**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lattice Optimizing
time_limit_ms: 7000
memory_limit_mb: 1024
Description: Consider a grid graph with $n$ rows and $n$ columns. Let the cell in row $x$ and column $y$ be $(x,y)$. There exists a directed edge from $(x,y)$ to $(x+1,y)$, with non-negative integer value $d_{x,y}$, for all $1\le x < n, 1\le y \le n$, and there also exists a directed edge from $(x,y)$ to $(x,y+1)$, with non-negative integer value $r_{x,y}$, for all $1\le x \le n, 1\le y < n$.

Initially, you are at $(1,1)$, with an empty set $S$. You need to walk along the edges and eventually reach $(n,n)$. Whenever you pass an edge, its value will be inserted into $S$. Please maximize the MEX$^{\text{∗}}$ of $S$ when you reach $(n,n)$.

$^{\text{∗}}$The MEX (minimum excluded) of an array is the smallest non- negative integer that does not belong to the array. For instance:

  * The MEX of $[2,2,1]$ is $0$, because $0$ does not belong to the array.    * The MEX of $[3,1,0,1]$ is $2$, because $0$ and $1$ belong to the array, but $2$ does not.    * The MEX of $[0,3,1,2]$ is $4$, because $0, 1, 2$, and $3$ belong to the array, but $4$ does not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\le n\le20$) — the number of rows and columns.

Each of the next $n-1$ lines contains $n$ integers separated by single spaces — the matrix $d$ ($0\le d_{x,y}\le 2n-2$).

Each of the next $n$ lines contains $n-1$ integers separated by single spaces — the matrix $r$ ($0\le r_{x,y}\le 2n-2$).

It is guaranteed that the sum of all $n^3$ does not exceed $8000$.

Output: For each test case, print a single integer — the maximum MEX of $S$ when you reach $(n,n)$.

Examples:
input:
2
3
1 0 2
0 1 3
2 1
0 3
3 0
3
1 2 0
0 1 2
2 0
1 2
0 1
output:
3
2

input:
1
10
16 7 3 15 9 17 1 15 9 0
4 3 1 12 13 10 10 14 6 12
3 1 3 9 5 16 0 12 7 12
11 4 8 7 13 7 15 13 9 2
2 3 9 9 4 12 17 7 10 15
10 6 15 17 13 6 15 9 4 9
13 3 3 14 1 2 10 10 12 16
8 2 9 13 18 7 1 6 2 6
15 12 2 6 0 0 13 3 7 17
7 3 17 17 10 15 12 14 15
4 3 3 17 3 13 11 16 6
16 17 7 7 12 5 2 4 10
18 9 9 3 5 9 1 16 7
1 0 4 2 10 10 12 2 1
4 14 15 16 15 5 8 4 18
7 18 10 11 2 0 14 8 18
2 17 6 0 9 6 13 5 11
5 15 7 11 6 3 17 14 5
1 3 16 16 13 1 0 13 11
output:
14

Note: In the first test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/70956fd41a5289db10b3b0bb41d0efae)

In the second test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/c6a0ac2a80551ddd517e35658fa66438)

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

public class Main {
    private static BufferedReader br;
    private static StringTokenizer tokenizer;
    private static String next() throws IOException{
        if (br == null) br = new BufferedReader(new InputStreamReader(System.in));
        if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());
        while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }
    private static long nextLong() throws IOException{
        return Long.parseLong(next());
    }

    private static StringBuilder sb = new StringBuilder();

    private static int ans;
    private static int[][] d;
    private static int[][] r;
    private static int n;
    private static int l;
    private static long[][] groups = new long[21][1 << 20];
    private static int[] indexes = new int[21];
    private static void dfs1(int x, int y, long mark) {
        if (x + y == l) {
            for (long tm = mark; tm > 0; tm = (tm - 1) & mark)  groups[x][indexes[x]++] = tm;
            return;
        }

        if (x < n - 1) dfs1(x + 1, y, mark | (1L << d[x][y]));
        if (y < n - 1) dfs1(x, y + 1, mark | (1L << r[x][y]));
    }
    private static void dfs2(int x, int y, long mark) {
        if (x + y == l) {
            for (long v = (1L << (ans + 1)) - 1;;v = (v << 1) + 1) {
                long w = (~mark) & v;
                int a = 0, b = indexes[x];
                while (a < b) {
                    int mid = a + ((b - a) >> 1);
                    if (groups[x][mid] >= w) {
                        b = mid;
                    } else {
                        a = mid + 1;
                    }
                }

                if (a == indexes[x] || groups[x][a] > w) break;
                ++ans;
            }
            return;
        }

        if (x > 0) dfs2(x - 1, y, mark | (1L << d[x - 1][y]));
        if (y > 0) dfs2(x, y - 1, mark | (1L << r[x][y - 1]));
    }

    private static void slove() throws IOException{
        n = nextInt();
        d = new int[n - 1][n];
        r = new int[n][n - 1];
        for (int i = 0; i < d.length; ++i) {
            for (int j = 0; j < d[0].length; ++j) d[i][j] = nextInt();
        }
        for (int i = 0; i < r.length; ++i) {
            for (int j = 0; j < r[0].length; ++j) r[i][j] = nextInt();
        }
        l = 2 * (n - 1) / 3;
        Arrays.fill(indexes, 1);
        ans = 0;
        dfs1(0, 0, 0L);

        for (int i = 0; i <= l; ++i) {
            Arrays.sort(groups[i], 0, indexes[i]);
            int pre = 0;
            for (int j = 1; j < indexes[i]; ++j) {
                if (groups[i][j] != groups[i][pre]) groups[i][++pre] = groups[i][j];
            }
            indexes[i] = pre + 1;
        }
        dfs2(n - 1, n - 1, 0L);
        sb.append(ans).append("\n");
    }

    public static void main(String[] args)  throws IOException {
        int tt = nextInt();
        sb = new StringBuilder();
        for (int i = 0; i < tt; ++i) slove();
        System.out.println(sb);
    }
}

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

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

Think step by step.

Response:
```plaintext
1
2
0 1
1 0
1
0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lattice Optimizing
time_limit_ms: 7000
memory_limit_mb: 1024
Description: Consider a grid graph with $n$ rows and $n$ columns. Let the cell in row $x$ and column $y$ be $(x,y)$. There exists a directed edge from $(x,y)$ to $(x+1,y)$, with non-negative integer value $d_{x,y}$, for all $1\le x < n, 1\le y \le n$, and there also exists a directed edge from $(x,y)$ to $(x,y+1)$, with non-negative integer value $r_{x,y}$, for all $1\le x \le n, 1\le y < n$.

Initially, you are at $(1,1)$, with an empty set $S$. You need to walk along the edges and eventually reach $(n,n)$. Whenever you pass an edge, its value will be inserted into $S$. Please maximize the MEX$^{\text{∗}}$ of $S$ when you reach $(n,n)$.

$^{\text{∗}}$The MEX (minimum excluded) of an array is the smallest non- negative integer that does not belong to the array. For instance:

  * The MEX of $[2,2,1]$ is $0$, because $0$ does not belong to the array.    * The MEX of $[3,1,0,1]$ is $2$, because $0$ and $1$ belong to the array, but $2$ does not.    * The MEX of $[0,3,1,2]$ is $4$, because $0, 1, 2$, and $3$ belong to the array, but $4$ does not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\le n\le20$) — the number of rows and columns.

Each of the next $n-1$ lines contains $n$ integers separated by single spaces — the matrix $d$ ($0\le d_{x,y}\le 2n-2$).

Each of the next $n$ lines contains $n-1$ integers separated by single spaces — the matrix $r$ ($0\le r_{x,y}\le 2n-2$).

It is guaranteed that the sum of all $n^3$ does not exceed $8000$.

Output: For each test case, print a single integer — the maximum MEX of $S$ when you reach $(n,n)$.

Examples:
input:
2
3
1 0 2
0 1 3
2 1
0 3
3 0
3
1 2 0
0 1 2
2 0
1 2
0 1
output:
3
2

input:
1
10
16 7 3 15 9 17 1 15 9 0
4 3 1 12 13 10 10 14 6 12
3 1 3 9 5 16 0 12 7 12
11 4 8 7 13 7 15 13 9 2
2 3 9 9 4 12 17 7 10 15
10 6 15 17 13 6 15 9 4 9
13 3 3 14 1 2 10 10 12 16
8 2 9 13 18 7 1 6 2 6
15 12 2 6 0 0 13 3 7 17
7 3 17 17 10 15 12 14 15
4 3 3 17 3 13 11 16 6
16 17 7 7 12 5 2 4 10
18 9 9 3 5 9 1 16 7
1 0 4 2 10 10 12 2 1
4 14 15 16 15 5 8 4 18
7 18 10 11 2 0 14 8 18
2 17 6 0 9 6 13 5 11
5 15 7 11 6 3 17 14 5
1 3 16 16 13 1 0 13 11
output:
14

Note: In the first test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/70956fd41a5289db10b3b0bb41d0efae)

In the second test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/c6a0ac2a80551ddd517e35658fa66438)

Buggy Code:
// Problem: G. Оптимизация решетки
// Contest: Codeforces - EPIC Institute of Technology Round August 2024 (Div. 1 + Div. 2)
// URL: https://codeforces.com/contest/2002/problem/G
// Memory Limit: 1024 MB
// Time Limit: 7000 ms
//
// Powered by CP Editor (https://cpeditor.org)

#include <iostream>
#include <fstream>
#include <vector>
#include <algorithm>
#include <set>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <utility>
#include <limits>
#include <map>
#include <queue>

using namespace std;

#define ll long long
#define F first
#define S second
#define YES cout << "YES\n"
#define NO cout << "NO\n"
#define Yes cout << "Yes\n"
#define No cout << "No\n"
#define yes cout << "yes\n"
#define no cout << "no\n"
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define fix(a) fixed << setprecision (a)
#define pb(a) push_back(a)
#define mp(a, b) make_pair (a, b)

ll gcd (ll a, ll b) {
	if (b == 0)
		return a;
	return gcd (b, a % b);
}

ll lcm (ll a, ll b) {
	return a * b / gcd (a, b);
}

ll binpow (ll a, ll n) {
	if (n == 0)
		return 1;
	if (n % 2 == 1)
		return binpow (a, n - 1) * a;
	else {
		ll b = binpow (a, n / 2);
		return b * b;
	}
}

bool isPrime(ll num) {
    if (num <= 1) {
        return false;
    }
    if (num <= 3) {
        return true;
    }
    if (num % 2 == 0 || num % 3 == 0) {
        return false;
    }
    for (ll i = 5; i * i <= num; i += 6) {
        if (num % i == 0 || num % (i + 2) == 0) {
            return false;
        }
    }
    return true;
}

void solve () {
	ll n;
	cin >> n;
	vector <vector<ll>> d (n - 1, vector <ll> (n));
	vector <vector<ll>> r (n, vector <ll> (n - 1));
	for (ll i = 0; i < n - 1; i++)
		for (ll j = 0; j < n; j++)
			cin >> d[i][j];
	for (ll i = 0; i < n; i++)
		for (ll j = 0; j < n - 1; j++)
			cin >> r[i][j];
	ll mx = 0;
	for (ll i = 0; i < binpow (2, 2 * n - 2); i++) {
		ll x = 0, y = 0;
		ll m = i;
		string s = "";
		ll cnt = 0;
		for (ll j = 0; j < 2 * n - 2; j++) {
			cnt += m % 2;
			s += char (m % 2 + '0');
			m /= 2;
		}
		if (cnt != n - 1)
			continue;
		// cout << s << endl;
		set <ll> st;
		for (auto ch : s) {
			// cout << x << " " << y << endl;
			if (ch == '1') {
				// cout << d[x][y] << " ";
				st.insert(d[x][y]);
				x++;
			} else {
				// cout << r[x][y] << " ";
				st.insert(r[x][y]);
				y++;
			}
		}
		ll cur = 0;
		for (auto it : st) {
			if (cur != it)
				break;
			cur++;
		}
		mx = max (mx, cur);
		// cout << cur << endl;
		// cout << endl;
	}
	cout << mx << endl;
	// cout << endl;
}

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	ll tt = 1;
	cin >> tt;
	while (tt--)
		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 expose the bug in the provided code, we need to focus on edge cases and scenarios that can highlight the issues in the algorithm. The code appears to be trying to find the maximum MEX by generating all possible paths from (1,1) to (n,n) and then calculating the MEX for each path. However, there are a few potential issues:

1. **Path Generation**: The code uses a binary number to represent the path, but it might not correctly handle all valid paths.
2. **MEX Calculation**: The MEX calculation might be incorrect if the set of values collected along the path is not properly managed.

To create a challenging test input, we can focus on a small grid where the optimal path is not immediately obvious and can be tricky to handle. Let's consider a 4x4 grid with specific edge values that could expose these issues.

```plaintext
1
4
1 0 2 3
0 1 2 3
2 1 0 3
3 2 1 0
1 2 3 0
0 1 2 3
3 0 1 2
2 3 0 1
```

### Explanation:
- **Grid Size**: 4x4, which is small enough to be manageable but large enough to have multiple paths.
- **Edge Values**:
  - The values in the `d` matrix are chosen such that they form a pattern that could lead to multiple paths with different MEX values.
  - The values in the `r` matrix are chosen similarly, ensuring that the paths can have varying MEX values.

This test input will challenge the algorithm to correctly generate all valid paths and compute the MEX for each path. If the code has issues with path generation or MEX calculation, this input should expose them.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lattice Optimizing
time_limit_ms: 7000
memory_limit_mb: 1024
Description: Consider a grid graph with $n$ rows and $n$ columns. Let the cell in row $x$ and column $y$ be $(x,y)$. There exists a directed edge from $(x,y)$ to $(x+1,y)$, with non-negative integer value $d_{x,y}$, for all $1\le x < n, 1\le y \le n$, and there also exists a directed edge from $(x,y)$ to $(x,y+1)$, with non-negative integer value $r_{x,y}$, for all $1\le x \le n, 1\le y < n$.

Initially, you are at $(1,1)$, with an empty set $S$. You need to walk along the edges and eventually reach $(n,n)$. Whenever you pass an edge, its value will be inserted into $S$. Please maximize the MEX$^{\text{∗}}$ of $S$ when you reach $(n,n)$.

$^{\text{∗}}$The MEX (minimum excluded) of an array is the smallest non- negative integer that does not belong to the array. For instance:

  * The MEX of $[2,2,1]$ is $0$, because $0$ does not belong to the array.    * The MEX of $[3,1,0,1]$ is $2$, because $0$ and $1$ belong to the array, but $2$ does not.    * The MEX of $[0,3,1,2]$ is $4$, because $0, 1, 2$, and $3$ belong to the array, but $4$ does not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\le n\le20$) — the number of rows and columns.

Each of the next $n-1$ lines contains $n$ integers separated by single spaces — the matrix $d$ ($0\le d_{x,y}\le 2n-2$).

Each of the next $n$ lines contains $n-1$ integers separated by single spaces — the matrix $r$ ($0\le r_{x,y}\le 2n-2$).

It is guaranteed that the sum of all $n^3$ does not exceed $8000$.

Output: For each test case, print a single integer — the maximum MEX of $S$ when you reach $(n,n)$.

Examples:
input:
2
3
1 0 2
0 1 3
2 1
0 3
3 0
3
1 2 0
0 1 2
2 0
1 2
0 1
output:
3
2

input:
1
10
16 7 3 15 9 17 1 15 9 0
4 3 1 12 13 10 10 14 6 12
3 1 3 9 5 16 0 12 7 12
11 4 8 7 13 7 15 13 9 2
2 3 9 9 4 12 17 7 10 15
10 6 15 17 13 6 15 9 4 9
13 3 3 14 1 2 10 10 12 16
8 2 9 13 18 7 1 6 2 6
15 12 2 6 0 0 13 3 7 17
7 3 17 17 10 15 12 14 15
4 3 3 17 3 13 11 16 6
16 17 7 7 12 5 2 4 10
18 9 9 3 5 9 1 16 7
1 0 4 2 10 10 12 2 1
4 14 15 16 15 5 8 4 18
7 18 10 11 2 0 14 8 18
2 17 6 0 9 6 13 5 11
5 15 7 11 6 3 17 14 5
1 3 16 16 13 1 0 13 11
output:
14

Note: In the first test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/70956fd41a5289db10b3b0bb41d0efae)

In the second test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/c6a0ac2a80551ddd517e35658fa66438)

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

using namespace std;
using lli = long long;

vector<vector<lli>> calc_masks(vector<vector<int>> &ds, vector<vector<int>> &ls, int x) {
	int n = ls.size();

	vector<vector<vector<lli>>> pos(2, vector<vector<lli>>(n));

	pos[0][0].push_back(0);

	vector<vector<lli>> ans;
	for (int i = 0; i < n; ++i) {
		int cur = i%2, old = 1 - i%2;
		for (int j = 0; j < n; ++j) if (i + j <= x) {
			if (i > 0) for (lli m: pos[old][j])
				pos[cur][j].push_back(m | (1LL << ds[i-1][j]));
			if (j > 0) for (lli m: pos[cur][j-1])
				pos[cur][j].push_back(m | (1LL << ls[i][j-1]));

			if (i + j == x) ans.push_back(pos[cur][j]);
		}

		for (int j = 0; j < n; ++j)
			pos[old][j].clear();
	}

	return ans;
}

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

	while (t--) {
		int n;
		cin >> n;

		vector<vector<int>> ds(n - 1, vector<int>(n));
		for (auto &xs: ds)
			for (int &d: xs)
				cin >> d;

		vector<vector<int>> ls(n, vector<int>(n - 1));
		for (auto &xs: ls)
			for (int &l: xs)
				cin >> l;

		int x = 2*(2*n - 2)/3;
		int y = 2*n - 2 - x;

		auto masks = calc_masks(ds, ls, x);

		// cerr << "->large\n";
		// for (auto &ms: masks) {
		// 	for (lli m: ms)
		// 		cerr << m << " ";
		// 	cerr << "\n";
		// }

		vector<vector<int>> d1s(n - 1, vector<int>(n));
		for (int i = 0; i < n-1; ++i)
			for (int j = 0; j < n; ++j)
				d1s[i][j] = ds[n - i - 2][n - j - 1];

		vector<vector<int>> l1s(n, vector<int>(n - 1));
		for (int i = 0; i < n; ++i)
			for (int j = 0; j < n-1; ++j)
				l1s[i][j] = ls[n - i - 1][n - j - 2];

		auto masks_small = calc_masks(d1s, l1s, y);

		reverse(masks_small.begin(), masks_small.end());

		// cerr << "->small\n";
		// for (auto &ms: masks_small) {
		// 	for (lli m: ms)
		// 		cerr << m << " ";
		// 	cerr << "\n";
		// }

		int best = 0;
		for (int i = 0; i < masks_small.size(); ++i) {
			sort(masks_small[i].begin(), masks_small[i].end());
			masks_small[i].erase(unique(masks_small[i].begin(), masks_small[i].end()), masks_small[i].end());

			vector<lli> submasks{0};
			for (lli mask: masks_small[i])
				for (lli sub = mask; sub > 0; sub = (sub - 1) & mask)
					submasks.push_back(sub);

			sort(submasks.begin(), submasks.end());
			submasks.erase(unique(submasks.begin(), submasks.end()), submasks.end());

			for (lli mask: masks[i])
				while (best < 2*n - 2 && find(submasks.begin(), submasks.end(),
					   (~mask) & ((1LL << (best + 1)) - 1)) != submasks.end())
					++best;
		}

		cout << best << "\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
1
3
0 0 0
0 0 0
0 0
0 0
3
0 0 0
0 0 0
0 0
0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lattice Optimizing
time_limit_ms: 7000
memory_limit_mb: 1024
Description: Consider a grid graph with $n$ rows and $n$ columns. Let the cell in row $x$ and column $y$ be $(x,y)$. There exists a directed edge from $(x,y)$ to $(x+1,y)$, with non-negative integer value $d_{x,y}$, for all $1\le x < n, 1\le y \le n$, and there also exists a directed edge from $(x,y)$ to $(x,y+1)$, with non-negative integer value $r_{x,y}$, for all $1\le x \le n, 1\le y < n$.

Initially, you are at $(1,1)$, with an empty set $S$. You need to walk along the edges and eventually reach $(n,n)$. Whenever you pass an edge, its value will be inserted into $S$. Please maximize the MEX$^{\text{∗}}$ of $S$ when you reach $(n,n)$.

$^{\text{∗}}$The MEX (minimum excluded) of an array is the smallest non- negative integer that does not belong to the array. For instance:

  * The MEX of $[2,2,1]$ is $0$, because $0$ does not belong to the array.    * The MEX of $[3,1,0,1]$ is $2$, because $0$ and $1$ belong to the array, but $2$ does not.    * The MEX of $[0,3,1,2]$ is $4$, because $0, 1, 2$, and $3$ belong to the array, but $4$ does not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\le n\le20$) — the number of rows and columns.

Each of the next $n-1$ lines contains $n$ integers separated by single spaces — the matrix $d$ ($0\le d_{x,y}\le 2n-2$).

Each of the next $n$ lines contains $n-1$ integers separated by single spaces — the matrix $r$ ($0\le r_{x,y}\le 2n-2$).

It is guaranteed that the sum of all $n^3$ does not exceed $8000$.

Output: For each test case, print a single integer — the maximum MEX of $S$ when you reach $(n,n)$.

Examples:
input:
2
3
1 0 2
0 1 3
2 1
0 3
3 0
3
1 2 0
0 1 2
2 0
1 2
0 1
output:
3
2

input:
1
10
16 7 3 15 9 17 1 15 9 0
4 3 1 12 13 10 10 14 6 12
3 1 3 9 5 16 0 12 7 12
11 4 8 7 13 7 15 13 9 2
2 3 9 9 4 12 17 7 10 15
10 6 15 17 13 6 15 9 4 9
13 3 3 14 1 2 10 10 12 16
8 2 9 13 18 7 1 6 2 6
15 12 2 6 0 0 13 3 7 17
7 3 17 17 10 15 12 14 15
4 3 3 17 3 13 11 16 6
16 17 7 7 12 5 2 4 10
18 9 9 3 5 9 1 16 7
1 0 4 2 10 10 12 2 1
4 14 15 16 15 5 8 4 18
7 18 10 11 2 0 14 8 18
2 17 6 0 9 6 13 5 11
5 15 7 11 6 3 17 14 5
1 3 16 16 13 1 0 13 11
output:
14

Note: In the first test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/70956fd41a5289db10b3b0bb41d0efae)

In the second test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/c6a0ac2a80551ddd517e35658fa66438)

Buggy Code:
def calculate_mex(collected):
    mex = 0
    while mex in collected:
        mex += 1
    return mex

def dfs(x, y, n, down, right, collected, visited):
    if (x, y) == (n - 1, n - 1):
        return calculate_mex(collected)

    if visited[x][y]:
        return -1  # Already visited this cell

    visited[x][y] = True
    max_mex = -1

    if x < n - 1:  # Move down
        new_collected = collected | {down[x][y]}
        max_mex = max(max_mex, dfs(x + 1, y, n, down, right, new_collected, visited))

    if y < n - 1:  # Move right
        new_collected = collected | {right[x][y]}
        max_mex = max(max_mex, dfs(x, y + 1, n, down, right, new_collected, visited))

    visited[x][y] = False  # Backtrack
    return max_mex

def maximum_mex(n, down, right):
    visited = [[False] * n for _ in range(n)]
    return dfs(0, 0, n, down, right, set(), visited)

def main():
    import sys
    input = sys.stdin.read
    data = input().splitlines()

    index = 0
    t = int(data[index])
    index += 1
    results = []

    for _ in range(t):
        n = int(data[index])
        index += 1

        down = []
        for i in range(n - 1):
            row = list(map(int, data[index].split()))
            down.append(row)
            index += 1

        right = []
        for i in range(n):
            row = list(map(int, data[index].split()))
            right.append(row)
            index += 1

        max_mex = maximum_mex(n, down, right)
        results.append(max_mex)

    print("\n".join(map(str, results)))

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 expose the bug in the given buggy code, we need to focus on edge cases that can highlight issues in the DFS traversal and the MEX calculation. Here are some key points to consider:

1. **Edge Cases for MEX Calculation**: The MEX calculation should handle cases where the set of collected numbers might already contain all small integers, making the MEX a higher value.
2. **DFS Traversal Issues**: The DFS traversal should ensure that all possible paths are considered, and backtracking is correctly handled.

A good test input would be one where the grid contains values that are dense in the range of small integers, making it challenging to find a path that maximizes the MEX.

Let's construct a small-scale test input:

```plaintext
1
4
0 1 2 3
1 2 3 0
2 3 0 1
3 0 1 2
0 1 2
1 2 3
2 3 0
3 0 1
```

### Explanation:
- **Grid Size**: 4x4
- **Down Matrix**: The values are arranged in a way that each row contains the integers 0, 1, 2, 3 in a cyclic manner. This ensures that the set of collected values will be dense in the range of small integers.
- **Right Matrix**: Similarly, the values are arranged in a way that each column contains the integers 0, 1, 2, 3 in a cyclic manner.

### Expected Output:
The expected output for this test input is 4. This is because the optimal path will collect all integers from 0 to 3, making the MEX 4.

This test input should expose the bug in the DFS traversal and MEX calculation, as the set of collected values will be dense, and the MEX calculation should correctly identify the smallest non-negative integer not in the set.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lattice Optimizing
time_limit_ms: 7000
memory_limit_mb: 1024
Description: Consider a grid graph with $n$ rows and $n$ columns. Let the cell in row $x$ and column $y$ be $(x,y)$. There exists a directed edge from $(x,y)$ to $(x+1,y)$, with non-negative integer value $d_{x,y}$, for all $1\le x < n, 1\le y \le n$, and there also exists a directed edge from $(x,y)$ to $(x,y+1)$, with non-negative integer value $r_{x,y}$, for all $1\le x \le n, 1\le y < n$.

Initially, you are at $(1,1)$, with an empty set $S$. You need to walk along the edges and eventually reach $(n,n)$. Whenever you pass an edge, its value will be inserted into $S$. Please maximize the MEX$^{\text{∗}}$ of $S$ when you reach $(n,n)$.

$^{\text{∗}}$The MEX (minimum excluded) of an array is the smallest non- negative integer that does not belong to the array. For instance:

  * The MEX of $[2,2,1]$ is $0$, because $0$ does not belong to the array.    * The MEX of $[3,1,0,1]$ is $2$, because $0$ and $1$ belong to the array, but $2$ does not.    * The MEX of $[0,3,1,2]$ is $4$, because $0, 1, 2$, and $3$ belong to the array, but $4$ does not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\le n\le20$) — the number of rows and columns.

Each of the next $n-1$ lines contains $n$ integers separated by single spaces — the matrix $d$ ($0\le d_{x,y}\le 2n-2$).

Each of the next $n$ lines contains $n-1$ integers separated by single spaces — the matrix $r$ ($0\le r_{x,y}\le 2n-2$).

It is guaranteed that the sum of all $n^3$ does not exceed $8000$.

Output: For each test case, print a single integer — the maximum MEX of $S$ when you reach $(n,n)$.

Examples:
input:
2
3
1 0 2
0 1 3
2 1
0 3
3 0
3
1 2 0
0 1 2
2 0
1 2
0 1
output:
3
2

input:
1
10
16 7 3 15 9 17 1 15 9 0
4 3 1 12 13 10 10 14 6 12
3 1 3 9 5 16 0 12 7 12
11 4 8 7 13 7 15 13 9 2
2 3 9 9 4 12 17 7 10 15
10 6 15 17 13 6 15 9 4 9
13 3 3 14 1 2 10 10 12 16
8 2 9 13 18 7 1 6 2 6
15 12 2 6 0 0 13 3 7 17
7 3 17 17 10 15 12 14 15
4 3 3 17 3 13 11 16 6
16 17 7 7 12 5 2 4 10
18 9 9 3 5 9 1 16 7
1 0 4 2 10 10 12 2 1
4 14 15 16 15 5 8 4 18
7 18 10 11 2 0 14 8 18
2 17 6 0 9 6 13 5 11
5 15 7 11 6 3 17 14 5
1 3 16 16 13 1 0 13 11
output:
14

Note: In the first test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/70956fd41a5289db10b3b0bb41d0efae)

In the second test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/c6a0ac2a80551ddd517e35658fa66438)

Buggy Code:
def find_mex(elements):
    mex = 0
    while mex in elements:
        mex += 1
    return mex

def explore_path(x, y, size, down_values, right_values, elements_collected, has_visited):
    if (x, y) == (size - 1, size - 1):
        return find_mex(elements_collected)

    if has_visited[x][y]:
        return -1  # This cell has been visited before

    has_visited[x][y] = True
    max_mex_value = -1

    if x < size - 1:  # Move downwards
        updated_elements = elements_collected | {down_values[x][y]}
        max_mex_value = max(max_mex_value, explore_path(x + 1, y, size, down_values, right_values, updated_elements, has_visited))

    if y < size - 1:  # Move rightwards
        updated_elements = elements_collected | {right_values[x][y]}
        max_mex_value = max(max_mex_value, explore_path(x, y + 1, size, down_values, right_values, updated_elements, has_visited))

    has_visited[x][y] = False  # Backtrack
    return max_mex_value

def compute_maximum_mex(size, down_values, right_values):
    has_visited = [[False] * size for _ in range(size)]
    return explore_path(0, 0, size, down_values, right_values, set(), has_visited)

def main():
    import sys
    input_data = sys.stdin.read
    lines = input_data().splitlines()

    current_index = 0
    num_tests = int(lines[current_index])
    current_index += 1
    results = []

    for _ in range(num_tests):
        size = int(lines[current_index])
        current_index += 1

        down_values = []
        for _ in range(size - 1):
            row = list(map(int, lines[current_index].split()))
            down_values.append(row)
            current_index += 1

        right_values = []
        for _ in range(size):
            row = list(map(int, lines[current_index].split()))
            right_values.append(row)
            current_index += 1

        max_mex_value = compute_maximum_mex(size, down_values, right_values)
        results.append(max_mex_value)

    print("\n".join(map(str, results)))

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
```plaintext
1
2
0 1
1
0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lattice Optimizing
time_limit_ms: 7000
memory_limit_mb: 1024
Description: Consider a grid graph with $n$ rows and $n$ columns. Let the cell in row $x$ and column $y$ be $(x,y)$. There exists a directed edge from $(x,y)$ to $(x+1,y)$, with non-negative integer value $d_{x,y}$, for all $1\le x < n, 1\le y \le n$, and there also exists a directed edge from $(x,y)$ to $(x,y+1)$, with non-negative integer value $r_{x,y}$, for all $1\le x \le n, 1\le y < n$.

Initially, you are at $(1,1)$, with an empty set $S$. You need to walk along the edges and eventually reach $(n,n)$. Whenever you pass an edge, its value will be inserted into $S$. Please maximize the MEX$^{\text{∗}}$ of $S$ when you reach $(n,n)$.

$^{\text{∗}}$The MEX (minimum excluded) of an array is the smallest non- negative integer that does not belong to the array. For instance:

  * The MEX of $[2,2,1]$ is $0$, because $0$ does not belong to the array.    * The MEX of $[3,1,0,1]$ is $2$, because $0$ and $1$ belong to the array, but $2$ does not.    * The MEX of $[0,3,1,2]$ is $4$, because $0, 1, 2$, and $3$ belong to the array, but $4$ does not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\le n\le20$) — the number of rows and columns.

Each of the next $n-1$ lines contains $n$ integers separated by single spaces — the matrix $d$ ($0\le d_{x,y}\le 2n-2$).

Each of the next $n$ lines contains $n-1$ integers separated by single spaces — the matrix $r$ ($0\le r_{x,y}\le 2n-2$).

It is guaranteed that the sum of all $n^3$ does not exceed $8000$.

Output: For each test case, print a single integer — the maximum MEX of $S$ when you reach $(n,n)$.

Examples:
input:
2
3
1 0 2
0 1 3
2 1
0 3
3 0
3
1 2 0
0 1 2
2 0
1 2
0 1
output:
3
2

input:
1
10
16 7 3 15 9 17 1 15 9 0
4 3 1 12 13 10 10 14 6 12
3 1 3 9 5 16 0 12 7 12
11 4 8 7 13 7 15 13 9 2
2 3 9 9 4 12 17 7 10 15
10 6 15 17 13 6 15 9 4 9
13 3 3 14 1 2 10 10 12 16
8 2 9 13 18 7 1 6 2 6
15 12 2 6 0 0 13 3 7 17
7 3 17 17 10 15 12 14 15
4 3 3 17 3 13 11 16 6
16 17 7 7 12 5 2 4 10
18 9 9 3 5 9 1 16 7
1 0 4 2 10 10 12 2 1
4 14 15 16 15 5 8 4 18
7 18 10 11 2 0 14 8 18
2 17 6 0 9 6 13 5 11
5 15 7 11 6 3 17 14 5
1 3 16 16 13 1 0 13 11
output:
14

Note: In the first test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/70956fd41a5289db10b3b0bb41d0efae)

In the second test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/c6a0ac2a80551ddd517e35658fa66438)

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

public class Main {
    private static BufferedReader br;
    private static StringTokenizer tokenizer;
    private static String next() throws IOException{
        if (br == null) br = new BufferedReader(new InputStreamReader(System.in));
        if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());
        while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }
    private static long nextLong() throws IOException{
        return Long.parseLong(next());
    }

    private static StringBuilder sb = new StringBuilder();

    private static int ans;
    private static int[][] d;
    private static int[][] r;
    private static int n;
    private static int l;
    private static Set<Long>[] groups = new HashSet[20];
    private static void dfs1(int x, int y, long mark) {
        if (x + y == l) {
            for (long tm = mark; tm > 0; tm = (tm - 1) & mark)  groups[x].add(tm);
            return;
        }

        if (x < n - 1) dfs1(x + 1, y, mark | (1L << d[x][y]));
        if (y < n - 1) dfs1(x, y + 1, mark | (1L << r[x][y]));
    }
    private static void dfs2(int x, int y, long mark) {
        if (x + y == l) {
            Set<Long> set = groups[x];
            long v = (1L << (ans + 1)) - 1;
            while ((mark | v) == mark || set.contains(v ^ (v & mark))) {
                ++ans;
                v = (v << 1) + 1;
            }
            return;
        }

        if (x > 0) dfs2(x - 1, y, mark | (1L << d[x - 1][y]));
        if (y > 0) dfs2(x, y - 1, mark | (1L << r[x][y - 1]));
    }

    private static void slove() throws IOException{
        n = nextInt();
        d = new int[n - 1][n];
        r = new int[n][n - 1];
        for (int i = 0; i < d.length; ++i) {
            for (int j = 0; j < d[0].length; ++j) d[i][j] = nextInt();
        }
        for (int i = 0; i < r.length; ++i) {
            for (int j = 0; j < r[0].length; ++j) r[i][j] = nextInt();
        }
        l = (int)(2.0/3*n+2);
        for (int i = 0; i < groups.length; i++) groups[i] = new HashSet<>();
        ans = 0;
        dfs1(0, 0, 0L);
        dfs2(n - 1, n - 1, 0L);
        sb.append(ans).append("\n");
    }

    public static void main(String[] args)  throws IOException {
        int tt = nextInt();
        sb = new StringBuilder();
        for (int i = 0; i < tt; ++i) slove();
        System.out.println(sb);
    }
}

**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
0 1 2
1 0 1
2 1 0
0 2 1
1 0 2
1 2 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lattice Optimizing
time_limit_ms: 7000
memory_limit_mb: 1024
Description: Consider a grid graph with $n$ rows and $n$ columns. Let the cell in row $x$ and column $y$ be $(x,y)$. There exists a directed edge from $(x,y)$ to $(x+1,y)$, with non-negative integer value $d_{x,y}$, for all $1\le x < n, 1\le y \le n$, and there also exists a directed edge from $(x,y)$ to $(x,y+1)$, with non-negative integer value $r_{x,y}$, for all $1\le x \le n, 1\le y < n$.

Initially, you are at $(1,1)$, with an empty set $S$. You need to walk along the edges and eventually reach $(n,n)$. Whenever you pass an edge, its value will be inserted into $S$. Please maximize the MEX$^{\text{∗}}$ of $S$ when you reach $(n,n)$.

$^{\text{∗}}$The MEX (minimum excluded) of an array is the smallest non- negative integer that does not belong to the array. For instance:

  * The MEX of $[2,2,1]$ is $0$, because $0$ does not belong to the array.    * The MEX of $[3,1,0,1]$ is $2$, because $0$ and $1$ belong to the array, but $2$ does not.    * The MEX of $[0,3,1,2]$ is $4$, because $0, 1, 2$, and $3$ belong to the array, but $4$ does not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\le n\le20$) — the number of rows and columns.

Each of the next $n-1$ lines contains $n$ integers separated by single spaces — the matrix $d$ ($0\le d_{x,y}\le 2n-2$).

Each of the next $n$ lines contains $n-1$ integers separated by single spaces — the matrix $r$ ($0\le r_{x,y}\le 2n-2$).

It is guaranteed that the sum of all $n^3$ does not exceed $8000$.

Output: For each test case, print a single integer — the maximum MEX of $S$ when you reach $(n,n)$.

Examples:
input:
2
3
1 0 2
0 1 3
2 1
0 3
3 0
3
1 2 0
0 1 2
2 0
1 2
0 1
output:
3
2

input:
1
10
16 7 3 15 9 17 1 15 9 0
4 3 1 12 13 10 10 14 6 12
3 1 3 9 5 16 0 12 7 12
11 4 8 7 13 7 15 13 9 2
2 3 9 9 4 12 17 7 10 15
10 6 15 17 13 6 15 9 4 9
13 3 3 14 1 2 10 10 12 16
8 2 9 13 18 7 1 6 2 6
15 12 2 6 0 0 13 3 7 17
7 3 17 17 10 15 12 14 15
4 3 3 17 3 13 11 16 6
16 17 7 7 12 5 2 4 10
18 9 9 3 5 9 1 16 7
1 0 4 2 10 10 12 2 1
4 14 15 16 15 5 8 4 18
7 18 10 11 2 0 14 8 18
2 17 6 0 9 6 13 5 11
5 15 7 11 6 3 17 14 5
1 3 16 16 13 1 0 13 11
output:
14

Note: In the first test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/70956fd41a5289db10b3b0bb41d0efae)

In the second test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/c6a0ac2a80551ddd517e35658fa66438)

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

public class Main {
    private static BufferedReader br;
    private static StringTokenizer tokenizer;
    private static String next() throws IOException{
        if (br == null) br = new BufferedReader(new InputStreamReader(System.in));
        if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());
        while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }
    private static long nextLong() throws IOException{
        return Long.parseLong(next());
    }

    static StringBuilder sb = new StringBuilder();

    static int ans;
    static int[][] d;
    static int[][] r;
    static int n;
    static int l;
    static int[][] direction = {{1,0},{0,1}};
    static Set<Long>[] groups = new HashSet[20];
    static int getKey(int x, int y) {
        return 41 * x + y;
    }
    static void dfs1(int x, int y, long mark, int l) {
        if (x + y == l) {
            Set<Long> set = groups[x];
            for (long tm = mark; tm > 0; tm = (tm - 1) & mark) set.add(tm);
            return;
        }

        if (x < n - 1) dfs1(x + 1, y, mark | (1L << d[x][y]), l);
        if (y < n - 1) dfs1(x, y + 1, mark | (1L << r[x][y]), l);
    }
    static void dfs2(int x, int y, long mark, int fx, int fy) {
        if (x == n - 1 && y == n - 1) {
            int key = getKey(fx, fy);
            Set<Long> set = groups[fx];
            long v = 1;
            int l = 0;
            while ((mark | v) == mark || set.contains(v ^ (v & mark))) {
                ++l;
                v |= (1L << l);
            }
            ans = Math.max(ans, l);
            return;
        }

        if (x < n - 1) dfs2(x + 1, y, mark | (1L << d[x][y]), fx, fy);
        if (y < n - 1) dfs2(x, y + 1, mark | (1L << r[x][y]), fx, fy);
    }

    static void slove() throws IOException{
        n = nextInt();
        d = new int[n - 1][n];
        r = new int[n][n - 1];
        for (int i = 0; i < d.length; ++i) {
            for (int j = 0; j < d[0].length; ++j) d[i][j] = nextInt();
        }
        for (int i = 0; i < r.length; ++i) {
            for (int j = 0; j < r[0].length; ++j) r[i][j] = nextInt();
        }
        l = (int)(2.0/3*n);
        for (int i = 0; i < groups.length; i++) groups[i] = new HashSet<Long>();
        ans = 0;
        dfs1(0, 0, 0L, l);
        for (int i = 0; i <= l; ++i) dfs2(i, l - i, 0L, i, l - i);
        sb.append(ans).append("\n");
    }

    public static void main(String[] args)  throws IOException {
        int tt = nextInt();
        sb = new StringBuilder();
        for (int i = 0; i < tt; ++i) slove();
        System.out.println(sb);
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. Here are the steps to think through this:

1. **Understand the Bug**: The code uses a combination of DFS to explore paths and bit manipulation to track the set of numbers. The main issue could be in how it handles the MEX calculation or in the way it explores paths.

2. **Edge Cases**:
   - **Small Grid**: Test with the smallest grid size to ensure the basic logic works.
   - **Zero Values**: Test with a grid that has many zeros, which could cause issues in MEX calculation.
   - **High Values**: Test with values that are close to the upper limit of `2n-2`.
   - **Symmetric Grid**: Test with a symmetric grid where the values are the same in both directions.
   - **Path Constraints**: Test with a grid where certain paths are forced to be taken due to high values on other paths.

3. **Generate Test Input**:
   - We will create a small grid (e.g., 3x3) with values that are likely to cause issues in the MEX calculation.

Here is a small-scale test input that could expose the bug:

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

This input consists of a 3x3 grid where all values are zero. This should force the MEX calculation to be zero, and it will test how the code handles a grid with no positive values.

Let's break it down:
- **Number of Test Cases**: 1
- **Grid Size**: 3x3
- **Matrix `d`**: All zeros
- **Matrix `r`**: All zeros

This input should be simple enough to verify but challenging enough to expose any issues in the MEX calculation logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lattice Optimizing
time_limit_ms: 7000
memory_limit_mb: 1024
Description: Consider a grid graph with $n$ rows and $n$ columns. Let the cell in row $x$ and column $y$ be $(x,y)$. There exists a directed edge from $(x,y)$ to $(x+1,y)$, with non-negative integer value $d_{x,y}$, for all $1\le x < n, 1\le y \le n$, and there also exists a directed edge from $(x,y)$ to $(x,y+1)$, with non-negative integer value $r_{x,y}$, for all $1\le x \le n, 1\le y < n$.

Initially, you are at $(1,1)$, with an empty set $S$. You need to walk along the edges and eventually reach $(n,n)$. Whenever you pass an edge, its value will be inserted into $S$. Please maximize the MEX$^{\text{∗}}$ of $S$ when you reach $(n,n)$.

$^{\text{∗}}$The MEX (minimum excluded) of an array is the smallest non- negative integer that does not belong to the array. For instance:

  * The MEX of $[2,2,1]$ is $0$, because $0$ does not belong to the array.    * The MEX of $[3,1,0,1]$ is $2$, because $0$ and $1$ belong to the array, but $2$ does not.    * The MEX of $[0,3,1,2]$ is $4$, because $0, 1, 2$, and $3$ belong to the array, but $4$ does not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\le n\le20$) — the number of rows and columns.

Each of the next $n-1$ lines contains $n$ integers separated by single spaces — the matrix $d$ ($0\le d_{x,y}\le 2n-2$).

Each of the next $n$ lines contains $n-1$ integers separated by single spaces — the matrix $r$ ($0\le r_{x,y}\le 2n-2$).

It is guaranteed that the sum of all $n^3$ does not exceed $8000$.

Output: For each test case, print a single integer — the maximum MEX of $S$ when you reach $(n,n)$.

Examples:
input:
2
3
1 0 2
0 1 3
2 1
0 3
3 0
3
1 2 0
0 1 2
2 0
1 2
0 1
output:
3
2

input:
1
10
16 7 3 15 9 17 1 15 9 0
4 3 1 12 13 10 10 14 6 12
3 1 3 9 5 16 0 12 7 12
11 4 8 7 13 7 15 13 9 2
2 3 9 9 4 12 17 7 10 15
10 6 15 17 13 6 15 9 4 9
13 3 3 14 1 2 10 10 12 16
8 2 9 13 18 7 1 6 2 6
15 12 2 6 0 0 13 3 7 17
7 3 17 17 10 15 12 14 15
4 3 3 17 3 13 11 16 6
16 17 7 7 12 5 2 4 10
18 9 9 3 5 9 1 16 7
1 0 4 2 10 10 12 2 1
4 14 15 16 15 5 8 4 18
7 18 10 11 2 0 14 8 18
2 17 6 0 9 6 13 5 11
5 15 7 11 6 3 17 14 5
1 3 16 16 13 1 0 13 11
output:
14

Note: In the first test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/70956fd41a5289db10b3b0bb41d0efae)

In the second test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/c6a0ac2a80551ddd517e35658fa66438)

Buggy Code:
import sys
from collections import defaultdict

def solve():
    n = int(input())
    d = [[int(x) for x in input().split()] for _ in range(n-1)]
    r = [[int(x) for x in input().split()] for _ in range(n)]

    dp = [[defaultdict(bool) for _ in range(n)] for _ in range(n)]
    dp[0][0][0] = True

    for i in range(n):
        for j in range(n):
            if i > 0:
                for x in dp[i-1][j]:
                    dp[i][j][x | (1 << d[i-1][j])] = True
            if j > 0:
                for x in dp[i][j-1]:
                    dp[i][j][x | (1 << r[i][j-1])] = True

    ans = 0
    for x in dp[n-1][n-1]:
        for i in range(2*n):
            if (x >> i) & 1 == 0:
                ans = max(ans, i)
                break

    print(ans)

if __name__ == '__main__':
    t = int(input())
    for _ in range(t):
        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 given buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The key issues in the buggy code are:

1. The bitwise operations used to track the MEX values are not correctly handling the MEX calculation.
2. The code does not correctly handle the case where the MEX can be a value that is not in the range of the bitwise operations.

To create a test input that exposes these issues, we can use a small grid where the values of the edges are carefully chosen to create a situation where the MEX calculation is incorrect. Here is a step-by-step approach to generate such a test input:

1. **Choose a small grid size**: A 3x3 grid is sufficient to demonstrate the issue.
2. **Set edge values**: Set the edge values such that the MEX calculation is challenging. For example, use values that are powers of 2 to make the bitwise operations more interesting.
3. **Ensure the MEX is not immediately obvious**: Use values that cover a range of possible MEX values, but not all, to make the MEX calculation non-trivial.

Here is the test input:

```plaintext
1
3
1 0 2
0 1 3
2 1
0 3
3 0
```

Explanation:
- The grid size is 3x3.
- The down edges (d) are: `1 0 2` for the first row, `0 1 3` for the second row.
- The right edges (r) are: `2 1` for the first column, `0 3` for the second column, `3 0` for the third column.

This input is designed to:
- Test the bitwise operations by using values that are powers of 2.
- Ensure that the MEX calculation is not trivial and requires correct handling of the bitwise operations.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lattice Optimizing
time_limit_ms: 7000
memory_limit_mb: 1024
Description: Consider a grid graph with $n$ rows and $n$ columns. Let the cell in row $x$ and column $y$ be $(x,y)$. There exists a directed edge from $(x,y)$ to $(x+1,y)$, with non-negative integer value $d_{x,y}$, for all $1\le x < n, 1\le y \le n$, and there also exists a directed edge from $(x,y)$ to $(x,y+1)$, with non-negative integer value $r_{x,y}$, for all $1\le x \le n, 1\le y < n$.

Initially, you are at $(1,1)$, with an empty set $S$. You need to walk along the edges and eventually reach $(n,n)$. Whenever you pass an edge, its value will be inserted into $S$. Please maximize the MEX$^{\text{∗}}$ of $S$ when you reach $(n,n)$.

$^{\text{∗}}$The MEX (minimum excluded) of an array is the smallest non- negative integer that does not belong to the array. For instance:

  * The MEX of $[2,2,1]$ is $0$, because $0$ does not belong to the array.    * The MEX of $[3,1,0,1]$ is $2$, because $0$ and $1$ belong to the array, but $2$ does not.    * The MEX of $[0,3,1,2]$ is $4$, because $0, 1, 2$, and $3$ belong to the array, but $4$ does not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\le n\le20$) — the number of rows and columns.

Each of the next $n-1$ lines contains $n$ integers separated by single spaces — the matrix $d$ ($0\le d_{x,y}\le 2n-2$).

Each of the next $n$ lines contains $n-1$ integers separated by single spaces — the matrix $r$ ($0\le r_{x,y}\le 2n-2$).

It is guaranteed that the sum of all $n^3$ does not exceed $8000$.

Output: For each test case, print a single integer — the maximum MEX of $S$ when you reach $(n,n)$.

Examples:
input:
2
3
1 0 2
0 1 3
2 1
0 3
3 0
3
1 2 0
0 1 2
2 0
1 2
0 1
output:
3
2

input:
1
10
16 7 3 15 9 17 1 15 9 0
4 3 1 12 13 10 10 14 6 12
3 1 3 9 5 16 0 12 7 12
11 4 8 7 13 7 15 13 9 2
2 3 9 9 4 12 17 7 10 15
10 6 15 17 13 6 15 9 4 9
13 3 3 14 1 2 10 10 12 16
8 2 9 13 18 7 1 6 2 6
15 12 2 6 0 0 13 3 7 17
7 3 17 17 10 15 12 14 15
4 3 3 17 3 13 11 16 6
16 17 7 7 12 5 2 4 10
18 9 9 3 5 9 1 16 7
1 0 4 2 10 10 12 2 1
4 14 15 16 15 5 8 4 18
7 18 10 11 2 0 14 8 18
2 17 6 0 9 6 13 5 11
5 15 7 11 6 3 17 14 5
1 3 16 16 13 1 0 13 11
output:
14

Note: In the first test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/70956fd41a5289db10b3b0bb41d0efae)

In the second test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/c6a0ac2a80551ddd517e35658fa66438)

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define lll __int128
#define pc __builtin_popcount
#define pr pair<int,int>
#define pb push_back
#define mp make_pair
#define x first
#define y second
using namespace std;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
ll rint(ll l,ll r){return uniform_int_distribution<ll>(l,r)(rnd);}
const int maxn=30;
int T,n,d,ans,a[maxn][maxn],b[maxn][maxn];
unordered_set<ll> f[maxn];
void read(int &X)
{
	X=0;
	char cr=getchar();
	while(cr<'0'||cr>'9') cr=getchar();
	while(cr>='0'&&cr<='9') X=(X<<3)+(X<<1)+cr-'0',cr=getchar();
}
void dfs1(int x,int y,ll val)
{
	if(x+y==d)
	{
		for(ll i=val;i;i=(i-1)&val) f[x].insert(i);
		return;
	}
	dfs1(x+1,y,val|(1ll<<a[x][y])),dfs1(x,y+1,val|(1ll<<b[x][y]));
}
void dfs2(int x,int y,ll val)
{
	if(x+y==d)
	{
		while(1)
		{
			ll tmp=(1ll<<(ans+1))-1;
			if(((val&tmp)^tmp)==0||f[x].count((val&tmp)^tmp)) ans++;
			else break;
		}
		return;
	}
	if(x) dfs2(x-1,y,val|(1ll<<a[x-1][y]));
	if(y) dfs2(x,y-1,val|(1ll<<b[x][y-1]));
}
int main()
{
//	freopen("20.in","r",stdin);
//	freopen(".out","w",stdout);
	read(T);
	while(T--)
	{
		read(n),d=(n<<1)/3,ans=0;
		if(n>15) d=(n<<1)/3-2;
		for(int i=0;i<n;i++) f[i].clear();
		for(int i=0;i<n-1;i++) for(int j=0;j<n;j++) read(a[i][j]);
		for(int i=0;i<n;i++) for(int j=0;j<n-1;j++) read(b[i][j]);
		dfs1(0,0,0),dfs2(n-1,n-1,0),printf("%d\n",ans);
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
4
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lattice Optimizing
time_limit_ms: 7000
memory_limit_mb: 1024
Description: Consider a grid graph with $n$ rows and $n$ columns. Let the cell in row $x$ and column $y$ be $(x,y)$. There exists a directed edge from $(x,y)$ to $(x+1,y)$, with non-negative integer value $d_{x,y}$, for all $1\le x < n, 1\le y \le n$, and there also exists a directed edge from $(x,y)$ to $(x,y+1)$, with non-negative integer value $r_{x,y}$, for all $1\le x \le n, 1\le y < n$.

Initially, you are at $(1,1)$, with an empty set $S$. You need to walk along the edges and eventually reach $(n,n)$. Whenever you pass an edge, its value will be inserted into $S$. Please maximize the MEX$^{\text{∗}}$ of $S$ when you reach $(n,n)$.

$^{\text{∗}}$The MEX (minimum excluded) of an array is the smallest non- negative integer that does not belong to the array. For instance:

  * The MEX of $[2,2,1]$ is $0$, because $0$ does not belong to the array.    * The MEX of $[3,1,0,1]$ is $2$, because $0$ and $1$ belong to the array, but $2$ does not.    * The MEX of $[0,3,1,2]$ is $4$, because $0, 1, 2$, and $3$ belong to the array, but $4$ does not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\le n\le20$) — the number of rows and columns.

Each of the next $n-1$ lines contains $n$ integers separated by single spaces — the matrix $d$ ($0\le d_{x,y}\le 2n-2$).

Each of the next $n$ lines contains $n-1$ integers separated by single spaces — the matrix $r$ ($0\le r_{x,y}\le 2n-2$).

It is guaranteed that the sum of all $n^3$ does not exceed $8000$.

Output: For each test case, print a single integer — the maximum MEX of $S$ when you reach $(n,n)$.

Examples:
input:
2
3
1 0 2
0 1 3
2 1
0 3
3 0
3
1 2 0
0 1 2
2 0
1 2
0 1
output:
3
2

input:
1
10
16 7 3 15 9 17 1 15 9 0
4 3 1 12 13 10 10 14 6 12
3 1 3 9 5 16 0 12 7 12
11 4 8 7 13 7 15 13 9 2
2 3 9 9 4 12 17 7 10 15
10 6 15 17 13 6 15 9 4 9
13 3 3 14 1 2 10 10 12 16
8 2 9 13 18 7 1 6 2 6
15 12 2 6 0 0 13 3 7 17
7 3 17 17 10 15 12 14 15
4 3 3 17 3 13 11 16 6
16 17 7 7 12 5 2 4 10
18 9 9 3 5 9 1 16 7
1 0 4 2 10 10 12 2 1
4 14 15 16 15 5 8 4 18
7 18 10 11 2 0 14 8 18
2 17 6 0 9 6 13 5 11
5 15 7 11 6 3 17 14 5
1 3 16 16 13 1 0 13 11
output:
14

Note: In the first test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/70956fd41a5289db10b3b0bb41d0efae)

In the second test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/c6a0ac2a80551ddd517e35658fa66438)

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

public class Main {
    private static BufferedReader br;
    private static StringTokenizer tokenizer;
    private static String next() throws IOException{
        if (br == null) br = new BufferedReader(new InputStreamReader(System.in));
        if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());
        while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }
    private static long nextLong() throws IOException{
        return Long.parseLong(next());
    }

    private static StringBuilder sb = new StringBuilder();

    private static int ans;
    private static int[][] d;
    private static int[][] r;
    private static int n;
    private static int l;
    private static Set<Long>[] groups = new HashSet[20];
    private static void dfs1(int x, int y, long mark) {
        if (x + y == l) {
            groups[x].add(mark);
            return;
        }

        if (x < n - 1) {
            dfs1(x + 1, y, mark);
            if(((mark >> d[x][y]) & 1) == 0) dfs1(x + 1, y, mark | (1L << d[x][y]));
        }
        if (y < n - 1) {
            dfs1(x, y + 1, mark);
            if(((mark >> r[x][y]) & 1) == 0) dfs1(x, y + 1, mark | (1L << r[x][y]));
        }
    }
    private static void dfs2(int x, int y, long mark) {
        if (x + y == l) {
            Set<Long> set = groups[x];
            long v = (1L << (ans + 1)) - 1;
            while ((mark | v) == mark || set.contains(v ^ (v & mark))) {
                ++ans;
                v = (v << 1) + 1;
            }
            return;
        }

        if (x > 0) dfs2(x - 1, y, mark | (1L << d[x - 1][y]));
        if (y > 0) dfs2(x, y - 1, mark | (1L << r[x][y - 1]));
    }

    private static void slove() throws IOException{
        n = nextInt();
        d = new int[n - 1][n];
        r = new int[n][n - 1];
        for (int i = 0; i < d.length; ++i) {
            for (int j = 0; j < d[0].length; ++j) d[i][j] = nextInt();
        }
        for (int i = 0; i < r.length; ++i) {
            for (int j = 0; j < r[0].length; ++j) r[i][j] = nextInt();
        }
        l = (int)(2.0/3*n);
        for (int i = 0; i < groups.length; i++) groups[i] = new HashSet<>();
        ans = 0;
        dfs1(0, 0, 0L);
        dfs2(n - 1, n - 1, 0L);
        sb.append(ans).append("\n");
    }

    public static void main(String[] args)  throws IOException {
        int tt = nextInt();
        sb = new StringBuilder();
        for (int i = 0; i < tt; ++i) slove();
        System.out.println(sb);
    }
}

**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, let's consider the following aspects:

1. **Edge Cases:**
   - Small grid sizes (minimum and maximum limits).
   - Grids with all edges having the same value.
   - Grids with a pattern that could cause the algorithm to miss the optimal path.

2. **Optimal Path Complexity:**
   - Grids where the optimal path is not straightforward and requires careful selection of edges.
   - Grids where the MEX value is high, requiring the algorithm to correctly handle large values.

3. **Boundary Conditions:**
   - Grids where the values of edges are at the boundary of the allowed range.
   - Grids where the values of edges are zero, which can affect the MEX calculation.

Given these considerations, a small-scale test input that could expose bugs in the code is:

```plaintext
3
3
0 0 0
0 0 0
0 0
0 0
3
2 2 2
2 2 2
2 2
2 2
3
1 2 3
4 5 6
7 8
9 10
```

### Explanation:
1. **First Test Case:**
   - All edges have a value of 0.
   - The optimal path will collect all zeros, and the MEX will be 1.

2. **Second Test Case:**
   - All edges have a value of 2.
   - The optimal path will collect all 2s, and the MEX will be 0.

3. **Third Test Case:**
   - The values are in a sequence that could challenge the algorithm to find the optimal path.
   - The optimal path should maximize the MEX, which in this case should be 1.

This input will test the algorithm's ability to handle uniform values, zero values, and a sequence of values that require careful path selection.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lattice Optimizing
time_limit_ms: 7000
memory_limit_mb: 1024
Description: Consider a grid graph with $n$ rows and $n$ columns. Let the cell in row $x$ and column $y$ be $(x,y)$. There exists a directed edge from $(x,y)$ to $(x+1,y)$, with non-negative integer value $d_{x,y}$, for all $1\le x < n, 1\le y \le n$, and there also exists a directed edge from $(x,y)$ to $(x,y+1)$, with non-negative integer value $r_{x,y}$, for all $1\le x \le n, 1\le y < n$.

Initially, you are at $(1,1)$, with an empty set $S$. You need to walk along the edges and eventually reach $(n,n)$. Whenever you pass an edge, its value will be inserted into $S$. Please maximize the MEX$^{\text{∗}}$ of $S$ when you reach $(n,n)$.

$^{\text{∗}}$The MEX (minimum excluded) of an array is the smallest non- negative integer that does not belong to the array. For instance:

  * The MEX of $[2,2,1]$ is $0$, because $0$ does not belong to the array.    * The MEX of $[3,1,0,1]$ is $2$, because $0$ and $1$ belong to the array, but $2$ does not.    * The MEX of $[0,3,1,2]$ is $4$, because $0, 1, 2$, and $3$ belong to the array, but $4$ does not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\le n\le20$) — the number of rows and columns.

Each of the next $n-1$ lines contains $n$ integers separated by single spaces — the matrix $d$ ($0\le d_{x,y}\le 2n-2$).

Each of the next $n$ lines contains $n-1$ integers separated by single spaces — the matrix $r$ ($0\le r_{x,y}\le 2n-2$).

It is guaranteed that the sum of all $n^3$ does not exceed $8000$.

Output: For each test case, print a single integer — the maximum MEX of $S$ when you reach $(n,n)$.

Examples:
input:
2
3
1 0 2
0 1 3
2 1
0 3
3 0
3
1 2 0
0 1 2
2 0
1 2
0 1
output:
3
2

input:
1
10
16 7 3 15 9 17 1 15 9 0
4 3 1 12 13 10 10 14 6 12
3 1 3 9 5 16 0 12 7 12
11 4 8 7 13 7 15 13 9 2
2 3 9 9 4 12 17 7 10 15
10 6 15 17 13 6 15 9 4 9
13 3 3 14 1 2 10 10 12 16
8 2 9 13 18 7 1 6 2 6
15 12 2 6 0 0 13 3 7 17
7 3 17 17 10 15 12 14 15
4 3 3 17 3 13 11 16 6
16 17 7 7 12 5 2 4 10
18 9 9 3 5 9 1 16 7
1 0 4 2 10 10 12 2 1
4 14 15 16 15 5 8 4 18
7 18 10 11 2 0 14 8 18
2 17 6 0 9 6 13 5 11
5 15 7 11 6 3 17 14 5
1 3 16 16 13 1 0 13 11
output:
14

Note: In the first test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/70956fd41a5289db10b3b0bb41d0efae)

In the second test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/c6a0ac2a80551ddd517e35658fa66438)

Buggy Code:
#include <bits/stdc++.h>
#define endl '\n'
using namespace std;

struct TRIE{
    TRIE(){
        ptr = 2;
        root = 1;
    }

    void clear(){
        ptr = 2;
        root = 1;
        trie[root][0] = 0;
        trie[root][1] = 0;
    }

    void add(long long X, int pos){
        int head = root;
        for(int i = 0; i <= 38; i ++){
            int bit = X & 1;
            if(trie[head][bit] == 0){
                trie[head][bit] = ptr++;
                trie[ptr - 1][0] = 0;
                trie[ptr - 1][1] = 0;
            }
            head = trie[head][bit];
            X /= 2;
        }
    }

    int check(long long C, int head, int d){
        int ans = 0;
        int b2 = C & 1;
        if(head == 0){
            return d - 1;
        }
        if(b2){
            return max(check(C / 2, trie[head][1], d + 1), check(C / 2, trie[head][0], d + 1));
        }
        else{
            return check(C / 2, trie[head][1], d + 1);
        }
    }

    int root;
    int ptr;
    int trie[20000000][2] = {0};
};
void solve(){
    int N;
    cin >> N;
    vector<vector<int>> cw(N - 1, vector<int>(N)), rw(N, vector<int>(N - 1));
    set<int> have;
    for(int i = 0; i < N - 1; i ++){
        for(int j = 0; j < N; j ++){
            cin >> cw[i][j];
            have.insert(cw[i][j]);
        }
    }
    for(int i = 0; i < N; i ++){
        for(int j = 0; j < N - 1; j ++){
            cin >> rw[i][j];
            have.insert(rw[i][j]);
        }
    }
    vector<vector<set<long long>>> gridtop(N, vector<set<long long>>(N)), gridbot(N, vector<set<long long>>(N));
    gridtop[0][0].insert(0);
    gridbot.back().back().insert(0);
    for(int i = 0; i < N; i ++){
        for(int j = 0; j < N && j + i < min(N * 2, N + 3); j ++){
            if(i){
                for(auto k : gridtop[i - 1][j]){
                    gridtop[i][j].insert((k | (1ll << cw[i - 1][j])));
                }
            }
            if(j){
                for(auto k : gridtop[i][j - 1]){
                    gridtop[i][j].insert((k | (1ll << rw[i][j - 1])));
                }
            }
        }
    }
    for(int i = N - 1; i >= 0; i --){
        for(int j = N - 1; j >= 0 && j + i >= min(N + 2, N * 2 - 2); j --){
            if(i < N - 1){
                for(auto k : gridbot[i + 1][j]){
                    gridbot[i][j].insert((k | (1ll << cw[i][j])));
                }
            }
            if(j < N - 1){
                for(auto k : gridbot[i][j + 1]){
                    gridbot[i][j].insert((k | (1ll << rw[i][j])));
                }
            }
        }
    }
    int mex = 0;
    while(have.count(mex)){
        mex ++;
    }
    // cout << "TOP GRID" << endl;
    // for(auto i : gridtop){
    //     for(auto j : i){
    //         cout << setw(5) << j.size() << " ";
    //     }
    //     cout << endl;
    // }
    // cout << "BOTTOM GRID" << endl;
    // for(auto i : gridbot){
    //     for(auto j : i){
    //         cout << setw(5) << j.size() << " ";
    //     }
    //     cout << endl;
    // }
    TRIE trie;
    int ans = 0;
    vector<int> order(N);
    iota(order.begin(), order.end(), 0);
    random_shuffle(order.begin(), order.end());
    for(int i : order){
        for(int j : order){
            if(i + j != min(N * 2 - 2, N + 2) || ans == mex){continue;}
            for(auto ii : gridtop[i][j]){
                trie.add(ii, ii);
            }
            vector<int> cc;
            for(int ii : gridbot[i][j]){
                cc.push_back(ii);
            }
            random_shuffle(cc.begin(), cc.end());
            for(auto ii : cc){
                ans = max(ans, trie.check(ii, 1, 0));
                if(ans == mex){
                    break;
                }
            }
            trie.clear();
        }
    }
    cout << ans << endl;
}

signed main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int N = 1;
    cin >> N;
    while(N--)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 test input that exposes the bug in the given buggy code, we need to focus on edge cases or scenarios that can cause the algorithm to fail. The key aspects to consider are:

1. **Edge Cases:**
   - Small values of \( n \) (e.g., \( n = 2 \)).
   - Large values of \( n \) (e.g., \( n = 20 \)).
   - Special values in the matrices \( d \) and \( r \) that can lead to incorrect MEX calculations.

2. **Optimal Path Analysis:**
   - Ensure that the path taken in the grid maximizes the MEX value.
   - Consider scenarios where the MEX value is influenced by the order in which values are added to the set \( S \).

3. **Edge Values:**
   - Values in the matrices \( d \) and \( r \) should be chosen to maximize the MEX value.
   - Include values that are close to the maximum possible value \( 2n-2 \) to test the boundary conditions.

Given these considerations, let's create a small-scale test input that can expose the bug in the code.

### Test Input
```plaintext
1
4
0 1 2 3
1 2 3 0
2 3 0 1
3 0 1 2
0 1 2 3
1 2 3 0
2 3 0 1
3 0 1 2
```

### Explanation
- **Grid Size:** \( n = 4 \)
- **Matrices \( d \) and \( r \):**
  - The values are chosen such that they form a pattern that can influence the MEX calculation.
  - The values are small and range from 0 to 3, which is within the allowed range \( 0 \le d_{x,y}, r_{x,y} \le 2n-2 \).

### Expected Output
The expected output for this test input should be the maximum MEX value that can be achieved by traversing the grid from \((1,1)\) to \((4,4)\). Given the pattern in the matrices, the MEX value should be 4, as the set \( S \) can contain the values 0, 1, 2, and 3, and the smallest non-negative integer not in the set is 4.

### Why This Test Input?
- **Pattern in Values:** The values in the matrices form a repeating pattern, which can help in identifying if the algorithm correctly handles the MEX calculation.
- **Small Grid Size:** The grid size is small enough to be manageable and easy to verify the correctness of the output.
- **Edge Case:** The values are chosen to be close to the boundary of the allowed range, which can help in exposing any issues with the boundary conditions in the algorithm.

This test input should help in exposing any bugs in the given buggy code by testing the MEX calculation and the path traversal logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lattice Optimizing
time_limit_ms: 7000
memory_limit_mb: 1024
Description: Consider a grid graph with $n$ rows and $n$ columns. Let the cell in row $x$ and column $y$ be $(x,y)$. There exists a directed edge from $(x,y)$ to $(x+1,y)$, with non-negative integer value $d_{x,y}$, for all $1\le x < n, 1\le y \le n$, and there also exists a directed edge from $(x,y)$ to $(x,y+1)$, with non-negative integer value $r_{x,y}$, for all $1\le x \le n, 1\le y < n$.

Initially, you are at $(1,1)$, with an empty set $S$. You need to walk along the edges and eventually reach $(n,n)$. Whenever you pass an edge, its value will be inserted into $S$. Please maximize the MEX$^{\text{∗}}$ of $S$ when you reach $(n,n)$.

$^{\text{∗}}$The MEX (minimum excluded) of an array is the smallest non- negative integer that does not belong to the array. For instance:

  * The MEX of $[2,2,1]$ is $0$, because $0$ does not belong to the array.    * The MEX of $[3,1,0,1]$ is $2$, because $0$ and $1$ belong to the array, but $2$ does not.    * The MEX of $[0,3,1,2]$ is $4$, because $0, 1, 2$, and $3$ belong to the array, but $4$ does not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\le n\le20$) — the number of rows and columns.

Each of the next $n-1$ lines contains $n$ integers separated by single spaces — the matrix $d$ ($0\le d_{x,y}\le 2n-2$).

Each of the next $n$ lines contains $n-1$ integers separated by single spaces — the matrix $r$ ($0\le r_{x,y}\le 2n-2$).

It is guaranteed that the sum of all $n^3$ does not exceed $8000$.

Output: For each test case, print a single integer — the maximum MEX of $S$ when you reach $(n,n)$.

Examples:
input:
2
3
1 0 2
0 1 3
2 1
0 3
3 0
3
1 2 0
0 1 2
2 0
1 2
0 1
output:
3
2

input:
1
10
16 7 3 15 9 17 1 15 9 0
4 3 1 12 13 10 10 14 6 12
3 1 3 9 5 16 0 12 7 12
11 4 8 7 13 7 15 13 9 2
2 3 9 9 4 12 17 7 10 15
10 6 15 17 13 6 15 9 4 9
13 3 3 14 1 2 10 10 12 16
8 2 9 13 18 7 1 6 2 6
15 12 2 6 0 0 13 3 7 17
7 3 17 17 10 15 12 14 15
4 3 3 17 3 13 11 16 6
16 17 7 7 12 5 2 4 10
18 9 9 3 5 9 1 16 7
1 0 4 2 10 10 12 2 1
4 14 15 16 15 5 8 4 18
7 18 10 11 2 0 14 8 18
2 17 6 0 9 6 13 5 11
5 15 7 11 6 3 17 14 5
1 3 16 16 13 1 0 13 11
output:
14

Note: In the first test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/70956fd41a5289db10b3b0bb41d0efae)

In the second test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/c6a0ac2a80551ddd517e35658fa66438)

Buggy Code:
//#include <bits/stdc++.h>
#include <vector>
#include <iostream>
#include <algorithm>
#include <unordered_set>

#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2")
using namespace std;

#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

const int INFi = 1e9 + 7;
const ll INFll = 1e18 + 7;

const int mod = 998244353;
//const int mod = 1e9 + 7;
const int maxn = 10007;

int n;
ll d[20][20], r[20][20];
vector<ll> res[2][21][21];

void solve() {
    cin >> n;
    ll max_el = 0;
    for (int i = 0; i < n - 1; ++i) {
        for (int j = 0; j < n; ++j) {
            cin >> d[i][j];

            d[i][j] = 1LL << d[i][j];
            max_el = max_el | d[i][j];
        }
    }
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n - 1; ++j) {
            cin >> r[i][j];

            r[i][j] = 1LL << r[i][j];
            max_el = max_el | r[i][j];
        }
    }
    for (int i = 0; i < 2 * n; ++i) {
        if ((max_el & (1 << i)) == 0) {
            max_el = (1 << (i + 1)) - 1;
            break;
        }
    }
    int k = (2 * n - 2 + 2) / 3;
    res[0][0][0].clear();
    res[0][0][0].push_back(0);
    for (int sum = 1; sum <= k; ++sum) {
        for (int x = 0; x <= sum; ++x) {
            int y = sum - x;
            res[0][x][y].clear();
            int kek = 0;
            if (x > 0) {
                kek += res[0][x - 1][y].size();
            }
            if (y > 0) {
                kek += res[0][x][y - 1].size();
            }
            res[0][x][y].reserve(kek);
            if (x > 0) {
                for (ll mask: res[0][x - 1][y]) {
                    res[0][x][y].push_back(mask | d[x - 1][y]);
                }
            }
            if (y > 0) {
                for (ll mask: res[0][x][y - 1]) {
                    res[0][x][y].push_back(mask | r[x][y - 1]);
                }
            }
        }
    }
    res[1][n - 1][n - 1].clear();
    res[1][n - 1][n - 1].push_back(0);
    for (int i = 0; i < n; ++i) {
        res[1][i][n].clear();
        res[1][n][i].clear();
    }
    for (int sum = 2 * n - 3; sum >= k; --sum) {
        for (int x = 0; x < n; ++x) {
            int y = sum - x;
            if (y < 0 || y >= n) {
                continue;
            }
            res[1][x][y].clear();
            res[1][x][y].reserve(res[1][x + 1][y].size() + res[1][x][y + 1].size());
            for (ll mask: res[1][x + 1][y]) {
                res[1][x][y].push_back(mask | d[x][y]);
            }
            for (ll mask: res[1][x][y + 1]) {
                res[1][x][y].push_back(mask | r[x][y]);
            }
            std::vector<ll>().swap(res[1][x][y + 1]);
        }
    }

    ll ans = 1;

    for (int x = 0; x <= k; ++x) {
        int y = k - x;
        unordered_set<ll> pos;
        pos.insert(0);
        ll p = 0;
        sort(all(res[0][x][y]));
        for (ll mask: res[0][x][y]) {
            if (mask == p) {
                continue;
            }
            p =mask;
            for (ll i = mask; i > 0; i = (i - 1) & mask) {
                pos.insert(i);
            }
        }
        for (ll mask: res[1][x][y]) {
            while (pos.find((~mask) & ans) != pos.end()) {
                ans = (ans << 1) + 1;
            }
            if (ans == max_el) {
                break;
            }
        }
    }
    int cnt = 0;
    for (int i = 0; i < 2 * n; ++i) {
        cnt += (ans & (1 << i)) != 0;
    }
    cout << cnt - 1 << '\n';
}

int main() {
//    stress();
//    prec();
  //  freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
    cin.tie(0);
    //cout.tie(0);
    ios_base::sync_with_stdio(false);
    int t = 1;
    cin >> t;
    for (int i = 0; i < t; i++) {
        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 provided buggy code, let's consider the following points:

1. **Edge Cases**: The algorithm involves bitwise operations and dynamic programming. Edge cases can include:
   - Small grid sizes (e.g., 2x2, 3x3).
   - Grids with all values being 0.
   - Grids with values that are powers of 2.
   - Grids with values that are not powers of 2 but close to the maximum allowed values.

2. **Complex Scenarios**: The algorithm should handle scenarios where the path choices are critical. For example:
   - Grids where the optimal path is not straightforward and requires careful selection of edges to maximize the MEX.
   - Grids where the values are distributed in a way that makes the bitwise operations more complex.

3. **Boundary Conditions**: Test the limits of the input constraints:
   - Maximum grid size (20x20).
   - Maximum values for edges (2n-2).

Given these considerations, let's create a small-scale test input that is complete and valid:

```plaintext
1
3
0 1 2
1 0 2
2 1 0
1 2 0
2 1 0
0 1 2
```

**Explanation**:
- **Grid Size**: 3x3, which is small enough to be manageable but large enough to test the algorithm's logic.
- **Edge Values**:
  - The values in the `d` matrix are 0, 1, and 2, which are small and easy to handle but can still test the bitwise operations.
  - The values in the `r` matrix are also 0, 1, and 2, which are small and can create interesting paths.

This test input will help expose any issues in the bitwise operations and dynamic programming logic used in the buggy code. The values are chosen to ensure that the MEX calculation is non-trivial and requires careful handling of the edge values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lattice Optimizing
time_limit_ms: 7000
memory_limit_mb: 1024
Description: Consider a grid graph with $n$ rows and $n$ columns. Let the cell in row $x$ and column $y$ be $(x,y)$. There exists a directed edge from $(x,y)$ to $(x+1,y)$, with non-negative integer value $d_{x,y}$, for all $1\le x < n, 1\le y \le n$, and there also exists a directed edge from $(x,y)$ to $(x,y+1)$, with non-negative integer value $r_{x,y}$, for all $1\le x \le n, 1\le y < n$.

Initially, you are at $(1,1)$, with an empty set $S$. You need to walk along the edges and eventually reach $(n,n)$. Whenever you pass an edge, its value will be inserted into $S$. Please maximize the MEX$^{\text{∗}}$ of $S$ when you reach $(n,n)$.

$^{\text{∗}}$The MEX (minimum excluded) of an array is the smallest non- negative integer that does not belong to the array. For instance:

  * The MEX of $[2,2,1]$ is $0$, because $0$ does not belong to the array.    * The MEX of $[3,1,0,1]$ is $2$, because $0$ and $1$ belong to the array, but $2$ does not.    * The MEX of $[0,3,1,2]$ is $4$, because $0, 1, 2$, and $3$ belong to the array, but $4$ does not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\le n\le20$) — the number of rows and columns.

Each of the next $n-1$ lines contains $n$ integers separated by single spaces — the matrix $d$ ($0\le d_{x,y}\le 2n-2$).

Each of the next $n$ lines contains $n-1$ integers separated by single spaces — the matrix $r$ ($0\le r_{x,y}\le 2n-2$).

It is guaranteed that the sum of all $n^3$ does not exceed $8000$.

Output: For each test case, print a single integer — the maximum MEX of $S$ when you reach $(n,n)$.

Examples:
input:
2
3
1 0 2
0 1 3
2 1
0 3
3 0
3
1 2 0
0 1 2
2 0
1 2
0 1
output:
3
2

input:
1
10
16 7 3 15 9 17 1 15 9 0
4 3 1 12 13 10 10 14 6 12
3 1 3 9 5 16 0 12 7 12
11 4 8 7 13 7 15 13 9 2
2 3 9 9 4 12 17 7 10 15
10 6 15 17 13 6 15 9 4 9
13 3 3 14 1 2 10 10 12 16
8 2 9 13 18 7 1 6 2 6
15 12 2 6 0 0 13 3 7 17
7 3 17 17 10 15 12 14 15
4 3 3 17 3 13 11 16 6
16 17 7 7 12 5 2 4 10
18 9 9 3 5 9 1 16 7
1 0 4 2 10 10 12 2 1
4 14 15 16 15 5 8 4 18
7 18 10 11 2 0 14 8 18
2 17 6 0 9 6 13 5 11
5 15 7 11 6 3 17 14 5
1 3 16 16 13 1 0 13 11
output:
14

Note: In the first test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/70956fd41a5289db10b3b0bb41d0efae)

In the second test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/c6a0ac2a80551ddd517e35658fa66438)

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

public class Main {
    private static BufferedReader br;
    private static StringTokenizer tokenizer;
    private static String next() throws IOException{
        if (br == null) br = new BufferedReader(new InputStreamReader(System.in));
        if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());
        while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }
    private static long nextLong() throws IOException{
        return Long.parseLong(next());
    }

    private static StringBuilder sb = new StringBuilder();

    private static int ans;
    private static int[][] d;
    private static int[][] r;
    private static int n;
    private static int l;
    private static Set<Long>[] groups = new HashSet[20];
    private static void dfs1(int x, int y, long mark) {
        if (x + y == l) {
            groups[x].add(0L);
            for (long tm = mark; tm > 0; tm = (tm - 1) & mark)  groups[x].add(tm);
            return;
        }

        if (x < n - 1) dfs1(x + 1, y, mark | (1L << d[x][y]));
        if (y < n - 1) dfs1(x, y + 1, mark | (1L << r[x][y]));
    }
    private static void dfs2(int x, int y, long mark) {
        if (x + y == l) {
            Set<Long> set = groups[x];
            long v = (1L << (ans + 1)) - 1;
            while (set.contains((~mark) & v)) {
                ++ans;
                v = (v << 1) + 1;
            }
            return;
        }

        if (x > 0) dfs2(x - 1, y, mark | (1L << d[x - 1][y]));
        if (y > 0) dfs2(x, y - 1, mark | (1L << r[x][y - 1]));
    }

    private static void slove() throws IOException{
        n = nextInt();
        d = new int[n - 1][n];
        r = new int[n][n - 1];
        for (int i = 0; i < d.length; ++i) {
            for (int j = 0; j < d[0].length; ++j) d[i][j] = nextInt();
        }
        for (int i = 0; i < r.length; ++i) {
            for (int j = 0; j < r[0].length; ++j) r[i][j] = nextInt();
        }
        l = 2 * (n - 1) / 3;
        for (int i = 0; i < groups.length; i++) groups[i] = new HashSet<>();
        ans = 0;
        dfs1(0, 0, 0L);
        dfs2(n - 1, n - 1, 0L);
        sb.append(ans).append("\n");
    }

    public static void main(String[] args)  throws IOException {
        int tt = nextInt();
        sb = new StringBuilder();
        for (int i = 0; i < tt; ++i) slove();
        System.out.println(sb);
    }
}

**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 consider the following aspects:

1. **Edge Cases**: Test cases where the grid has specific patterns or values that can cause the algorithm to fail.
2. **Small Scale**: Ensure the test case is small enough to be manageable and valid according to the problem constraints.
3. **Specific Values**: Use values that can highlight the issues in the algorithm, such as values that might cause overflow or incorrect MEX calculations.

Given the nature of the problem and the provided code, a good test case would be one where the MEX calculation is not straightforward and the path choices are critical. Let's construct a test case with a 4x4 grid:

```plaintext
1
4
1 2 3 0
0 1 2 3
3 0 1 2
2 3 0 1
1 2 3
3 0 1
1 2 3
2 3 0
```

### Explanation:
- **Grid Size**: 4x4, which is within the constraints and manageable.
- **Matrix `d`**: Values are chosen such that they form a pattern that can challenge the MEX calculation.
- **Matrix `r`**: Values are chosen to complement the `d` matrix, ensuring that the path choices are not trivial.

This test case is designed to ensure that the algorithm must carefully consider the values it collects and the MEX calculation, which can expose issues in the buggy code's logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kevin and Combination Lock
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kevin is trapped in Lakeside Village by Grace. At the exit of the village, there is a combination lock that can only be unlocked if Kevin solves it.

The combination lock starts with an integer $$$ x $$$. Kevin can perform one of the following two operations zero or more times:

1. If $$$ x \neq 33 $$$, he can select two consecutive digits $$$ 3 $$$ from $$$ x $$$ and remove them simultaneously. For example, if $$$ x = 13\,323 $$$, he can remove the second and third $$$ 3 $$$, changing $$$ x $$$ to $$$ 123 $$$.
2. If $$$ x \geq 33 $$$, he can change $$$ x $$$ to $$$ x - 33 $$$. For example, if $$$ x = 99 $$$, he can choose this operation to change $$$ x $$$ to $$$ 99 - 33 = 66 $$$.

When the value of $$$ x $$$ on the combination lock becomes $$$ 0 $$$, Kevin can unlock the lock and escape from Lakeside Village. Please determine whether it is possible for Kevin to unlock the combination lock and escape.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$).

The only line of each test case contains a positive integer $$$x$$$ ($$$1\leq x\leq 10^9$$$).

Output: For each test case, output "YES" or "NO" (without quotes) in one line, representing whether Kevin can unlock the combination lock and escape. You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
165
6369
666
114514
133333332
output:
YES
YES
NO
NO
YES

Note: For the first test case, $$$165\xrightarrow{-33}132\xrightarrow{-33}99\xrightarrow{-33}66\xrightarrow{-33}33\xrightarrow{-33}0$$$.

For the second test case, $$$6369\xrightarrow{-33}6{\color{red}{33}}6\xrightarrow{\text{remove "33"}}66\xrightarrow{-33}33\xrightarrow{-33}0$$$.

For the third test case, it can be proven that, regardless of the operations performed, $$$666$$$ cannot be transformed into $$$0$$$.

Buggy Code:
#include <stdio.h>
#include <stdbool.h>
#include <string.h>
#include <stdlib.h>
bool check33(char str[])
{
    for (int i = 0; i < strlen(str) - 1; i++)
    {

        if (str[i == '3'] && str[i + 1] == '3')
        {
            return true;
        }
    }
    return false;
}
int main()
{

    int t;
    scanf("%d", &t);
    while (t--)
    {
        int n;
        scanf("%ld", &n);
        if (n < 33)
        {
            printf("NO\n");
            continue;
        }
        if (n == 33 || n % 33 == 0)
        {
            printf("YES\n");
            continue;
        }
        while (n > 33)

        {
            if (n % 33 == 0)
            {
                break;
            }
            char str[9];
            sprintf(str, "%ld", n);
            if (check33(str))
            {
                char l[9];
                int j = 0;
                for (int i = 0; i < strlen(str) - 1; i++)
                {
                    if (i = strlen(str) - 1)
                    {
                        l[j++] = str[i];
                        break;
                    }
                    if (str[i] == '3' && str[i + 1] == '3')
                    {
                        i++;
                        continue;
                    }

                    l[j++] = str[i];
                }
                l[j] = '\0';
                n = atoi(l);
            }
            else
            {
                n -= 33;
            }
        }
        if (n % 33 == 0)
        {
            printf("YES\n");
        }
        else
        {
            printf("NO\n");
        }
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
333
366
633
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kevin and Combination Lock
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kevin is trapped in Lakeside Village by Grace. At the exit of the village, there is a combination lock that can only be unlocked if Kevin solves it.

The combination lock starts with an integer $$$ x $$$. Kevin can perform one of the following two operations zero or more times:

1. If $$$ x \neq 33 $$$, he can select two consecutive digits $$$ 3 $$$ from $$$ x $$$ and remove them simultaneously. For example, if $$$ x = 13\,323 $$$, he can remove the second and third $$$ 3 $$$, changing $$$ x $$$ to $$$ 123 $$$.
2. If $$$ x \geq 33 $$$, he can change $$$ x $$$ to $$$ x - 33 $$$. For example, if $$$ x = 99 $$$, he can choose this operation to change $$$ x $$$ to $$$ 99 - 33 = 66 $$$.

When the value of $$$ x $$$ on the combination lock becomes $$$ 0 $$$, Kevin can unlock the lock and escape from Lakeside Village. Please determine whether it is possible for Kevin to unlock the combination lock and escape.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$).

The only line of each test case contains a positive integer $$$x$$$ ($$$1\leq x\leq 10^9$$$).

Output: For each test case, output "YES" or "NO" (without quotes) in one line, representing whether Kevin can unlock the combination lock and escape. You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
165
6369
666
114514
133333332
output:
YES
YES
NO
NO
YES

Note: For the first test case, $$$165\xrightarrow{-33}132\xrightarrow{-33}99\xrightarrow{-33}66\xrightarrow{-33}33\xrightarrow{-33}0$$$.

For the second test case, $$$6369\xrightarrow{-33}6{\color{red}{33}}6\xrightarrow{\text{remove "33"}}66\xrightarrow{-33}33\xrightarrow{-33}0$$$.

For the third test case, it can be proven that, regardless of the operations performed, $$$666$$$ cannot be transformed into $$$0$$$.

Buggy Code:
#include<iostream>
#include<map>
#include<vector>
#include<string>
#include<algorithm>
#include<cmath>
using namespace std;

int main(){
    int t;
    cin>>t;
    while(t--){
        long long int n;
        cin>>n;
        if(n>33){
            if(n%33!=0){
                int cond=0;
                while(n>33){
                    string s = to_string(n);
                    string snew="";
                    for(int i=0; i<s.length(); i++){
                        if((s[i]=='3' && s[i+1]=='3')&&(i+1<s.length())){
                            i++;continue;
                        }
                        snew = snew + s[i];
                    }
                    long long int ans = stoi(snew);
                    // cout<<ans<<endl;
                    if(ans%33==0){cout<<"YES"<<endl;cond=1;break;}
                    n = ans-33;
                }
                if(n%33==0){cout<<"YES"<<endl;}
                else if(cond==0){cout<<"NO"<<endl;}
            }
            else{cout<<"YES"<<endl;}
        }
        else{cout<<"NO"<<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
3
333333333
33333333
3333333
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kevin and Combination Lock
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kevin is trapped in Lakeside Village by Grace. At the exit of the village, there is a combination lock that can only be unlocked if Kevin solves it.

The combination lock starts with an integer $$$ x $$$. Kevin can perform one of the following two operations zero or more times:

1. If $$$ x \neq 33 $$$, he can select two consecutive digits $$$ 3 $$$ from $$$ x $$$ and remove them simultaneously. For example, if $$$ x = 13\,323 $$$, he can remove the second and third $$$ 3 $$$, changing $$$ x $$$ to $$$ 123 $$$.
2. If $$$ x \geq 33 $$$, he can change $$$ x $$$ to $$$ x - 33 $$$. For example, if $$$ x = 99 $$$, he can choose this operation to change $$$ x $$$ to $$$ 99 - 33 = 66 $$$.

When the value of $$$ x $$$ on the combination lock becomes $$$ 0 $$$, Kevin can unlock the lock and escape from Lakeside Village. Please determine whether it is possible for Kevin to unlock the combination lock and escape.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$).

The only line of each test case contains a positive integer $$$x$$$ ($$$1\leq x\leq 10^9$$$).

Output: For each test case, output "YES" or "NO" (without quotes) in one line, representing whether Kevin can unlock the combination lock and escape. You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
165
6369
666
114514
133333332
output:
YES
YES
NO
NO
YES

Note: For the first test case, $$$165\xrightarrow{-33}132\xrightarrow{-33}99\xrightarrow{-33}66\xrightarrow{-33}33\xrightarrow{-33}0$$$.

For the second test case, $$$6369\xrightarrow{-33}6{\color{red}{33}}6\xrightarrow{\text{remove "33"}}66\xrightarrow{-33}33\xrightarrow{-33}0$$$.

For the third test case, it can be proven that, regardless of the operations performed, $$$666$$$ cannot be transformed into $$$0$$$.

Buggy Code:
#include <iostream>
using namespace std;

bool canEscape(int x) {
    while (x > 0) {
        if (x % 33 == 0) {
            x -= 33; // Вычитаем 33, если делится
        } else {
            bool found = false;
            int prev_digit = -1, temp = x, new_x = 0, multiplier = 1;

            // Ищем пару "33" без преобразования в строку
            while (temp > 0) {
                int digit = temp % 10;
                temp /= 10;

                if (prev_digit == 3 && digit == 3 && !found) {
                    // Пропускаем пару "33"
                    found = true;
                    continue;
                }

                new_x += digit * multiplier;
                multiplier *= 10;
                prev_digit = digit;
            }

            if (!found) {
                return false; // Если пара "33" не найдена, невозможно сбежать
            }

            x = new_x; // Обновляем x после удаления пары "33"
        }
    }

    return x == 0; // Успех, если x стал 0
}

int main() {
    int t;
    cin >> t;
    while (t--) {
        int x;
        cin >> x;
        if (canEscape(x)) {
            cout << "YES" << endl;
        } else {
            cout << "NO" << 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
5
3333
333333
33333333
3333333333
1333333332
```