "\n";
    }
    return 0;
}
*/
/*
// 树形背包
vector<LL>g[N];
LL val[N], f[1010][1010], siz[N];
void dfs(LL x)
{
    f[x][1] = val[x];
    siz[x] = 1;
    for (LL i = 0; i < g[x].size(); i++)
    {
        LL y = g[x][i];
        dfs(y);
        for (LL j = siz[x]; j >= 1; j--)
            for (LL k = 0; k <= siz[y]; k++)
                f[x][j + k] = max(f[x][j + k], f[x][j] + f[y][k]);
        siz[x] += siz[y];
    }

}
// 数位统计
LL a[N], f[N], b[N];
LL ans1[N], ans2[N];
void slv(LL t, LL *Ans)
{
    LL tmp = t;
    LL len = 0;
    while (tmp)
    {
        b[++len] = tmp % 10;
        tmp /= 10;
    }
    for (LL i = len; i >= 1; i--)
    {
        for (LL j = 0; j < 10; j++)
            Ans[j] += f[i - 1] * b[i];
        for (LL j = 0; j < b[i]; j++)
            Ans[j] += a[i - 1];
        t -= a[i - 1] * b[i], Ans[b[i]] += t + 1;
        Ans[0] -= a[i - 1];

    }
}
int main()
{
    IOS;
    T = 1;
    //cin >> T;
    while (T--)
    {
        LL l, r;
        cin >> l >> r;
        a[0] = 1ll;
        for (LL i = 1; i <= 13; i++)
        {
            f[i] = f[i - 1] * 10 + a[i - 1];
            a[i] = 10ll * a[i - 1];
        }
        slv(r, ans1), slv(l - 1, ans2);
        for (LL i = 0; i < 10; i++)
            cout << ans1[i] - ans2[i] << " ";
    }
    return 0;
}
// 数位dp
LL f[1010][1010], num[8];
LL dfs(LL pos, LL last, bool lead, bool limit)
{
    LL Ans = 0;
    if (!pos)
        return 1;
    if (!lead && !limit && f[pos][last] != -1)
        return f[pos][last];
    LL up = (limit ? num[pos] : 9);
    for (LL i = 0; i <= up; i++)
    {
        if (i == 4 || (i == 2 && last == 6))
            continue;
        if (lead && i == 0)
            Ans += dfs(pos - 1, -2, true, limit && i == up);
        else
            Ans += dfs(pos - 1, i, false, limit && i == up);
    }
    if (!limit && !lead)
        f[pos][last] = Ans;
    return Ans;
}
LL slv(LL x)
{
    LL len = 0;
    while (x)
    {
        num[++len] = x % 10;
        x /= 10;
    }
    for (LL i = 0; i <= 20; i++)
        for (LL j = 0; j <= 20; j++)
            f[i][j] = -1;
    //cout << "??\n";
    return dfs(len, -2, true, true);
}
int main()
{
    IOS;
    T = 1;
    //cin >> T;
    while (1)
    {
        LL l, r;
        cin >> l >> r;
        if (!l && !r)
            break;
        cout << slv(r) - slv(l - 1) << "\n";
    }
    return 0;
}

LL f[11][65][1025], num[65], b;
LL dfs(LL pos, LL last, bool lead, bool limit)
{
    LL Ans = 0;
    if (!pos)
        return !last;
    if (!lead && !limit && f[b][pos][last] != -1)
        return f[b][pos][last];
    LL up = (limit ? num[pos] : b - 1);
    for (LL i = 0; i <= up; i++)
        Ans += dfs(pos - 1, (lead && !i) ? 0 : (last ^ (1 << i)), lead && !i, limit && (i == up));
    if (!limit && !lead)
        f[b][pos][last] = Ans;
    return Ans;
}
LL slv(LL x)
{
    LL len = 0;
    while (x)
    {
        num[++len] = x % b;
        x /= b;
    }
    //cout << "??\n";
    return dfs(len, 0, true, true);
}
int main()
{
    memset(f, -1, sizeof f);
    IOS;
    T = 1;
    cin >> T;
    while (T--)
    {
        LL l, r;
        cin >> b >> l >> r;
        cout << slv(r) - slv(l - 1) << "\n";
    }
    return 0;
}

map<LL, LL>mp;
LL find(LL x)
{
    if (x == mp[x])
        return x;
    else
        return mp[x] = find(mp[x]);
}
void merge(LL x, LL y)
{
    mp[find(x)] = find(y);
}
*/
// a.insert(upper_bound(a.begin(), a.end(), x), x); 插入的数均已排列
/*
struct nd
{
    LL v, id;
    friend bool operator < (const nd& a, const nd& b)
    {
        return a.v > b.v;
    }
}
priority_queue<nd>q;
*/
/*//区间dp
        memset(f, INF, sizeof f);
        for (LL len = 2; len <= n; len++)// 枚举区间长度
            for (LL l = 1; l + len - 1 <= n; l++)
            {
                LL r = l + len - 1;
                for (LL k = l; k < r; k++)// L ~ k  k + 1 ~ R  =>  k为断点
                    f[l][r] = min(f[l][r], f[l][k] + f[k + 1][r]);
            }
        cout << f[1][n] << "\n";
  //关灯问题
        for (LL len = 2; len <= n; len++)
            for (LL l = 1; l + len - 1 <= n; l++)
            {
                LL r = l + len - 1;
                f[l][r][0] = min(f[l + 1][r][0] + (a[l + 1].x - a[l].x) * (a[l].v + a[n].v - a[r].v),
                    f[l + 1][r][1] + (a[r].x - a[l].x) * (a[l].v + a[n].v - a[r].v));

                f[l][r][1] = min(f[l][r - 1][0] + (a[r].x - a[l].x) * (a[l - 1].v + a[n].v - a[r - 1].v),
                    f[l][r - 1][1] + (a[r].x - a[r - 1].x) * (a[l - 1].v + a[n].v - a[r - 1].v));
            }
        cout << min(f[1][n][0], f[1][n][1]) << "\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:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios that might cause the algorithm to fail. The key points to focus on are:

1. **Small `n` values**: Since the code uses dynamic programming with state compression, small `n` values (e.g., `n = 2` or `n = 3`) can help ensure the algorithm's logic is correct without overwhelming the system.
2. **Uniform or near-uniform values**: If the values in the cube are too uniform, it might be easier for the algorithm to find a valid solution, potentially masking bugs.
3. **Large values in specific positions**: Placing large values in positions that should be chosen can help identify issues with the selection logic.
4. **Symmetry**: Symmetrical values can sometimes cause issues in algorithms that do not handle symmetry correctly.

Given these considerations, let's create a test input for `n = 3` with values that are designed to challenge the algorithm:

```plaintext
3
1 2 3
4 5 6
7 8 9
1 1 1
2 2 2
3 3 3
4 3 0
2 1 4
9 8 9
```

This input is designed to:
- Have a mix of small and large values.
- Include a zero value to test if the algorithm can correctly handle it.
- Have values that are not uniformly distributed to test the selection logic.

Let's break down the input:
- The first layer of the cube has values that increase linearly.
- The second layer has uniform values.
- The third layer has a mix of values, including a zero.

This should provide a good test case to identify any issues in the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cube
time_limit_ms: 3000
memory_limit_mb: 1024
Description: You are given an $$$n\times n\times n$$$ big three-dimensional cube that contains $$$n^3$$$ numbers. You have to choose $$$n$$$ of those numbers so that their sum is as small as possible. It is, however, forbidden to choose two numbers that are located in the same plane. That is, if we identify the positions in the cube by three Cartesian coordinates, then choosing two numbers from positions $$$(x,y,z)$$$ and $$$(x',y',z')$$$ is forbidden if $$$x=x'$$$, $$$y=y'$$$, or $$$z=z'$$$.

Input: The input consists of the number $$$n$$$ followed by $$$n^3$$$ numbers in the cube. The numbers are presented as $$$n$$$ two-dimensional matrices, one for each layer of the cube. More precisely, there will be $$$n^2$$$ lines follow, each having $$$n$$$ numbers. For each $$$x, y, z$$$ ($$$1\le x, y, z\le n$$$), the number at the position $$$(x, y, z)$$$ is listed as the $$$z$$$-th number in the $$$((x-1)\times n+y)$$$-th line.

- $$$2 \leq n \leq 12$$$
- All numbers in the cube are integers between $$$0$$$ and $$$2\times 10^7$$$.

Output: The output consists of a single number. It is the minimum sum of $$$n$$$ numbers chosen from the cube according to the above rules.

Examples:
input:
3
1 2 3
4 5 6
7 8 9
1 1 1
2 2 2
3 3 3
4 3 0
2 1 4
9 8 9
output:
5

Note:

Buggy Code:
#include<iostream>
#include<map>
#include<utility>
#include<unordered_map>
#include<time.h>
using namespace std;
typedef pair<int, int> PII;
#define swap(_a, _b) { \
auto __c = _a; \
_a = _b; \
_b = __c; \
}

int space[250][10000];
int flagx[10000], flagy[10000], flagz[10000];
int x[10000], y[10000];
int ans = 0x3f3f3f3f;
int n;
unordered_map<int, int> mp;
unordered_map<int, int> fff;
int tt = 0;

void dfs(int z = 1, int now = 0, int s = 0) {
	if(now >= ans) return;
	if(fff[s] && now >= mp[s]) return;
	if(z == n + 1) {
		ans = min(ans, now);
		return;
	}
	if(tt > 1900000) return;
	tt++;
	fff[s] = 1;
	mp[s] = now;
	for(int i = 1; i <= n * n; i++) {
		if(flagx[x[i]]) continue;
		if(flagy[y[i]]) continue;
		flagx[x[i]] = 1;
		flagy[y[i]] = 1;
		dfs(z + 1, now + space[z][i], s + (1 << x[i]) + (1 << (n + 1 + y[i])));
		flagx[x[i]] = 0;
		flagy[y[i]] = 0;
	}
	return;
}

int main() {
	cin >> n;
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= n * n; j++) {
			cin >> space[i][j];
		}
	}
	int p = 1;
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= n; j++) {
			x[p] = j;
			y[p] = i;
			p++;
		}
	}
	srand(time(0));
	for(int i = 1, I = 30; i <= I; i++) {
		int a = rand() % n + 1;
		int b = rand() % n + 1;
		for(int j = 1; j <= n * n; j++) {
			swap(space[a][j], space[b][j]);
		}
	}
	for(int i = 1, I = 30; i <= I; i++) {
		int a = rand() % n + 1;
		int b = rand() % n + 1;
		for(int j = 1; j <= n; j++) {
			swap(space[j][a], space[j][b]);
		}
	}
	dfs();
	cout << ans << endl;
	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:
To generate a challenging test input that exposes the bug in the buggy code, we need to consider edge cases and scenarios that can maximize the failure probability in the faulty solution. Here are the steps to think through this:

1. **Understand the Bug**: The buggy code uses random shuffling and depth-first search (DFS) to find the minimum sum. The main issue is that the random shuffling can lead to suboptimal solutions, especially for certain structured inputs.

2. **Edge Cases**:
   - **Small n**: For small values of \( n \), the number of possible combinations is manageable, but the random shuffling might not explore all combinations effectively.
   - **Structured Data**: Inputs with a specific pattern or structure can highlight the inefficiency of the random shuffling.

3. **Challenging Input**:
   - **Input with a Clear Minimum Sum**: Create an input where the minimum sum is clear but the random shuffling might not find it easily.
   - **Input with High Variance**: Create an input where the numbers are highly varied, making it difficult for the random shuffling to find the optimal solution.

Given these considerations, let's create a small-scale test input with \( n = 3 \) that is structured to expose the bug:

```plaintext
3
0 1 2
3 4 5
6 7 8
9 10 11
12 13 14
15 16 17
18 19 20
21 22 23
24 25 26
```

This input is structured such that the minimum sum can be achieved by selecting the smallest numbers from different planes. The numbers are arranged in a way that the optimal solution is clear but the random shuffling might not always find it.

Explanation:
- The first layer contains the smallest numbers.
- The second and third layers contain progressively larger numbers.

The optimal solution is to select the numbers \(0, 9, 18\) from different planes, which gives a minimum sum of \(27\).

This test input should be challenging for the buggy code because the random shuffling might not always explore the optimal solution effectively.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cube
time_limit_ms: 3000
memory_limit_mb: 1024
Description: You are given an $$$n\times n\times n$$$ big three-dimensional cube that contains $$$n^3$$$ numbers. You have to choose $$$n$$$ of those numbers so that their sum is as small as possible. It is, however, forbidden to choose two numbers that are located in the same plane. That is, if we identify the positions in the cube by three Cartesian coordinates, then choosing two numbers from positions $$$(x,y,z)$$$ and $$$(x',y',z')$$$ is forbidden if $$$x=x'$$$, $$$y=y'$$$, or $$$z=z'$$$.

Input: The input consists of the number $$$n$$$ followed by $$$n^3$$$ numbers in the cube. The numbers are presented as $$$n$$$ two-dimensional matrices, one for each layer of the cube. More precisely, there will be $$$n^2$$$ lines follow, each having $$$n$$$ numbers. For each $$$x, y, z$$$ ($$$1\le x, y, z\le n$$$), the number at the position $$$(x, y, z)$$$ is listed as the $$$z$$$-th number in the $$$((x-1)\times n+y)$$$-th line.

- $$$2 \leq n \leq 12$$$
- All numbers in the cube are integers between $$$0$$$ and $$$2\times 10^7$$$.

Output: The output consists of a single number. It is the minimum sum of $$$n$$$ numbers chosen from the cube according to the above rules.

Examples:
input:
3
1 2 3
4 5 6
7 8 9
1 1 1
2 2 2
3 3 3
4 3 0
2 1 4
9 8 9
output:
5

Note:

Buggy Code:
def min_sum_from_cube(n, numbers):
    # Creating the 3D cube
    cube = [[[0] * n for _ in range(n)] for _ in range(n)]
    index = 0
    for z in range(n):
        for x in range(n):
            for y in range(n):
                cube[x][y][z] = numbers[index]
                index += 1

    # DP initialization: use a large number for initial states
    inf = float('inf')
    dp = [[inf] * (1 << n) for _ in range(n)]  # dp[z][mask]

    # Base case: selecting from layer 0
    for x in range(n):
        for y in range(n):
            dp[0][1 << x] = min(dp[0][1 << x], cube[x][y][0])

    # Dynamic programming to fill in the dp array
    for z in range(1, n):
        for mask in range(1 << n):
            if dp[z - 1][mask] < inf:  # We can only update if previous state is valid
                for x in range(n):
                    if (mask & (1 << x)) == 0:  # x-plane not yet taken
                        for y in range(n):
                            new_mask = mask | (1 << x)
                            dp[z][new_mask] = min(dp[z][new_mask], dp[z - 1][mask] + cube[x][y][z])

    # Final result is minimum for the last layer with all planes taken
    result = inf
    final_mask = (1 << n) - 1
    for z in range(n):
        result = min(result, dp[z][final_mask])

    return result

# Main function to read input and call the calculation
def main():
    import sys
    input = sys.stdin.read
    data = list(map(int, input().split()))

    n = data[0]
    numbers = data[1:]

    result = min_sum_from_cube(n, numbers)
    print(result)

# This prevents the main code from running if imported
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:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that can stress the algorithm. Here are the steps to think through this:

1. **Edge Cases for `n`**: Since `n` can be as small as 2 or as large as 12, we should consider both ends of the spectrum. However, for a small-scale test, we can start with a small `n` like 4.

2. **Balanced Distribution of Numbers**: We need to ensure that the numbers are distributed in a way that can challenge the algorithm. For example, having numbers that are very close in value can make the selection process more difficult.

3. **Forcing Specific Constraints**: We should create a test case where the algorithm might incorrectly choose numbers from the same plane, which is forbidden.

4. **Minimizing the Sum**: The test case should be designed such that the correct minimum sum is not easily achieved, but the incorrect algorithm might still produce an incorrect result.

Given these considerations, let's construct a test input for `n = 4`:

```plaintext
4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
17 18 19 20
21 22 23 24
25 26 27 28
29 30 31 32
33 34 35 36
37 38 39 40
41 42 43 44
45 46 47 48
49 50 51 52
53 54 55 56
57 58 59 60
61 62 63 64
```

This test input is designed to be simple yet challenging:
- The numbers are in a strictly increasing order, which can make it difficult for the algorithm to correctly choose the minimum sum without violating the plane constraints.
- The structure of the numbers ensures that the correct minimum sum is not immediately obvious, forcing the algorithm to carefully consider the constraints.

Let's verify the correctness of this test input:
- The correct minimum sum should be the sum of the smallest numbers in each layer, ensuring no two numbers are in the same plane.

For `n = 4`, the correct minimum sum should be:
- Layer 0: 1
- Layer 1: 5
- Layer 2: 9
- Layer 3: 13

Sum = 1 + 5 + 9 + 13 = 28

This test input should help in exposing any bugs in the provided algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cube
time_limit_ms: 3000
memory_limit_mb: 1024
Description: You are given an $$$n\times n\times n$$$ big three-dimensional cube that contains $$$n^3$$$ numbers. You have to choose $$$n$$$ of those numbers so that their sum is as small as possible. It is, however, forbidden to choose two numbers that are located in the same plane. That is, if we identify the positions in the cube by three Cartesian coordinates, then choosing two numbers from positions $$$(x,y,z)$$$ and $$$(x',y',z')$$$ is forbidden if $$$x=x'$$$, $$$y=y'$$$, or $$$z=z'$$$.

Input: The input consists of the number $$$n$$$ followed by $$$n^3$$$ numbers in the cube. The numbers are presented as $$$n$$$ two-dimensional matrices, one for each layer of the cube. More precisely, there will be $$$n^2$$$ lines follow, each having $$$n$$$ numbers. For each $$$x, y, z$$$ ($$$1\le x, y, z\le n$$$), the number at the position $$$(x, y, z)$$$ is listed as the $$$z$$$-th number in the $$$((x-1)\times n+y)$$$-th line.

- $$$2 \leq n \leq 12$$$
- All numbers in the cube are integers between $$$0$$$ and $$$2\times 10^7$$$.

Output: The output consists of a single number. It is the minimum sum of $$$n$$$ numbers chosen from the cube according to the above rules.

Examples:
input:
3
1 2 3
4 5 6
7 8 9
1 1 1
2 2 2
3 3 3
4 3 0
2 1 4
9 8 9
output:
5

Note:

Buggy Code:
def solve_cube_min_sum(n, cube):
    """
    Solve minimum sum problem using optimized backtracking with aggressive pruning
    and sorted position lists
    """
    min_sum = float('inf')

    # Create list of all positions with their values
    positions = []
    for x in range(n):
        for y in range(n):
            for z in range(n):
                positions.append((cube[x][y][z], x, y, z))

    # Sort positions by value for better pruning
    positions.sort()  # Sort by value to try smallest values first

    def backtrack(depth, curr_sum, used_x, used_y, used_z, start_idx):
        nonlocal min_sum

        # Early pruning: if current sum + best possible remaining choices exceeds min_sum
        remaining = n - depth
        if remaining > 0:
            # If we can't beat current min_sum even with remaining smallest values
            if curr_sum + sum(positions[i][0] for i in range(start_idx, start_idx + remaining)) >= min_sum:
                return

        if depth == n:
            min_sum = curr_sum
            return

        # Try positions in sorted order
        for i in range(start_idx, len(positions)):
            val, x, y, z = positions[i]

            # Skip if coordinates already used
            if (used_x & (1 << x)) or (used_y & (1 << y)) or (used_z & (1 << z)):
                continue

            # Current sum + val already exceeds min_sum, no point checking further
            if curr_sum + val >= min_sum:
                break  # Since positions are sorted, all further values will be larger

            # Add current position and recurse
            backtrack(depth + 1,
                     curr_sum + val,
                     used_x | (1 << x),
                     used_y | (1 << y),
                     used_z | (1 << z),
                     i + 1)

    backtrack(0, 0, 0, 0, 0, 0)
    return min_sum

# Read input
n = int(input())
cube = [[[0 for _ in range(n)] for _ in range(n)] for _ in range(n)]

# Read the cube layer by layer
for x in range(n):
    for y in range(n):
        row = list(map(int, input().split()))
        for z in range(n):
            cube[x][y][z] = row[z]

# Solve and print result
print(solve_cube_min_sum(n, cube))

**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 potential bugs in the given algorithm, we need to consider edge cases and scenarios that can stress the algorithm. Here are the steps to think through this:

1. **Small Scale**: Since the problem specifies \(2 \leq n \leq 12\), we should use a smaller value of \(n\) to make the test input manageable. Let's choose \(n = 3\).

2. **Edge Cases**:
   - **High Variance in Values**: Use a mix of very small and very large values to test the algorithm's ability to find the minimum sum.
   - **Duplicate Values**: Include some duplicate values to test how the algorithm handles ties.
   - **Highly Constrained Positions**: Ensure that some of the smallest values are in positions that are highly constrained (i.e., they share the same plane with other small values).

3. **Construct the Test Input**:
   - We will place small values in positions that are highly constrained to see if the algorithm can correctly avoid choosing values from the same plane.
   - We will also include some large values to ensure the algorithm does not choose them if a smaller sum is possible.

Here is the test input:

```plaintext
3
1 10000000 2
10000000 1 10000000
10000000 10000000 1
1 10000000 10000000
10000000 1 10000000
10000000 10000000 1
10000000 1 10000000
1 10000000 10000000
10000000 10000000 1
```

### Explanation:
- The cube is designed such that the smallest values (1) are placed in positions that are highly constrained. For example, the first layer has a 1 at (1,1,1), and the second layer has a 1 at (1,2,1), and the third layer has a 1 at (1,3,1). These positions are all in the same x-plane.
- The algorithm must avoid choosing these positions together and instead find the correct combination of values from different planes to achieve the minimum sum.

This test input should effectively expose any issues with the algorithm's ability to handle constraints and find the optimal solution.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cube
time_limit_ms: 3000
memory_limit_mb: 1024
Description: You are given an $$$n\times n\times n$$$ big three-dimensional cube that contains $$$n^3$$$ numbers. You have to choose $$$n$$$ of those numbers so that their sum is as small as possible. It is, however, forbidden to choose two numbers that are located in the same plane. That is, if we identify the positions in the cube by three Cartesian coordinates, then choosing two numbers from positions $$$(x,y,z)$$$ and $$$(x',y',z')$$$ is forbidden if $$$x=x'$$$, $$$y=y'$$$, or $$$z=z'$$$.

Input: The input consists of the number $$$n$$$ followed by $$$n^3$$$ numbers in the cube. The numbers are presented as $$$n$$$ two-dimensional matrices, one for each layer of the cube. More precisely, there will be $$$n^2$$$ lines follow, each having $$$n$$$ numbers. For each $$$x, y, z$$$ ($$$1\le x, y, z\le n$$$), the number at the position $$$(x, y, z)$$$ is listed as the $$$z$$$-th number in the $$$((x-1)\times n+y)$$$-th line.

- $$$2 \leq n \leq 12$$$
- All numbers in the cube are integers between $$$0$$$ and $$$2\times 10^7$$$.

Output: The output consists of a single number. It is the minimum sum of $$$n$$$ numbers chosen from the cube according to the above rules.

Examples:
input:
3
1 2 3
4 5 6
7 8 9
1 1 1
2 2 2
3 3 3
4 3 0
2 1 4
9 8 9
output:
5

Note:

Buggy Code:
import sys

def solve():
    n = int(input())
    cube = [ [ list(map(int, input().split())) for _ in range(n)] for _ in range(n)]
    dp = [sys.maxsize] * (1 << n)
    dp[0] = 0
    for z in range(n):
        new_dp = dp[:]
        for mask in range(1 << n):
            for x in range(n):
                if mask & (1 << x) == 0:
                    for y in range(n):
                        new_dp[mask | (1 << x)] = min(new_dp[mask | (1 << x)], dp[mask] + cube[z][x][y])
        dp = new_dp
    print(dp[-1])
solve()

**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 that could cause the algorithm to fail. The bug in the code is likely related to how it handles the constraints of choosing numbers from different planes without violating the rules.

### Step-by-Step Reasoning:

1. **Understanding the Bug**:
   - The code uses a dynamic programming (DP) approach to find the minimum sum of \( n \) numbers.
   - It iterates through each layer of the cube and updates the DP table.
   - The bug might be in how it handles the constraints of not choosing numbers from the same plane.

2. **Edge Cases to Consider**:
   - **Small \( n \)**: Test with the smallest possible \( n \) (i.e., \( n = 2 \)) to see if the algorithm handles the simplest case correctly.
   - **Uniform Values**: Test with a cube where all values are the same to see if the algorithm can handle uniform input.
   - **Large Values**: Test with large values to see if the algorithm can handle the upper limit of the input range.
   - **Special Patterns**: Test with a cube that has a specific pattern to ensure the algorithm correctly avoids choosing numbers from the same plane.

3. **Challenging Input**:
   - For \( n = 3 \), create a cube where the values are such that the minimum sum path is not straightforward. This will help expose any issues in the DP logic.

### Chosen Test Input:
For \( n = 3 \), we will create a cube where the values are strategically placed to make the minimum sum path non-trivial. Specifically, we will place values in such a way that the algorithm must carefully choose the numbers to avoid the same plane constraint.

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

### Explanation:
- The first layer has values increasing from 1 to 9.
- The second layer has values decreasing from 9 to 1.
- The third layer has values increasing from 1 to 3 in each row.

This input will challenge the algorithm to correctly identify the minimum sum path while avoiding the same plane constraint. The correct output for this input should be 6, which is the sum of the smallest numbers from each layer while adhering to the constraints.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cube
time_limit_ms: 3000
memory_limit_mb: 1024
Description: You are given an $$$n\times n\times n$$$ big three-dimensional cube that contains $$$n^3$$$ numbers. You have to choose $$$n$$$ of those numbers so that their sum is as small as possible. It is, however, forbidden to choose two numbers that are located in the same plane. That is, if we identify the positions in the cube by three Cartesian coordinates, then choosing two numbers from positions $$$(x,y,z)$$$ and $$$(x',y',z')$$$ is forbidden if $$$x=x'$$$, $$$y=y'$$$, or $$$z=z'$$$.

Input: The input consists of the number $$$n$$$ followed by $$$n^3$$$ numbers in the cube. The numbers are presented as $$$n$$$ two-dimensional matrices, one for each layer of the cube. More precisely, there will be $$$n^2$$$ lines follow, each having $$$n$$$ numbers. For each $$$x, y, z$$$ ($$$1\le x, y, z\le n$$$), the number at the position $$$(x, y, z)$$$ is listed as the $$$z$$$-th number in the $$$((x-1)\times n+y)$$$-th line.

- $$$2 \leq n \leq 12$$$
- All numbers in the cube are integers between $$$0$$$ and $$$2\times 10^7$$$.

Output: The output consists of a single number. It is the minimum sum of $$$n$$$ numbers chosen from the cube according to the above rules.

Examples:
input:
3
1 2 3
4 5 6
7 8 9
1 1 1
2 2 2
3 3 3
4 3 0
2 1 4
9 8 9
output:
5

Note:

Buggy Code:
import java.util.Scanner;

public class Main{
    static final int MAXN = 12;
    static final int INF = (int) 1e9;
    static int[][][] cube = new int[MAXN][MAXN][MAXN];
    static int[][][] dp = new int[MAXN][1 << MAXN][1 << MAXN];

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();

        // Read the cube
        for (int x = 0; x < n; x++) {
            for (int y = 0; y < n; y++) {
                for (int z = 0; z < n; z++) {
                    cube[x][y][z] = sc.nextInt();
                }
            }
        }

        // Initialize dp table with -1
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < (1 << n); j++) {
                for (int k = 0; k < (1 << n); k++) {
                    dp[i][j][k] = -1;
                }
            }
        }

        // Start the recursive function
        System.out.println(solve(n, 0, 0, 0, 0));
    }

    private static int solve(int n, int depth, int usedX, int usedY, int usedZ) {
        if (depth == n) {
            return 0;
        }
        if (dp[depth][usedX][usedY] != -1) {
            return dp[depth][usedX][usedY];
        }

        int minSum = INF;
        for (int x = 0; x < n; x++) {
            for (int y = 0; y < n; y++) {
                for (int z = 0; z < n; z++) {
                    if (((usedX >> x) & 1) == 0 && ((usedY >> y) & 1) == 0 && ((usedZ >> z) & 1) == 0) {
                        int newUsedX = usedX | (1 << x);
                        int newUsedY = usedY | (1 << y);
                        int newUsedZ = usedZ | (1 << z);
                        minSum = Math.min(minSum, cube[x][y][z] + solve(n, depth + 1, newUsedX, newUsedY, newUsedZ));
                    }
                }
            }
        }

        dp[depth][usedX][usedY] = minSum;
        return minSum;
    }
}

// IhSzCDtZr5F49vEeNX6

**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 given code, we need to focus on edge cases or scenarios that maximize the failure probability. The bug in the code is related to the dynamic programming (DP) table initialization and the state transitions. Specifically, the DP table is not properly initialized, and the state transitions might not handle all cases correctly.

### Steps to Generate the Test Input:
1. **Small Scale**: Since the input should be small-scale, we can use \( n = 3 \).
2. **Edge Cases**: We should consider cases where the numbers in the cube are such that the algorithm might incorrectly choose numbers that violate the constraints (i.e., choosing numbers from the same plane).
3. **High Contrast**: Use numbers that have a high contrast in values to make it clear if the algorithm is making the wrong choices.

### Test Input:
We will create a 3x3x3 cube where the numbers are arranged in such a way that the correct solution is obvious but the buggy code might fail to find it.

```plaintext
3
0 1 2
3 4 5
6 7 8
9 10 11
12 13 14
15 16 17
18 19 20
21 22 23
24 25 26
```

### Explanation:
- The cube is filled with numbers from 0 to 26.
- The correct solution should be the sum of the smallest numbers in different planes, which are 0, 9, and 18.
- The sum of these numbers is \( 0 + 9 + 18 = 27 \).

This test input is designed to expose the bug in the DP table initialization and state transitions. If the code is correctly implemented, it should output 27. If it fails, it will likely output a higher number due to incorrect state transitions or initialization.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cube
time_limit_ms: 3000
memory_limit_mb: 1024
Description: You are given an $$$n\times n\times n$$$ big three-dimensional cube that contains $$$n^3$$$ numbers. You have to choose $$$n$$$ of those numbers so that their sum is as small as possible. It is, however, forbidden to choose two numbers that are located in the same plane. That is, if we identify the positions in the cube by three Cartesian coordinates, then choosing two numbers from positions $$$(x,y,z)$$$ and $$$(x',y',z')$$$ is forbidden if $$$x=x'$$$, $$$y=y'$$$, or $$$z=z'$$$.

Input: The input consists of the number $$$n$$$ followed by $$$n^3$$$ numbers in the cube. The numbers are presented as $$$n$$$ two-dimensional matrices, one for each layer of the cube. More precisely, there will be $$$n^2$$$ lines follow, each having $$$n$$$ numbers. For each $$$x, y, z$$$ ($$$1\le x, y, z\le n$$$), the number at the position $$$(x, y, z)$$$ is listed as the $$$z$$$-th number in the $$$((x-1)\times n+y)$$$-th line.

- $$$2 \leq n \leq 12$$$
- All numbers in the cube are integers between $$$0$$$ and $$$2\times 10^7$$$.

Output: The output consists of a single number. It is the minimum sum of $$$n$$$ numbers chosen from the cube according to the above rules.

Examples:
input:
3
1 2 3
4 5 6
7 8 9
1 1 1
2 2 2
3 3 3
4 3 0
2 1 4
9 8 9
output:
5

Note:

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

public class C_Cube {
    static class Position {
        int x, y, z;
        Position(int x, int y, int z) {
            this.x = x;
            this.y = y;
            this.z = z;
        }
        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Position position = (Position) o;
            return x == position.x && y == position.y && z == position.z;
        }
        @Override
        public int hashCode() {
            return Objects.hash(x, y, z);
        }
    }
    static class Cell {
        Position pos;
        long value;
        Cell(Position pos, long value) {
            this.pos = pos;
            this.value = value;
        }
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long[][][] cube = new long[n][n][n];
        List<Cell> allCells = new ArrayList<>();
        for (int x = 0; x < n; x++) {
            for (int y = 0; y < n; y++) {
                for (int z = 0; z < n; z++) {
                    cube[x][y][z] = sc.nextLong();
                    allCells.add(new Cell(new Position(x, y, z), cube[x][y][z]));
                }
            }
        }
        Collections.sort(allCells, (a, b) -> Long.compare(a.value, b.value));
        long result = solve(n, allCells);
        System.out.println(result);
    }
    private static long solve(int n, List<Cell> sortedCells) {
        long minSum = Long.MAX_VALUE;
        int totalCells = sortedCells.size();
        for (int i = 0; i < Math.min(totalCells, n * n); i++) {
            Cell firstCell = sortedCells.get(i);
            List<Cell> validCells = new ArrayList<>();
            for (Cell cell : sortedCells) {
                if (!sharesPlane(firstCell.pos, cell.pos)) {
                    validCells.add(cell);
                }
            }
            if (validCells.size() >= n - 1) {
                long currentSum = firstCell.value;
                Set<Position> usedPositions = new HashSet<>();
                usedPositions.add(firstCell.pos);
                for (Cell cell : validCells) {
                    if (usedPositions.size() == n) break;

                    boolean canUse = true;
                    for (Position pos : usedPositions) {
                        if (sharesPlane(cell.pos, pos)) {
                            canUse = false;
                            break;
                        }
                    }
                    if (canUse) {
                        currentSum += cell.value;
                        usedPositions.add(cell.pos);
                    }
                }
                if (usedPositions.size() == n) {
                    minSum = Math.min(minSum, currentSum);
                }
            }
        }
        return minSum;
    }

    private static boolean sharesPlane(Position p1, Position p2) {
        if (p1.equals(p2)) return true;
        return p1.x == p2.x || p1.y == p2.y || p1.z == p2.z;
    }

    static class CoordinatePoint{
        int x;
        int y;
        public CoordinatePoint(int x, int y){
            this.x = x;
            this.y = y;
        }
    }
    public static String sortString(String inputString){
        char tempArray[] = inputString.toCharArray();
        Arrays.sort(tempArray);
        return new String(tempArray);
    }
    static boolean possible(long h[], long x){
        int n = h.length;
        long a[] = new long[n];
        for (int i = 0; i < n; i++)
            a[i] = h[i];
        for (int i = n - 1; i >= 0; i--){
            if (a[i] < x){
                return false;
            }
            else{
                long d = (a[i] - x) / 3;
                d = Math.min(d, h[i] / 3);

                a[i] -= 3 * d;
                if (i > 1)
                    a[i - 1] += d;
                if (i > 1)
                    a[i - 2] += 2 * d;
            }
        }
        return true;
    }
    static class pair{
        char c;
        int f = 0;
        pair(char ff, int ss){
            c = ff;
            f = ss;
        }
    }
    static long fast_pow(long a, long p, long mod){
        long res = 1;
        while (p > 0) {
            if (p % 2 == 0) {
                a = ((a % mod) * (a % mod)) % mod;
                p /= 2;
            } else {
                res = ((res % mod) * (a % mod)) % mod;
                p--;
            }
        }
        return res;
    }
    static void takeInput(int a[], int n){
        Scanner in = new Scanner(System.in);
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
        }
    }

    static void longInput(long a[], int n){
        Scanner in = new Scanner(System.in);
        for (int i = 0; i < n; i++) {
            a[i] = in.nextLong();
        }
    }

    static int powerOfTwo(long n) {
        int power = Long.numberOfTrailingZeros(n);
        return power;
    }

    static class node{
        int i;
        int j;

        node(int ii, int jj) {
            i = ii;
            j = jj;
        }
    }
    static boolean isEven(long num){
        return (num & 1) == 0;
    }

    static long kadens(int ii, int jj, long arr[]){
        long sum = 0;
        long max = -(int) 1e18;
        for (int i = ii; i <= jj; i++) {
            sum += arr[i];
            max = Math.max(max, sum);
            if (sum < 0)
                sum = 0;
        }
        return max;
    }

    public static void dfs(int v, boolean[] visited, List<List<Integer>> adjList){
        visited[v] = true;
        System.out.print(v + " ");

        for (int neighbor : adjList.get(v)) {
            if (!visited[neighbor]) {
                dfs(neighbor, visited, adjList);
            }
        }
    }

    public static void perdfs(int vertices, List<List<Integer>> adjList){
        boolean[] visited = new boolean[vertices];
        for (int i = 0; i < vertices; i++) {
            if (!visited[i]) {
                dfs(i, visited, adjList);
                System.out.println();
            }
        }
    }

    public static void bfs(int start, List<List<Integer>> adjList){
        boolean[] visited = new boolean[adjList.size()];
        Queue<Integer> queue = new LinkedList<>();
        visited[start] = true;
        queue.add(start);

        while (!queue.isEmpty()) {
            int v = queue.poll();
            System.out.print(v + " ");

            for (int neighbor : adjList.get(v)) {
                if (!visited[neighbor]) {
                    visited[neighbor] = true;
                    queue.add(neighbor);
                }
            }
        }
        System.out.println();
    }

    static boolean[] visited = new boolean[1000000];
    public static void performBFS(int vertices, List<List<Integer>> adjList){
        for (int i = 0; i < vertices; i++) {
            if (!visited[i]) {
                bfs(i, adjList);
            }
        }
    }

    static final int MAX_CAKES = 5010;
    static long[][] dp = new long[MAX_CAKES][MAX_CAKES];
    static long[] cakeCount = new long[MAX_CAKES];

    static class FastReaader{
        BufferedReader br;
        StringTokenizer st;

        public FastReaader() {
            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;
        }
    }

    static long remainder(long x, long y, long k){
        long remainder = x % y;
        long diff = y - remainder;

        if (k < diff) {
            return x + k;
        }
        if (k == diff) {
            x += k;
            while (x % y == 0) {
                x /= y;
            }
            return x;
        }
        while (true) {
            long step = diff;
            if (x == 1) {
                break;
            }
            if (k >= step) {
                k -= step;
                x += step;
                while (x % y == 0) {
                    x /= y;
                }
                diff = y - x % y;
            } else {
                x += k;
                k = 0;
                break;
            }
        }
        if (k != 0) {
            return k % (y - 1) + 1;
        }
        return x;
    }

    public static int knapsack(int[] weights, int[] values, int capacity){
        int n = weights.length;
        int[][] dp = new int[n + 1][capacity + 1];

        for (int i = 1; i <= n; i++) {
            for (int w = 0; w <= capacity; w++) {
                if (weights[i - 1] <= w) {
                    dp[i][w] = Math.max(dp[i - 1][w], dp[i - 1][w - weights[i - 1]] + values[i - 1]);
                } else {
                    dp[i][w] = dp[i - 1][w];
                }
            }
        }

        return dp[n][capacity];
    }

    static long calculateOptimalCakes(int currentIndex, int remainingCakes){
        if (remainingCakes < 0)
            return -2;
        if (currentIndex == 0)
            return remainingCakes != 0 ? -2 : 0;
        if (dp[currentIndex][remainingCakes] != -1)
            return dp[currentIndex][remainingCakes];

        switch ((int) cakeCount[currentIndex]) {
            case 0:
                dp[currentIndex][remainingCakes] = calculateOptimalCakes(currentIndex - 1, remainingCakes);
                break;
            default:
                dp[currentIndex][remainingCakes] = calculateOptimalCakes(currentIndex - 1, remainingCakes - 1);
                break;
        }

        if (calculateOptimalCakes(currentIndex - 1, remainingCakes) != -2 &&
                calculateOptimalCakes(currentIndex - 1, remainingCakes) + cakeCount[currentIndex] <= remainingCakes) {

            if (dp[currentIndex][remainingCakes] == -2 ||
                    calculateOptimalCakes(currentIndex - 1, remainingCakes)
                            + cakeCount[currentIndex] < dp[currentIndex][remainingCakes]) {

                dp[currentIndex][remainingCakes] = calculateOptimalCakes(currentIndex - 1, remainingCakes)
                        + cakeCount[currentIndex];
            }
        }

        return dp[currentIndex][remainingCakes];
    }

    public static int[] mergeSortedArrays(int[] arr1, int[] arr2){
        int n1 = arr1.length;
        int n2 = arr2.length;
        int[] mergedArray = new int[n1 + n2];

        int i = 0, j = 0, k = 0;
        while (i < n1 && j < n2) {
            if (arr1[i] <= arr2[j]) {
                mergedArray[k++] = arr1[i++];
            } else {
                mergedArray[k++] = arr2[j++];
            }
        }
        while (i < n1) {
            mergedArray[k++] = arr1[i++];
        }
        while (j < n2) {
            mergedArray[k++] = arr2[j++];
        }

        return mergedArray;
    }

    static class FastReader{
        BufferedReader bufferedReader;
        StringTokenizer stringTokenizer;

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

        String next() {
            while (stringTokenizer == null || !stringTokenizer.hasMoreElements()) {
                try {
                    stringTokenizer = new StringTokenizer(bufferedReader.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return stringTokenizer.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 = bufferedReader.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    public static long ncr(int n, int r){
        if (r >n -r) {
            r = n -r;
        }
        long ans =1;
        for (int i =0; i< r;i++) {
            ans *= (n -i);
            ans /= (i +1);
        }
        return ans;
    }
    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 * b / gcd(a, b);
    }

}

**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, we need to consider edge cases and scenarios that are likely to cause the algorithm to fail. The primary issues in the buggy code include:

1. **Inefficient Selection of Cells**: The code does not efficiently handle the selection of cells to ensure the minimum sum while adhering to the constraints.
2. **Handling of Large Numbers**: The code may have issues with large numbers, especially when sorting and comparing.
3. **Edge Cases**: The code may not handle edge cases, such as when all numbers are the same or when the numbers are distributed in a way that makes it difficult to find a valid selection.

Given these points, a small-scale test input that could expose these issues is:

```plaintext
3
0 1 1
1 0 1
1 1 0
1 1 1
1 1 1
1 1 1
1 1 1
1 1 1
1 1 1
```

### Explanation:
- **Matrix Structure**: The cube is structured such that the first layer has a diagonal of 0s and 1s, and the rest of the layers are filled with 1s. This setup ensures that the minimum sum is achievable by selecting the 0s from the first layer.
- **Expected Output**: The minimum sum should be 0 + 0 + 0 = 0.
- **Challenge**: The code needs to correctly identify the 0s and ensure that they are selected from different planes. If the code fails to handle this, it might select higher values, leading to an incorrect result.

This test input is designed to be small and valid while challenging the buggy code to handle the selection of minimum values correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cube
time_limit_ms: 3000
memory_limit_mb: 1024
Description: You are given an $$$n\times n\times n$$$ big three-dimensional cube that contains $$$n^3$$$ numbers. You have to choose $$$n$$$ of those numbers so that their sum is as small as possible. It is, however, forbidden to choose two numbers that are located in the same plane. That is, if we identify the positions in the cube by three Cartesian coordinates, then choosing two numbers from positions $$$(x,y,z)$$$ and $$$(x',y',z')$$$ is forbidden if $$$x=x'$$$, $$$y=y'$$$, or $$$z=z'$$$.

Input: The input consists of the number $$$n$$$ followed by $$$n^3$$$ numbers in the cube. The numbers are presented as $$$n$$$ two-dimensional matrices, one for each layer of the cube. More precisely, there will be $$$n^2$$$ lines follow, each having $$$n$$$ numbers. For each $$$x, y, z$$$ ($$$1\le x, y, z\le n$$$), the number at the position $$$(x, y, z)$$$ is listed as the $$$z$$$-th number in the $$$((x-1)\times n+y)$$$-th line.

- $$$2 \leq n \leq 12$$$
- All numbers in the cube are integers between $$$0$$$ and $$$2\times 10^7$$$.

Output: The output consists of a single number. It is the minimum sum of $$$n$$$ numbers chosen from the cube according to the above rules.

Examples:
input:
3
1 2 3
4 5 6
7 8 9
1 1 1
2 2 2
3 3 3
4 3 0
2 1 4
9 8 9
output:
5

Note:

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

typedef long long ll;
typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;

#define F first
#define S second
#define pb push_back
#define nl "\n"
ll n ;
ll dp[1 << 12][1 << 12];
ll arr[12*12][12] ;

ll rec(ll mask_y , ll mask_z )
{
    ll var = __builtin_popcount(mask_y);
    ll ans = LLONG_MAX ;
    // cout<<var<<'\n';
    if (var >= n)
    {
        return 0 ;
    }
    for(ll i = (var)*n ; i < (var + 1 )*n ; i++)
    {
        if ( (((1 << (i % n )) & mask_y) != 0))
        {
            continue ;
        }
        for(ll j = 0 ;j < n ; j++)
        {
            if ( (((1 << j ) & mask_z) == 0))
            {
                ans = min(ans , arr[i][j] + rec((mask_y | (1 << (i % n) )) , (mask_z | ( 1 << j )))) ;
            }
        }
    }
    return dp[mask_y][mask_z] = ans ;
}
// {x , y , z } == {plane , line , number }
void solve() {
    cin>>n;
    ll var = (1ll << (n+1) ) ;
    // cout<<var*var<<'\n';
    for (int i = 0; i < n*n; ++i)
      {
          for(int j = 0 ; j < n ; j++)
          {
            cin>>arr[i][j] ;
          }
      }
      ll ans = rec(0ll , 0ll) ;

      cout<<ans<<'\n' ;
      return ;
}

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

    // Uncomment the following lines for multiple test cases
    // ll t;
    // cin >> t;
    // while (t--)
    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:
To create a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. Here are some key considerations:

1. **Edge Cases:**
   - **Small n:** Test with the minimum value of \( n \) (i.e., \( n = 2 \)).
   - **Large n:** Test with the maximum value of \( n \) (i.e., \( n = 12 \)).
   - **Uniform Distribution:** Test with a cube where all numbers are the same.
   - **High Contrast:** Test with a cube where the numbers have a high contrast (e.g., very small and very large numbers).

2. **Scenarios to Maximize Failure Probability:**
   - **Zeroes in the Cube:** Test with a cube that contains many zeroes, which should make it easy to find the minimum sum.
   - **Concentrated Low Values:** Test with a cube where low values are concentrated in specific planes, lines, or columns, making it challenging to select the minimum sum without violating the constraints.

Given these considerations, let's create a test input for \( n = 3 \) with a high contrast of values to make it challenging for the algorithm to find the correct minimum sum.

### Test Input:
```plaintext
3
1 2 3
4 5 6
7 8 9
10000000 10000000 10000000
10000000 10000000 10000000
10000000 10000000 10000000
1 1 1
2 2 2
3 3 3
```

### Explanation:
- The first layer contains numbers from 1 to 9.
- The second layer contains very large numbers (10,000,000).
- The third layer contains numbers from 1 to 3.

This input should challenge the algorithm to correctly identify the minimum sum while adhering to the constraints. The correct minimum sum should be 3 (1 + 1 + 1), but the algorithm must avoid selecting numbers from the same plane, line, or column.