**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Novice's Mistake
time_limit_ms: 3000
memory_limit_mb: 256
Description: One of the first programming problems by K1o0n looked like this: "Noobish_Monk has $$$n$$$ $$$(1 \le n \le 100)$$$ friends. Each of them gave him $$$a$$$ $$$(1 \le a \le 10000)$$$ apples for his birthday. Delighted with such a gift, Noobish_Monk returned $$$b$$$ $$$(1 \le b \le \min(10000, a \cdot n))$$$ apples to his friends. How many apples are left with Noobish_Monk?"

K1o0n wrote a solution, but accidentally considered the value of $$$n$$$ as a string, so the value of $$$n \cdot a - b$$$ was calculated differently. Specifically:

- when multiplying the string $$$n$$$ by the integer $$$a$$$, he will get the string $$$s=\underbrace{n + n + \dots + n + n}_{a\ \text{times}}$$$
- when subtracting the integer $$$b$$$ from the string $$$s$$$, the last $$$b$$$ characters will be removed from it. If $$$b$$$ is greater than or equal to the length of the string $$$s$$$, it will become empty.

Learning about this, ErnKor became interested in how many pairs $$$(a, b)$$$ exist for a given $$$n$$$, satisfying the constraints of the problem, on which K1o0n's solution gives the correct answer.

"The solution gives the correct answer" means that it outputs a non-empty string, and this string, when converted to an integer, equals the correct answer, i.e., the value of $$$n \cdot a - b$$$.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$)  — the number of test cases.

For each test case, a single line of input contains an integer $$$n$$$ ($$$1 \le n \le 100$$$).

It is guaranteed that in all test cases, $$$n$$$ is distinct.

Output: For each test case, output the answer in the following format:

In the first line, output the integer $$$x$$$ — the number of bad tests for the given $$$n$$$.

In the next $$$x$$$ lines, output two integers $$$a_i$$$ and $$$b_i$$$ — such integers that K1o0n's solution on the test "$$$n$$$ $$$a_i$$$ $$$b_i$$$" gives the correct answer.

Examples:
input:
3
2
3
10
output:
3
20 18
219 216
2218 2214
1
165 162
1
1262 2519

Note: In the first example, $$$a = 20$$$, $$$b = 18$$$ are suitable, as "$$$\text{2}$$$" $$$\cdot 20 - 18 =$$$ "$$$\text{22222222222222222222}$$$"$$$- 18 = 22 = 2 \cdot 20 - 18$$$

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

public class E_Novice_s_Mistake {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

    static FastReader vt = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        try {
            int testCases = vt.nextInt();
            // int testCases = 1;
            while (testCases-- > 0) {
                // write code here
                solve();
            }
            out.flush();
        } catch (Exception e) {
            return;
        }
    }

    static void solve() {
        int n = vt.nextInt();
        int len = findlen(n);
        ArrayList<int[]> ans = new ArrayList<>();
        for (int i = 1; i <= 10000; i++) {
            int totlen = len * i;
            int res = n * i;
            int req = findlen(res);
            int diff = totlen - req;
            if (diff <= 0 || diff > Math.min(10000, n * i)) {
                continue;
            }
            String s = String.valueOf(res - diff);
            String ns = String.valueOf(n);
            boolean ok = true;
            if (totlen - diff != s.length()) {
                if (totlen - diff - 1 == s.length()) {
                    s = String.valueOf(res - diff - 1);
                } else {
                    continue;
                }
            }
            for (int j = 0; j < s.length(); j++) {
                if (s.charAt(j) != ns.charAt(j % ns.length())) {
                    ok = false;
                    break;
                }
            }
            if (ok) {
                ans.add(new int[] { i, diff });
            }
        }
        out.println(ans.size());
        for (int[] a : ans) {
            out.println(a[0] + " " + a[1]);
        }
    }

    static int findlen(int n) {
        int len = 1;
        if (n >= 10) {
            len++;
        }
        if (n >= 100) {
            len++;
        }
        if (n >= 1000) {
            len++;
        }
        if (n >= 10000) {
            len++;
        }
        if (n >= 100000) {
            len++;
        }
        if (n >= 1000000) {
            len++;
        }
        if (n >= 10000000) {
            len++;
        }
        return len;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases where the faulty string manipulation and integer arithmetic can lead to incorrect results. Specifically, we need to test cases where the string manipulation results in a string that, when converted to an integer, does not match the correct arithmetic result.

Here are the steps to generate such a test input:

1. **Edge Cases for `n`**:
   - Test with `n = 1` to see if the code handles the smallest possible value correctly.
   - Test with `n = 99` and `n = 100` to check the behavior at the upper limits of the input range.

2. **Edge Cases for `a` and `b`**:
   - Test with `a` and `b` values that are close to the upper limits (10000 and 10000 * n respectively).
   - Test with `a` and `b` values that result in a string that, when converted to an integer, does not match the expected arithmetic result.

3. **Test Cases**:
   - For `n = 1`, we can test with `a = 10000` and `b = 9999` to see if the string manipulation results in a non-empty string.
   - For `n = 99`, we can test with `a = 100` and `b = 9801` to see if the string manipulation results in a non-empty string.
   - For `n = 100`, we can test with `a = 100` and `b = 10000` to see if the string manipulation results in an empty string.

Here is the test input:

```plaintext
3
1
99
100
```

This test input will help to identify the issues in the buggy code by covering edge cases and ensuring that the string manipulation and integer arithmetic are correctly handled.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Billetes MX (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, it is guaranteed that $q = 0$. You can make hacks only if both versions of the problem are solved.

An integer grid $A$ with $p$ rows and $q$ columns is called beautiful if:

  * All elements of the grid are integers between $0$ and $2^{30}-1$, and    * For any subgrid, the XOR of the values at the corners is equal to $0$. Formally, for any four integers $i_1$, $i_2$, $j_1$, $j_2$ ($1 \le i_1 < i_2 \le p$; $1 \le j_1 < j_2 \le q$), $A_{i_1, j_1} \oplus A_{i_1, j_2} \oplus A_{i_2, j_1} \oplus A_{i_2, j_2} = 0$, where $\oplus$ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

There is a partially filled integer grid $G$ with $n$ rows and $m$ columns where only $k$ cells are filled. Polycarp wants to know how many ways he can assign integers to the unfilled cells so that the grid is beautiful.

However, Monocarp thinks that this problem is too easy. Therefore, he will perform $q$ updates on the grid. In each update, he will choose an unfilled cell and assign an integer to it. Note that these updates are persistent. That is, changes made to the grid will apply when processing future updates.

For each of the $q + 1$ states of the grid, the initial state and after each of the $q$ queries, determine the number of ways Polycarp can assign integers to the unfilled cells so that the grid is beautiful. Since this number can be very large, you are only required to output their values modulo $10^9+7$.

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

The first line of each test case contains four integers $n$, $m$, $k$ and $q$ ($2 \le n, m \le 10^5$; $0 \le k \le 10^5$; $q = 0$) — the number of rows, the number of columns, the number of fixed cells, and the number of updates.

The following $k$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

The following $q$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

It is guaranteed that the pairs $(r,c)$ over all assignments are distinct.

It is guaranteed that the sum of $n$, $m$, $k$ and $q$ over all test cases does not exceed $10^5$ respectively.

Output: For each test case, output $q + 1$ lines. The $i$-th line of output should contain the answer of the $i$-th state of the grid modulo $10^9 + 7$.

Examples:
input:
2
3 3 8 0
2 1 6
3 2 12
1 2 6
2 2 0
1 3 10
1 1 0
2 3 12
3 1 10
2 5 2 0
1 1 10
1 2 30
output:
1
489373567

Note: In the first test case of the example, we have the following grid:

$0$| $6$| $10$   ---|---|---   $6$| $0$| $12$   $10$| $12$| $?$      It can be proven that the only valid value for tile $(3, 3)$ is $0$.

Buggy Code:
#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=1e5+10;
const int mod=1e9+7;
int T,n,m,k,r,c,v,col[N],ans,q,us;
bool f,vv[N],flag,ff;
struct node{
	int to,val;
};
vector<node>edge[N];
bool vis[N];
void dfs(int x)
{
	if(vv[x]==1)
	flag=1;
	vis[x]=1;
	for(int i=0;i<edge[x].size();i++)
	{
		node to=edge[x][i];
		if(vv[to.to]==1)
		{
			if(col[to.to]!=(to.val^col[x]^us))
			{
				if(ff=0)
				{
					us=to.val^col[x];
					ff=1;
					continue;
				}
				f=1;
				return;
			}
		}
		if(vis[to.to]==0)
		{
			col[to.to]=to.val^col[x];
			dfs(to.to);
		}
		else
		{
			if(col[to.to]!=(to.val^col[x]^us))
			{
				if(ff=0)
				{
					us=to.val^col[x];
					ff=1;
					continue;
				}
				f=1;
				return;
			}
		}
	}
	return;
}
int qpow(int a,int b)
{
	int s=1,base=a;
	while(b!=0)
	{
		if(b&1==1)
		{
			s*=base;
			s%=mod;
		}
		base*=base;
		base%=mod;
		b>>=1;
	}
	return s;
}
signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0),cout.tie(0);
	cin>>T;
	while(T--)
	{
		cin>>n>>m>>k>>q;
		for(int i=1;i<=n+m;i++)
		{
			vis[i]=0,col[i]=0,vv[i]=0;
			edge[i].clear();
		}
		ans=1;
		us=0;
		ff=0;
		for(int i=1;i<=k;i++)
		{
			cin>>r>>c>>v;
			if(r==1&&c==1)
			{
				ff=1;
				us=v;
				ans--;
			}
			else if(r==1)
			{
				col[c+n]=v;
				vv[c+n]=1;
			}
			else if(c==1)
			{
				vv[r]=1;
				col[r]=v;
			}
			else
			{
				edge[r].push_back(node{c+n,v});
				edge[c+n].push_back(node{r,v});
			}
		}
		f=0;
		for(int i=2;i<=n+m;i++)
		{
			if(i==n+1)
			continue;
			if(vis[i]==0)
			{
				flag=0;
				dfs(i);
				ans+=flag^1;
			}
		}
		if(f==1)
		{
			cout<<0<<'\n';
			continue;
		}
		cout<<qpow((1ll<<30),ans)<<'\n';
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
3 3 4 0
1 1 0
1 2 1
2 1 1
2 2 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Billetes MX (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, it is guaranteed that $q = 0$. You can make hacks only if both versions of the problem are solved.

An integer grid $A$ with $p$ rows and $q$ columns is called beautiful if:

  * All elements of the grid are integers between $0$ and $2^{30}-1$, and    * For any subgrid, the XOR of the values at the corners is equal to $0$. Formally, for any four integers $i_1$, $i_2$, $j_1$, $j_2$ ($1 \le i_1 < i_2 \le p$; $1 \le j_1 < j_2 \le q$), $A_{i_1, j_1} \oplus A_{i_1, j_2} \oplus A_{i_2, j_1} \oplus A_{i_2, j_2} = 0$, where $\oplus$ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

There is a partially filled integer grid $G$ with $n$ rows and $m$ columns where only $k$ cells are filled. Polycarp wants to know how many ways he can assign integers to the unfilled cells so that the grid is beautiful.

However, Monocarp thinks that this problem is too easy. Therefore, he will perform $q$ updates on the grid. In each update, he will choose an unfilled cell and assign an integer to it. Note that these updates are persistent. That is, changes made to the grid will apply when processing future updates.

For each of the $q + 1$ states of the grid, the initial state and after each of the $q$ queries, determine the number of ways Polycarp can assign integers to the unfilled cells so that the grid is beautiful. Since this number can be very large, you are only required to output their values modulo $10^9+7$.

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

The first line of each test case contains four integers $n$, $m$, $k$ and $q$ ($2 \le n, m \le 10^5$; $0 \le k \le 10^5$; $q = 0$) — the number of rows, the number of columns, the number of fixed cells, and the number of updates.

The following $k$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

The following $q$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

It is guaranteed that the pairs $(r,c)$ over all assignments are distinct.

It is guaranteed that the sum of $n$, $m$, $k$ and $q$ over all test cases does not exceed $10^5$ respectively.

Output: For each test case, output $q + 1$ lines. The $i$-th line of output should contain the answer of the $i$-th state of the grid modulo $10^9 + 7$.

Examples:
input:
2
3 3 8 0
2 1 6
3 2 12
1 2 6
2 2 0
1 3 10
1 1 0
2 3 12
3 1 10
2 5 2 0
1 1 10
1 2 30
output:
1
489373567

Note: In the first test case of the example, we have the following grid:

$0$| $6$| $10$   ---|---|---   $6$| $0$| $12$   $10$| $12$| $?$      It can be proven that the only valid value for tile $(3, 3)$ is $0$.

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

using LL = long long;
#define int LL // delete if MLE/TLE

using namespace std;
using vInt = vector<int>;
using vLL = vector<LL>;
using pLL = pair<LL, LL>;
using vPLL = vector<pLL>;
using pii = pair<int, int>;
using vPii = vector<pii>;
using matInt = vector<vInt>;
using matPii = vector<vPii>;
using matLL = vector<vLL>;
using vBool = vector<bool>;
using matBool = vector<vBool>;
using vStr = vector<string>;

#define all(x) (x).begin(), (x).end()
#define cyes cout << "YES" << '\n'
#define cno cout << "NO" << '\n'
#define for0(i, n) for(int (i)=0; (i)<(n); (i)++)
#define for1(i, n) for(int (i)=1; (i)<=(n); (i)++)
#define newInt(n) int n; cin >> n
#define newLL(n) LL n; cin >> n
#define newVInt(A, f, t) vInt A(t); for(int i=f; i<t; i++) cin >> A[i]
#define newVLL(A, f, t) vLL A(t); for(int i=f; i<t; i++) cin >> A[i]
#define endl '\n'

#define ff first
#define ss second

const int MOD = 1e9+7;

int modAdd(LL a, LL b){
    return (a+b)%MOD;
}
int modMul(LL a, LL b){
    return (a*b)%MOD;
}
int modExp(LL a, int k){
    LL ans = 1;
    while(k){
        if(k&1) ans = modMul(ans, a);
        a = modMul(a, a);
        k >>= 1;
    }
    return ans;
}

struct Disj{
    vInt disj;
    vInt xors;
    Disj(int N){
        disj.resize(N+1);
        xors.resize(N+1);
        for1(i, N){
            disj[i] = i;
            xors[i] = 0;
        }
    }
    int find(int u){
        if(disj[u] == u)
            return u;
        int v = find(disj[u]);
        xors[u] ^= xors[disj[u]];
        return disj[u] = v;
    }
    bool merge(int u, int v, int x){
        int uRoot = find(u);
        int vRoot = find(v);
        if(uRoot == vRoot)
            return xors[u] ^ xors[v] == x;
        disj[uRoot] = vRoot;
        xors[uRoot] = xors[u] ^ xors[v] ^ x;
        return true;
    }
};

void solve(){
    int N, M, K, Q;
    cin >> N >> M >> K >> Q;

    matPii RX(N+1);
    matPii CX(M+1);

    for(int i=1; i<=K; i++){
        int r, c, v;
        cin >> r >> c >> v;

        RX[r].push_back({c, v});
        CX[c].push_back({r, v});
    }

    for(int i=1; i<=N; i++)
        sort(all(RX[i]));

    for(int i=1; i<=M; i++)
        sort(all(CX[i]));

    Disj disj1(N), disj2(M);

    for(int i=1; i<=N; i++){
        for(int j=1; j<RX[i].size(); j++){
            int c1 = RX[i][j-1].ff;
            int c2 = RX[i][j].ff;
            int x1 = RX[i][j-1].ss;
            int x2 = RX[i][j].ss;

            if(!disj1.merge(c1, c2, x1 ^ x2)){
                cout << 0 << endl;
                return;
            }
        }
    }

    for(int i=1; i<=M; i++){
        for(int j=1; j<CX[i].size(); j++){
            int r1 = CX[i][j-1].ff;
            int r2 = CX[i][j].ff;
            int x1 = CX[i][j-1].ss;
            int x2 = CX[i][j].ss;

            if(!disj2.merge(r1, r2, x1 ^ x2)){
                cout << 0 << endl;
                return;
            }
        }
    }

    int ans = 1;
    int xv = modExp(2, 30);

    vInt chk1(N+1), chk2(M+1);

    int nr = 0, nc = 0;
    for(int i=1; i<=N; i++){
        int u = disj1.find(i);
        if(chk1[u] == 0){
            chk1[u] = 1;
            nr++;
        }
    }

    for(int i=1; i<=M; i++){
        int u = disj2.find(i);
        if(chk2[u] == 0){
            chk2[u] = 1;
            nc++;
        }
    }

    int tot = modExp(xv, nr+nc-2);

    cout << tot << endl;
}

int32_t main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    int T;
    cin >> T;
    while(T--)
        solve();

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
2
3 3 4 0
1 1 0
1 2 1
2 1 1
2 2 0
2 2 2 0
1 1 0
1 2 1
2 1 1
2 2 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Billetes MX (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, it is guaranteed that $q = 0$. You can make hacks only if both versions of the problem are solved.

An integer grid $A$ with $p$ rows and $q$ columns is called beautiful if:

  * All elements of the grid are integers between $0$ and $2^{30}-1$, and    * For any subgrid, the XOR of the values at the corners is equal to $0$. Formally, for any four integers $i_1$, $i_2$, $j_1$, $j_2$ ($1 \le i_1 < i_2 \le p$; $1 \le j_1 < j_2 \le q$), $A_{i_1, j_1} \oplus A_{i_1, j_2} \oplus A_{i_2, j_1} \oplus A_{i_2, j_2} = 0$, where $\oplus$ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

There is a partially filled integer grid $G$ with $n$ rows and $m$ columns where only $k$ cells are filled. Polycarp wants to know how many ways he can assign integers to the unfilled cells so that the grid is beautiful.

However, Monocarp thinks that this problem is too easy. Therefore, he will perform $q$ updates on the grid. In each update, he will choose an unfilled cell and assign an integer to it. Note that these updates are persistent. That is, changes made to the grid will apply when processing future updates.

For each of the $q + 1$ states of the grid, the initial state and after each of the $q$ queries, determine the number of ways Polycarp can assign integers to the unfilled cells so that the grid is beautiful. Since this number can be very large, you are only required to output their values modulo $10^9+7$.

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

The first line of each test case contains four integers $n$, $m$, $k$ and $q$ ($2 \le n, m \le 10^5$; $0 \le k \le 10^5$; $q = 0$) — the number of rows, the number of columns, the number of fixed cells, and the number of updates.

The following $k$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

The following $q$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

It is guaranteed that the pairs $(r,c)$ over all assignments are distinct.

It is guaranteed that the sum of $n$, $m$, $k$ and $q$ over all test cases does not exceed $10^5$ respectively.

Output: For each test case, output $q + 1$ lines. The $i$-th line of output should contain the answer of the $i$-th state of the grid modulo $10^9 + 7$.

Examples:
input:
2
3 3 8 0
2 1 6
3 2 12
1 2 6
2 2 0
1 3 10
1 1 0
2 3 12
3 1 10
2 5 2 0
1 1 10
1 2 30
output:
1
489373567

Note: In the first test case of the example, we have the following grid:

$0$| $6$| $10$   ---|---|---   $6$| $0$| $12$   $10$| $12$| $?$      It can be proven that the only valid value for tile $(3, 3)$ is $0$.

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

#define ft first
#define sd second

using namespace std;

using ll = long long;
using ld = long double;
using pp = pair<int, int>;

template <typename T, typename F>
istream& operator>>(istream& s, pair<T, F>& a) {
    s >> a.ft >> a.sd;
    return s;
}

template <typename T, typename F>
ostream& operator<<(ostream& s, pair<T, F>& a) {
    s << a.ft << ' ' << a.sd << (char)(10);
    return s;
}
template <typename T>
istream& operator>>(istream& s, vector<T>& a) {
    for (T& el : a) {
        s >> el;
    }
    return s;
}

template <typename T>
ostream& operator<<(ostream& s, vector<T>& a) {
    for (T el : a) {
        s << el << ' ';
    }
    s << (char)(10);
    return s;
}

bool ok = true;

vector<int> check_val, col;
vector<vector<pp>> graph;
int n, m, k, q;

void dfs(int v) {
    for (auto [to, w] : graph[v]) {
        if (check_val[to] == -1) {
            check_val[to] = check_val[v] ^ w;
            dfs(to);
        } else if (check_val[to] != (check_val[v] ^ w)) {
            ok = false;
        }
    }
}

void check() {
    check_val.assign(graph.size(), -1);
    for (int i = 0; i < graph.size(); i++) {
        if (i == 0 || i == n) {
            continue;
        }
        if (check_val[i] == -1) {
            check_val[i] = 0;
            dfs(i);
        }
    }
}

void dfs2(int v, int cl) {
    col[v] = cl;
    for (auto [to, w] : graph[v]) {
        if (col[to] == -1) {
            dfs2(to, cl);
        }
    }
}

const ll mod = 1e9 + 7, base = (1ll << 30) % mod;

void solve() {
    cin >> n >> m >> k >> q;
    graph.assign(n + m, {});
    bool set_corner = false;
    vector<int> set_nodes;
    for (int i = 0; i < k; i++) {
        int r, c, w;
        cin >> r >> c >> w;
        r--;
        c--;
        int a = r;
        int b = c + n;
        if (r == 0 && c == 0) {
            set_corner = true;
        } else if (r == 0) {
            set_nodes.push_back(b);
        } else if (c == 0) {
            set_nodes.push_back(a);
        } else {
            graph[a].emplace_back(b, w);
            graph[b].emplace_back(a, w);
        }
    }
    check();
    if (!ok) {
        cout << 0;
        return;
    }
    col.assign(graph.size(), -1);
    int cl = 0;
    for (int i = 0; i < graph.size(); i++) {
        if (i == 0 || i == n) {
            continue;
        }
        if (col[i] == -1) {
            dfs2(i, cl);
            cl++;
        }
    }
    vector<bool> is_marked(cl);
    for (int el : set_nodes) {
        is_marked[col[el]] = true;
    }
    ll ans = (set_corner ? 1 : base);
    for (auto x : is_marked) {
        if (!x) {
            ans = ans * base % mod;
        }
    }
    cout << ans << '\n';
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    ll t;
    cin >> t;
    while (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:
```plaintext
1
2 2 3 0
1 1 0
1 2 1
2 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Billetes MX (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, it is guaranteed that $q = 0$. You can make hacks only if both versions of the problem are solved.

An integer grid $A$ with $p$ rows and $q$ columns is called beautiful if:

  * All elements of the grid are integers between $0$ and $2^{30}-1$, and    * For any subgrid, the XOR of the values at the corners is equal to $0$. Formally, for any four integers $i_1$, $i_2$, $j_1$, $j_2$ ($1 \le i_1 < i_2 \le p$; $1 \le j_1 < j_2 \le q$), $A_{i_1, j_1} \oplus A_{i_1, j_2} \oplus A_{i_2, j_1} \oplus A_{i_2, j_2} = 0$, where $\oplus$ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

There is a partially filled integer grid $G$ with $n$ rows and $m$ columns where only $k$ cells are filled. Polycarp wants to know how many ways he can assign integers to the unfilled cells so that the grid is beautiful.

However, Monocarp thinks that this problem is too easy. Therefore, he will perform $q$ updates on the grid. In each update, he will choose an unfilled cell and assign an integer to it. Note that these updates are persistent. That is, changes made to the grid will apply when processing future updates.

For each of the $q + 1$ states of the grid, the initial state and after each of the $q$ queries, determine the number of ways Polycarp can assign integers to the unfilled cells so that the grid is beautiful. Since this number can be very large, you are only required to output their values modulo $10^9+7$.

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

The first line of each test case contains four integers $n$, $m$, $k$ and $q$ ($2 \le n, m \le 10^5$; $0 \le k \le 10^5$; $q = 0$) — the number of rows, the number of columns, the number of fixed cells, and the number of updates.

The following $k$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

The following $q$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

It is guaranteed that the pairs $(r,c)$ over all assignments are distinct.

It is guaranteed that the sum of $n$, $m$, $k$ and $q$ over all test cases does not exceed $10^5$ respectively.

Output: For each test case, output $q + 1$ lines. The $i$-th line of output should contain the answer of the $i$-th state of the grid modulo $10^9 + 7$.

Examples:
input:
2
3 3 8 0
2 1 6
3 2 12
1 2 6
2 2 0
1 3 10
1 1 0
2 3 12
3 1 10
2 5 2 0
1 1 10
1 2 30
output:
1
489373567

Note: In the first test case of the example, we have the following grid:

$0$| $6$| $10$   ---|---|---   $6$| $0$| $12$   $10$| $12$| $?$      It can be proven that the only valid value for tile $(3, 3)$ is $0$.

Buggy Code:
import sys
from collections import deque
input = sys.stdin.readline
def fg():
    return int(input())
def fgh():
    return [int(xx) for xx in input().split()]
def tt():
    sys.stdout.flush()
def sd():
    print('Yes')
def df():
    print('No')
M = 10 ** 9 + 7
for __ in range(fg()):
    n, m, k, q = fgh()
    u = 2 ** 30
    step = [1]
    for i in range(n + m):
        step.append(step[-1] * u % M)
    ks = [[i] for i in range(n + m)]
    z = [i for i in range(n + m)]
    d = [0] * (n + m)
    vvv = [0] * (n + m)
    T = [0] * (n + m)
    bd = n + m - 1
    if k == 0:
        print(step[bd])
    do = {}
    for _ in range(k + q):
        x, y, h = fgh()
        x -= 1
        y -= 1
        if bd == -1 and _ >= k - 1:
            print(0)
            continue
        if x + y == 0:
            if (1, 1) not in do:
                do[(1, 1)] = h
                bd -= 1
            else:
                if h != do[(1, 1)]:
                    bd = -1
        else:
            u = x
            v = y + n
            if x == 0:
                #print(x, y)
                #print(*vvv)
                #print(vvv[z[v]])
                #print(*z)
                #print(v)
                if vvv[z[v]] == 0:
                    bd -= 1
                    st = d[v] ^ h
                    for uu in ks[z[v]]:
                        d[uu] ^= st
                        z[uu] = v
                    vvv[v] = 1
                else:
                    if d[v] != h:
                        #print(222)
                        bd = -1
            elif y == 0:
                if vvv[z[u]] == 0:
                    bd -= 1
                    st = d[u] ^ h
                    for uu in ks[z[u]]:
                        d[uu] ^= st
                        z[uu] = u
                    vvv[u] = 1
                else:
                    if d[u] != h:
                        #print(111)
                        bd = -1
            else:
                if (z[u] == z[v]) and ((d[u] ^ d[v]) != h):
                    #print(444)
                    bd = -1
                elif z[u] != z[v]:
                    if len(ks[z[u]]) > len(ks[z[v]]):
                        u, v = v, u
                    if vvv[z[u]] + vvv[z[v]] <= 1:
                        bd -= 1
                    p = d[v] ^ h ^ d[u]
                    if vvv[z[u]] and vvv[z[v]] and p:
                        bd = -1
                        #print(333)
                        continue

                    if vvv[u] == 1 and vvv[v] == 0:
                        u, v = v, u
                    oo = z[u]
                    for j in ks[z[u]]:
                        z[j] = z[v]
                        d[j] ^= p
                        ks[z[v]].append(j)
                    if vvv[z[u]] and vvv[z[v]]:
                        if (1, 1) not in do:
                            do[(1, 1)] = d[v] ^ d[u] ^ h
                            bd -= 1
                            #print(do[(1, 1)])
                        elif (1, 1) in do and do[(1, 1)] != p:
                            bd = -1
                    vvv[z[v]] = max(vvv[oo], vvv[z[v]])
        #print(*z)
        #print(bd)
        if _ >= k - 1:
            if bd >= 0:
                print(step[bd])
            else:
                print(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, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are the steps to create such a test input:

1. **Edge Cases for Grid Size**: Use the minimum and maximum allowed grid sizes to test the boundaries.
2. **Edge Cases for Filled Cells**: Test with no filled cells, one filled cell, and multiple filled cells.
3. **Edge Cases for Updates**: Since \( q = 0 \) in this version, we only need to focus on the initial state.
4. **Consistency Checks**: Ensure that the input is consistent and valid according to the problem constraints.

Let's create a test input that includes these aspects:

- **Test Case 1**: A small grid with no filled cells.
- **Test Case 2**: A small grid with one filled cell.
- **Test Case 3**: A small grid with multiple filled cells.
- **Test Case 4**: A large grid with multiple filled cells.

Here is the test input:

```plaintext
4
2 2 0 0
2 2 1 0
1 1 5
2 2 3 0
1 1 5
1 2 3
2 1 7
100000 100000 3 0
1 1 1
1 2 2
2 1 3
```

### Explanation:
1. **Test Case 1**: A 2x2 grid with no filled cells. This tests the scenario where the grid is completely empty.
2. **Test Case 2**: A 2x2 grid with one filled cell. This tests the scenario where there is a single filled cell.
3. **Test Case 3**: A 2x2 grid with three filled cells. This tests the scenario where most cells are filled.
4. **Test Case 4**: A large 100000x100000 grid with three filled cells. This tests the scenario with a large grid size and a few filled cells to check the performance and correctness.

This test input should help in exposing any bugs in the provided code by covering a range of edge cases and scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Billetes MX (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, it is guaranteed that $q = 0$. You can make hacks only if both versions of the problem are solved.

An integer grid $A$ with $p$ rows and $q$ columns is called beautiful if:

  * All elements of the grid are integers between $0$ and $2^{30}-1$, and    * For any subgrid, the XOR of the values at the corners is equal to $0$. Formally, for any four integers $i_1$, $i_2$, $j_1$, $j_2$ ($1 \le i_1 < i_2 \le p$; $1 \le j_1 < j_2 \le q$), $A_{i_1, j_1} \oplus A_{i_1, j_2} \oplus A_{i_2, j_1} \oplus A_{i_2, j_2} = 0$, where $\oplus$ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

There is a partially filled integer grid $G$ with $n$ rows and $m$ columns where only $k$ cells are filled. Polycarp wants to know how many ways he can assign integers to the unfilled cells so that the grid is beautiful.

However, Monocarp thinks that this problem is too easy. Therefore, he will perform $q$ updates on the grid. In each update, he will choose an unfilled cell and assign an integer to it. Note that these updates are persistent. That is, changes made to the grid will apply when processing future updates.

For each of the $q + 1$ states of the grid, the initial state and after each of the $q$ queries, determine the number of ways Polycarp can assign integers to the unfilled cells so that the grid is beautiful. Since this number can be very large, you are only required to output their values modulo $10^9+7$.

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

The first line of each test case contains four integers $n$, $m$, $k$ and $q$ ($2 \le n, m \le 10^5$; $0 \le k \le 10^5$; $q = 0$) — the number of rows, the number of columns, the number of fixed cells, and the number of updates.

The following $k$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

The following $q$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

It is guaranteed that the pairs $(r,c)$ over all assignments are distinct.

It is guaranteed that the sum of $n$, $m$, $k$ and $q$ over all test cases does not exceed $10^5$ respectively.

Output: For each test case, output $q + 1$ lines. The $i$-th line of output should contain the answer of the $i$-th state of the grid modulo $10^9 + 7$.

Examples:
input:
2
3 3 8 0
2 1 6
3 2 12
1 2 6
2 2 0
1 3 10
1 1 0
2 3 12
3 1 10
2 5 2 0
1 1 10
1 2 30
output:
1
489373567

Note: In the first test case of the example, we have the following grid:

$0$| $6$| $10$   ---|---|---   $6$| $0$| $12$   $10$| $12$| $?$      It can be proven that the only valid value for tile $(3, 3)$ is $0$.

Buggy Code:
import sys
from collections import deque
input = sys.stdin.readline
def fg():
    return int(input())
def fgh():
    return [int(xx) for xx in input().split()]
def tt():
    sys.stdout.flush()
def sd():
    print('Yes')
def df():
    print('No')
M = 10 ** 9 + 7
for __ in range(fg()):
    n, m, k, q = fgh()
    u = 2 ** 30
    step = [1]
    for i in range(n + m):
        step.append(step[-1] * u % M)
    ks = [[i] for i in range(n + m)]
    z = [i for i in range(n + m)]
    d = [0] * (n + m)
    vvv = [0] * (n + m)
    T = [0] * (n + m)
    bd = n + m - 1
    if k == 0:
        print(step[bd])
    do = {}
    for _ in range(k + q):
        x, y, h = fgh()
        x -= 1
        y -= 1
        if bd == -1 and _ >= k - 1:
            print(0)
            continue
        if x + y == 0:
            if (0, 0) not in do:
                bd -= 1
            else:
                if bd != do[(0, 0)]:
                    bd = -1
        else:
            u = x
            v = y + n
            if x == 0:
                #print(x, y)
                #print(*vvv)
                #print(vvv[z[v]])
                #print(*z)
                #print(v)
                if vvv[z[v]] == 0:
                    bd -= 1
                    st = d[v] ^ h
                    for uu in ks[z[v]]:
                        d[uu] ^= st
                        z[uu] = v
                    vvv[v] = 1
                else:
                    if d[v] != h:
                        #print(222)
                        bd = -1
            elif y == 0:
                if vvv[z[u]] == 0:
                    bd -= 1
                    st = d[u] ^ h
                    for uu in ks[z[u]]:
                        d[uu] ^= st
                        z[uu] = u
                    vvv[u] = 1
                else:
                    if d[u] != h:
                        #print(111)
                        bd = -1
            else:
                if (z[u] == z[v]) and ((d[u] ^ d[v]) != h):
                    #print(444)
                    bd = -1
                elif z[u] != z[v]:
                    if len(ks[z[u]]) > len(ks[z[v]]):
                        u, v = v, u
                    if vvv[z[u]] + vvv[z[v]] <= 1:
                        bd -= 1
                    p = d[v] ^ h ^ d[u]
                    if vvv[z[u]] and vvv[z[v]] and p:
                        bd = -1
                        #print(333)
                        continue
                    if vvv[z[u]] and vvv[z[v]]:
                        if (1, 1) not in do:
                            do[(1, 1)] = d[v] ^ d[u] ^ h
                        elif (1, 1) in do and do[(1, 1)]:
                            bd = -1
                    if vvv[u] == 1 and vvv[v] == 0:
                        u, v = v, u
                    oo = z[u]
                    for j in ks[z[u]]:
                        z[j] = z[v]
                        d[j] ^= p
                        ks[z[v]].append(j)
                    vvv[z[v]] = max(vvv[oo], vvv[z[v]])
        #print(*z)
        #print(bd)
        if _ >= k - 1:
            if bd >= 0:
                print(step[bd])
            else:
                print(0)

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

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

Think step by step.

Response:
```plaintext
1
3 3 1 0
1 1 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Billetes MX (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, it is guaranteed that $q = 0$. You can make hacks only if both versions of the problem are solved.

An integer grid $A$ with $p$ rows and $q$ columns is called beautiful if:

  * All elements of the grid are integers between $0$ and $2^{30}-1$, and    * For any subgrid, the XOR of the values at the corners is equal to $0$. Formally, for any four integers $i_1$, $i_2$, $j_1$, $j_2$ ($1 \le i_1 < i_2 \le p$; $1 \le j_1 < j_2 \le q$), $A_{i_1, j_1} \oplus A_{i_1, j_2} \oplus A_{i_2, j_1} \oplus A_{i_2, j_2} = 0$, where $\oplus$ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

There is a partially filled integer grid $G$ with $n$ rows and $m$ columns where only $k$ cells are filled. Polycarp wants to know how many ways he can assign integers to the unfilled cells so that the grid is beautiful.

However, Monocarp thinks that this problem is too easy. Therefore, he will perform $q$ updates on the grid. In each update, he will choose an unfilled cell and assign an integer to it. Note that these updates are persistent. That is, changes made to the grid will apply when processing future updates.

For each of the $q + 1$ states of the grid, the initial state and after each of the $q$ queries, determine the number of ways Polycarp can assign integers to the unfilled cells so that the grid is beautiful. Since this number can be very large, you are only required to output their values modulo $10^9+7$.

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

The first line of each test case contains four integers $n$, $m$, $k$ and $q$ ($2 \le n, m \le 10^5$; $0 \le k \le 10^5$; $q = 0$) — the number of rows, the number of columns, the number of fixed cells, and the number of updates.

The following $k$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

The following $q$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

It is guaranteed that the pairs $(r,c)$ over all assignments are distinct.

It is guaranteed that the sum of $n$, $m$, $k$ and $q$ over all test cases does not exceed $10^5$ respectively.

Output: For each test case, output $q + 1$ lines. The $i$-th line of output should contain the answer of the $i$-th state of the grid modulo $10^9 + 7$.

Examples:
input:
2
3 3 8 0
2 1 6
3 2 12
1 2 6
2 2 0
1 3 10
1 1 0
2 3 12
3 1 10
2 5 2 0
1 1 10
1 2 30
output:
1
489373567

Note: In the first test case of the example, we have the following grid:

$0$| $6$| $10$   ---|---|---   $6$| $0$| $12$   $10$| $12$| $?$      It can be proven that the only valid value for tile $(3, 3)$ is $0$.

Buggy Code:
import sys
from collections import deque
input = sys.stdin.readline
def fg():
    return int(input())
def fgh():
    return [int(xx) for xx in input().split()]
def tt():
    sys.stdout.flush()
def sd():
    print('Yes')
def df():
    print('No')
M = 10 ** 9 + 7
for __ in range(fg()):
    n, m, k, q = fgh()
    u = 2 ** 30
    step = [1]
    for i in range(n + m):
        step.append(step[-1] * u % M)
    ks = [[i] for i in range(n + m)]
    z = [i for i in range(n + m)]
    d = [0] * (n + m)
    vvv = [0] * (n + m)
    T = [0] * (n + m)
    bd = n + m - 1
    if k == 0:
        print(step[bd])
    for _ in range(k + q):
        x, y, h = fgh()
        x -= 1
        y -= 1
        if bd == -1 and _ >= k - 1:
            print(0)
            continue
        if x + y == 0:
            bd -= 1
        else:
            u = x
            v = y + n
            if x == 0:
                #print(x, y)
                #print(*vvv)
                #print(vvv[z[v]])
                #print(*z)
                #print(v)
                if vvv[z[v]] == 0:
                    bd -= 1
                    st = d[v] ^ h
                    for uu in ks[z[v]]:
                        d[uu] ^= st
                        z[uu] = v
                    vvv[v] = 1
                else:
                    if d[v] != h:
                        #print(222)
                        bd = -1
            elif y == 0:
                if vvv[z[u]] == 0:
                    bd -= 1
                    st = d[u] ^ h
                    for uu in ks[z[u]]:
                        d[uu] ^= st
                        z[uu] = u
                    vvv[u] = 1
                else:
                    if d[u] != h:
                        #print(111)
                        bd = -1
            else:
                if (z[u] == z[v]) and ((d[u] ^ d[v]) != h):
                    #print(444)
                    bd = -1
                elif z[u] != z[v]:
                    if len(ks[z[u]]) > len(ks[z[v]]):
                        u, v = v, u
                    if vvv[z[u]] + vvv[z[v]] <= 1:
                        bd -= 1
                    p = d[v] ^ h ^ d[u]
                    if vvv[z[u]] and vvv[z[v]] and p:
                        bd = -1
                        #print(333)
                        continue
                    if vvv[u] == 1 and vvv[v] == 0:
                        u, v = v, u
                    oo = z[u]
                    for j in ks[z[u]]:
                        z[j] = z[v]
                        d[j] ^= p
                        ks[z[v]].append(j)
                    vvv[z[v]] = max(vvv[oo], vvv[z[v]])
        #print(*z)
        #print(bd)
        if _ >= k - 1:
            if bd >= 0:
                print(step[bd])
            else:
                print(0)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Billetes MX (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, it is guaranteed that $q = 0$. You can make hacks only if both versions of the problem are solved.

An integer grid $A$ with $p$ rows and $q$ columns is called beautiful if:

  * All elements of the grid are integers between $0$ and $2^{30}-1$, and    * For any subgrid, the XOR of the values at the corners is equal to $0$. Formally, for any four integers $i_1$, $i_2$, $j_1$, $j_2$ ($1 \le i_1 < i_2 \le p$; $1 \le j_1 < j_2 \le q$), $A_{i_1, j_1} \oplus A_{i_1, j_2} \oplus A_{i_2, j_1} \oplus A_{i_2, j_2} = 0$, where $\oplus$ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

There is a partially filled integer grid $G$ with $n$ rows and $m$ columns where only $k$ cells are filled. Polycarp wants to know how many ways he can assign integers to the unfilled cells so that the grid is beautiful.

However, Monocarp thinks that this problem is too easy. Therefore, he will perform $q$ updates on the grid. In each update, he will choose an unfilled cell and assign an integer to it. Note that these updates are persistent. That is, changes made to the grid will apply when processing future updates.

For each of the $q + 1$ states of the grid, the initial state and after each of the $q$ queries, determine the number of ways Polycarp can assign integers to the unfilled cells so that the grid is beautiful. Since this number can be very large, you are only required to output their values modulo $10^9+7$.

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

The first line of each test case contains four integers $n$, $m$, $k$ and $q$ ($2 \le n, m \le 10^5$; $0 \le k \le 10^5$; $q = 0$) — the number of rows, the number of columns, the number of fixed cells, and the number of updates.

The following $k$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

The following $q$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

It is guaranteed that the pairs $(r,c)$ over all assignments are distinct.

It is guaranteed that the sum of $n$, $m$, $k$ and $q$ over all test cases does not exceed $10^5$ respectively.

Output: For each test case, output $q + 1$ lines. The $i$-th line of output should contain the answer of the $i$-th state of the grid modulo $10^9 + 7$.

Examples:
input:
2
3 3 8 0
2 1 6
3 2 12
1 2 6
2 2 0
1 3 10
1 1 0
2 3 12
3 1 10
2 5 2 0
1 1 10
1 2 30
output:
1
489373567

Note: In the first test case of the example, we have the following grid:

$0$| $6$| $10$   ---|---|---   $6$| $0$| $12$   $10$| $12$| $?$      It can be proven that the only valid value for tile $(3, 3)$ is $0$.

Buggy Code:
import java.util.Scanner;

public class test {
    private static final int MOD = 1000000007;

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        int t = scanner.nextInt();

        for (int test = 0; test < t; test++) {
            int n = scanner.nextInt();
            int m = scanner.nextInt();
            int k = scanner.nextInt();
            int q = scanner.nextInt();

            int[][] grid = new int[n][m];
            boolean[][] fixed = new boolean[n][m];

            for (int i = 0; i < k; i++) {
                int r = scanner.nextInt() - 1;
                int c = scanner.nextInt() - 1;
                int v = scanner.nextInt();
                grid[r][c] = v;
                fixed[r][c] = true;
            }

            int freeCells = (n * m) - k;
            long ways = modExp(2, freeCells, MOD);

        }
        System.out.println(1);
        System.out.println(489373567);
        scanner.close();
    }

    private static long modExp(long base, int exp, int mod) {
        long result = 1;
        while (exp > 0) {
            if ((exp & 1) == 1) {
                result = (result * base) % mod;
            }
            base = (base * base) % mod;
            exp >>= 1;
        }
        return result;
    }
}

**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 scenarios that highlight the issue in the code. The buggy code currently does not handle the constraints and conditions of the problem correctly, particularly the requirement that the grid must be beautiful after each update.

### Analysis of the Buggy Code:
1. **Incorrect Calculation of Free Cells:**
   - The code calculates the number of free cells as `(n * m) - k`, but it does not consider the constraints that the grid must be beautiful. This means that not all free cells can be filled independently; the values must satisfy the XOR condition for all subgrids.

2. **No Handling of Updates:**
   - The code does not handle any updates, which is not a problem since `q = 0` in the easy version. However, it does not ensure that the initial grid is beautiful.

### Challenging Test Input:
To expose the bug, we can create a test case where the initial grid is not beautiful, and the number of free cells is such that the grid cannot be made beautiful by any assignment of values to the free cells.

### Test Case:
- **Test Case 1:**
  - A 3x3 grid with 4 fixed cells.
  - The fixed cells are chosen such that the XOR condition for subgrids is not satisfied.

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

### Explanation:
- **Grid:**
  ```
  0 1 ?
  2 3 ?
  ? ? ?
  ```

- **XOR Condition:**
  - For the subgrid with corners (1,1), (1,2), (2,1), (2,2):
    - `0 ⊕ 1 ⊕ 2 ⊕ 3 = 0 ⊕ 1 ⊕ 2 ⊕ 3 = 4 ≠ 0`
  - This subgrid does not satisfy the XOR condition, so the grid cannot be made beautiful by any assignment of values to the free cells.

- **Expected Output:**
  - The number of ways to assign integers to the unfilled cells so that the grid is beautiful should be `0` modulo `10^9 + 7`.

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

This test input will expose the bug in the buggy code because the code will incorrectly calculate the number of ways to assign integers to the unfilled cells, not considering the XOR condition for subgrids.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Billetes MX (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, it is guaranteed that $q = 0$. You can make hacks only if both versions of the problem are solved.

An integer grid $A$ with $p$ rows and $q$ columns is called beautiful if:

  * All elements of the grid are integers between $0$ and $2^{30}-1$, and    * For any subgrid, the XOR of the values at the corners is equal to $0$. Formally, for any four integers $i_1$, $i_2$, $j_1$, $j_2$ ($1 \le i_1 < i_2 \le p$; $1 \le j_1 < j_2 \le q$), $A_{i_1, j_1} \oplus A_{i_1, j_2} \oplus A_{i_2, j_1} \oplus A_{i_2, j_2} = 0$, where $\oplus$ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

There is a partially filled integer grid $G$ with $n$ rows and $m$ columns where only $k$ cells are filled. Polycarp wants to know how many ways he can assign integers to the unfilled cells so that the grid is beautiful.

However, Monocarp thinks that this problem is too easy. Therefore, he will perform $q$ updates on the grid. In each update, he will choose an unfilled cell and assign an integer to it. Note that these updates are persistent. That is, changes made to the grid will apply when processing future updates.

For each of the $q + 1$ states of the grid, the initial state and after each of the $q$ queries, determine the number of ways Polycarp can assign integers to the unfilled cells so that the grid is beautiful. Since this number can be very large, you are only required to output their values modulo $10^9+7$.

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

The first line of each test case contains four integers $n$, $m$, $k$ and $q$ ($2 \le n, m \le 10^5$; $0 \le k \le 10^5$; $q = 0$) — the number of rows, the number of columns, the number of fixed cells, and the number of updates.

The following $k$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

The following $q$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

It is guaranteed that the pairs $(r,c)$ over all assignments are distinct.

It is guaranteed that the sum of $n$, $m$, $k$ and $q$ over all test cases does not exceed $10^5$ respectively.

Output: For each test case, output $q + 1$ lines. The $i$-th line of output should contain the answer of the $i$-th state of the grid modulo $10^9 + 7$.

Examples:
input:
2
3 3 8 0
2 1 6
3 2 12
1 2 6
2 2 0
1 3 10
1 1 0
2 3 12
3 1 10
2 5 2 0
1 1 10
1 2 30
output:
1
489373567

Note: In the first test case of the example, we have the following grid:

$0$| $6$| $10$   ---|---|---   $6$| $0$| $12$   $10$| $12$| $?$      It can be proven that the only valid value for tile $(3, 3)$ is $0$.

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

public class BeautifulGrid {
    static final int MOD = 1_000_000_007;
    static int n, m, k;
    static int varCount;
    static Map<Long, Integer> varMap;
    static int[] parent;
    static int[] rank;
    static int[] parity;
    static boolean inconsistent;
    static int totalVariables;

    public static void main(String[] args) throws IOException {
        // Use BufferedReader and PrintWriter for efficient IO
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);

        int t = Integer.parseInt(br.readLine());
        for(int test = 0; test < t; test++) {
            String[] parts = br.readLine().split(" ");
            n = Integer.parseInt(parts[0]);
            m = Integer.parseInt(parts[1]);
            k = Integer.parseInt(parts[2]);
            int q = Integer.parseInt(parts[3]); // q=0 in this version

            varMap = new HashMap<>();
            int idx = 0;
            // Map variables for first row (excluding (1,1))
            for(int j = 1; j <= m; j++) {
                long key = ((long)1 << 32) | j;
                varMap.put(key, idx++);
            }
            // Map variables for first column (excluding (1,1))
            for(int i = 2; i <= n; i++) {
                long key = ((long)i << 32) | 1;
                varMap.put(key, idx++);
            }
            totalVariables = idx; // Total number of variables
            int[][] fixedCells = new int[k][3];
            for(int i = 0; i < k; i++) {
                parts = br.readLine().split(" ");
                int r = Integer.parseInt(parts[0]);
                int c = Integer.parseInt(parts[1]);
                int v = Integer.parseInt(parts[2]);
                fixedCells[i][0] = r;
                fixedCells[i][1] = c;
                fixedCells[i][2] = v;
            }

            long totalWays = 1;
            for(int bit = 0; bit < 30; bit++) {
                inconsistent = false;
                // Initialize Union-Find structures
                parent = new int[totalVariables + 1]; // +1 for virtual node
                rank = new int[totalVariables + 1];
                parity = new int[totalVariables + 1];
                for(int i = 0; i <= totalVariables; i++) {
                    parent[i] = i;
                    parity[i] = 0;
                }
                int freeVariables = totalVariables;
                // Virtual node index is totalVariables
                int virtualNode = totalVariables;

                // Process fixed cells
                for(int[] cell : fixedCells) {
                    int r = cell[0];
                    int c = cell[1];
                    int v = (cell[2] >> bit) & 1;
                    int varIdx = getVarIndex(r, c);
                    if(varIdx == -1) continue; // Not a variable
                    union(varIdx, virtualNode, v);
                    if(inconsistent) break;
                }
                if(inconsistent) {
                    totalWays = 0;
                    break;
                }
                // Process equations from fixed cells not in first row or first column
                for(int[] cell : fixedCells) {
                    int r = cell[0];
                    int c = cell[1];
                    if(r == 1 || c == 1) continue; // Already processed
                    int v = (cell[2] >> bit) & 1;
                    int var1 = getVarIndex(r - 1, c);
                    int var2 = getVarIndex(r, c - 1);
                    int var3 = getVarIndex(r - 1, c - 1);
                    int rhs = v;
                    // Equation: var1 ^ var2 ^ var3 = rhs
                    // So we can write: var1 ^ var2 ^ var3 ^ rhs = 0
                    // We can represent this as union of var1 and var2 with parity = var3 ^ rhs
                    if(var1 == -1 || var2 == -1 || var3 == -1) continue;
                    int combinedParity = parityValue(var3) ^ rhs;
                    union(var1, var2, combinedParity);
                    if(inconsistent) break;
                }
                if(inconsistent) {
                    totalWays = 0;
                    break;
                }
                // Count the number of components (excluding virtual node)
                int components = 0;
                for(int i = 0; i < totalVariables; i++) {
                    if(find(i) == i && find(i) != find(virtualNode)) {
                        components++;
                    }
                }
                totalWays = (totalWays * modPow(2, components, MOD)) % MOD;
            }
            out.println(totalWays);
        }
        out.flush();
    }

    static int getVarIndex(int r, int c) {
        long key = ((long)r << 32) | c;
        return varMap.getOrDefault(key, -1);
    }

    static int find(int x) {
        if(parent[x] != x) {
            int origParent = parent[x];
            parent[x] = find(parent[x]);
            parity[x] ^= parity[origParent];
        }
        return parent[x];
    }

    static void union(int x, int y, int val) {
        int px = find(x);
        int py = find(y);
        int parityX = parity[x];
        int parityY = parity[y];
        if(px == py) {
            if((parityX ^ parityY) != val) {
                inconsistent = true;
            }
        } else {
            parent[px] = py;
            parity[px] = parityX ^ parityY ^ val;
        }
    }

    static int parityValue(int x) {
        find(x);
        return parity[x];
    }

    static long modPow(long base, long exp, long mod) {
        long result = 1;
        base %= mod;
        while(exp > 0) {
            if((exp & 1) == 1) result = (result * base) % mod;
            base = (base * base) % mod;
            exp >>=1;
        }
        return result;
    }
}

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

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

Think step by step.

Response:
```plaintext
2
3 3 8 0
2 1 6
3 2 12
1 2 6
2 2 0
1 3 10
1 1 0
2 3 12
3 1 10
2 5 2 0
1 1 10
1 2 30
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Billetes MX (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, it is guaranteed that $q = 0$. You can make hacks only if both versions of the problem are solved.

An integer grid $A$ with $p$ rows and $q$ columns is called beautiful if:

  * All elements of the grid are integers between $0$ and $2^{30}-1$, and    * For any subgrid, the XOR of the values at the corners is equal to $0$. Formally, for any four integers $i_1$, $i_2$, $j_1$, $j_2$ ($1 \le i_1 < i_2 \le p$; $1 \le j_1 < j_2 \le q$), $A_{i_1, j_1} \oplus A_{i_1, j_2} \oplus A_{i_2, j_1} \oplus A_{i_2, j_2} = 0$, where $\oplus$ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

There is a partially filled integer grid $G$ with $n$ rows and $m$ columns where only $k$ cells are filled. Polycarp wants to know how many ways he can assign integers to the unfilled cells so that the grid is beautiful.

However, Monocarp thinks that this problem is too easy. Therefore, he will perform $q$ updates on the grid. In each update, he will choose an unfilled cell and assign an integer to it. Note that these updates are persistent. That is, changes made to the grid will apply when processing future updates.

For each of the $q + 1$ states of the grid, the initial state and after each of the $q$ queries, determine the number of ways Polycarp can assign integers to the unfilled cells so that the grid is beautiful. Since this number can be very large, you are only required to output their values modulo $10^9+7$.

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

The first line of each test case contains four integers $n$, $m$, $k$ and $q$ ($2 \le n, m \le 10^5$; $0 \le k \le 10^5$; $q = 0$) — the number of rows, the number of columns, the number of fixed cells, and the number of updates.

The following $k$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

The following $q$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

It is guaranteed that the pairs $(r,c)$ over all assignments are distinct.

It is guaranteed that the sum of $n$, $m$, $k$ and $q$ over all test cases does not exceed $10^5$ respectively.

Output: For each test case, output $q + 1$ lines. The $i$-th line of output should contain the answer of the $i$-th state of the grid modulo $10^9 + 7$.

Examples:
input:
2
3 3 8 0
2 1 6
3 2 12
1 2 6
2 2 0
1 3 10
1 1 0
2 3 12
3 1 10
2 5 2 0
1 1 10
1 2 30
output:
1
489373567

Note: In the first test case of the example, we have the following grid:

$0$| $6$| $10$   ---|---|---   $6$| $0$| $12$   $10$| $12$| $?$      It can be proven that the only valid value for tile $(3, 3)$ is $0$.

Buggy Code:
#include <bits/stdc++.h>
#define mk make_pair
#define fi first
#define se second
#define pb emplace_back
using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
inline int read()
{
	int s = 0, w = 1;
	char ch = getchar();
	while (!isdigit(ch))
	{
		if (ch == '-')
			w = -1;
		ch = getchar();
	}
	while (isdigit(ch))
	{
		s = s * 10 + ch - '0';
		ch = getchar();
	}
	return s * w;
}
const int mod = 1e9 + 7;
inline int Add(int x, int y) {
	return (x + y) >= mod ? (x + y - mod) : (x + y);
}
inline int Mul(int x, int y) {
	return 1ll * x * y % mod;
}
const int N = 4e5 + 10;
int T, n, m, k, q;
struct Node {
	int r, c, v;
} a[N];
int rv[N], cv[N];
int vis[N], f[N];
int tj[N];
inline int find(int x) {
	if(x == f[x]) return x;
	return f[x] = find(f[x]);
}
int low[N], dfn[N], inst[N], col[N], in[N], out[N], dfstime;
int node, top, st[N], scc;
vector <int> G[N];
void tarjan(int x) {
	low[x] = dfn[x] = ++dfstime; inst[x] = 1; st[++top] = x;
	for(auto v : G[x]) {
		if(!dfn[v]) {
			tarjan(v); low[x] = min(low[x], low[v]);
		}
		else if(inst[v]) low[x] = min(low[x], dfn[v]);
	}
	if(dfn[x] == low[x]) {
		int y = 0; ++scc;
		while(y = st[top--]) {
			col[y] = scc;
			inst[y] = 0;
			if(y == x) break;
		}
	}
}
inline void link(int x, int y) {
	G[x].push_back(y);
}
int main() {
	T = read(); while(T--) {
		n = read(); m = read(); k = read(); q = read();
		node = dfstime = top = 0;
		for(int i = 1; i <= n; ++i) {
			in[i] = ++node; out[i] = ++node;
			low[i] = dfn[i] = inst[i] = 0;
		}
		for(int i = 1; i <= m; ++i) {
			int nxt = i + n;
			in[nxt] = ++node; out[nxt] = ++node;
			low[nxt] = dfn[nxt] = inst[nxt] = 0;
		}
		int ok = 0, ov = 0;
		for(int i = 1; i <= k; ++i) {
			a[i].r = read(); a[i].c = read(); a[i].v = read();
			if(a[i].r == 1 && a[i].c == 1) ok = 1, ov = a[i].v;
		}
		int ans = 1;
		int ans0 = 0;
		for(int i = 0; i < 30; ++i) {
			int res = 0;
			for(int O = 0; O < 2; ++O) {
				if(ok && O != (ov >> i & 1)) continue;
				for(int ii = 1; ii <= node; ++ii) G[ii].clear();
				for(int ii = 1; ii <= node; ++ii) {
					dfn[ii] = low[ii] = inst[ii] = 0;
				}
				dfstime = top = scc = 0;
				for(int o = 1; o <= n; ++o) vis[o] = 0, f[o] = o;
				for(int o = 1; o <= m; ++o) vis[o + n] = 0, f[o + n] = o + n;
				vis[1] = vis[1 + n] = 1, f[1] = 1 + n;
				for(int j = 1; j <= k; ++j) {
					if(a[j].r == 1 && a[j].c == 1) continue;
					if(a[j].r == 1) vis[a[j].c + n] = 1;
					if(a[j].c == 1) vis[a[j].r] = 1;
					if(a[j].r != 1 && a[j].c != 1) {
						int nowv = a[j].v >> i & 1;
						nowv ^= O;
						int rp = a[j].r, cp = a[j].c + n;
						if(nowv == 0) {
							link(in[rp], in[cp]);
							link(in[cp], in[rp]);
							link(out[rp], out[cp]);
							link(out[cp], out[rp]);
						}
						else {
							link(in[rp], out[cp]);
							link(in[cp], out[rp]);
							link(out[rp], in[cp]);
							link(out[cp], in[rp]);
						}
						int frp = find(rp), fcp = find(cp);
						if(frp == fcp) continue;
						if(vis[frp] == 1) f[fcp] = frp;
						else f[frp] = fcp;
					}
				}
				for(int p = 1; p <= node; ++p) if(!dfn[p]) tarjan(p);
				int fk = 0;
				for(int o = 1; o <= n; ++o) if(col[in[o]] == col[out[o]]) fk = 1;
				for(int o = 1 + n; o <= m + n; ++o) if(col[in[o]] == col[out[o]]) fk = 1;
				if(fk) {
//					cout << "wdnmd ??? i = " << i << '\n';
					ans0 = 1;
					break;
				}
				for(int o = 1; o <= n; ++o) tj[o] = 0;
				for(int o = 1; o <= m; ++o) tj[o + n] = 0;
				int rans = 1;
				for(int o = 1; o <= n; ++o) {
					int fo = find(o);
					if(tj[fo]) continue;
					tj[fo] = 1;
					int num = 1;
					if(!vis[fo]) num = 2;
					rans = Mul(rans, num);
				}
				for(int o = 1; o <= m; ++o) {
					int fo = find(o + n);
					if(tj[fo]) continue;
					tj[fo] = 1;
					int num = 1;
					if(!vis[fo]) num = 2;
					rans = Mul(rans, num);
				}
				res = Add(res, rans);
				for(int ii = 1; ii <= node; ++ii) G[ii].clear();
				for(int ii = 1; ii <= node; ++ii) {
					dfn[ii] = low[ii] = inst[ii] = 0;
				}
				dfstime = top = scc = 0;
			}
			if(ans0) break;
			ans = Mul(ans, res);
//			cout << "i = " << i << " res = " << res << '\n';
		}
//		cout << "ans0 = " << ans0 << '\n';
		if(ans0) cout << 0 << '\n';
		else cout << ans << '\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
2
3 3 8 0
2 1 6
3 2 12
1 2 6
2 2 0
1 3 10
1 1 0
2 3 12
3 1 10
2 5 2 0
1 1 10
1 2 30
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Billetes MX (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, it is guaranteed that $q = 0$. You can make hacks only if both versions of the problem are solved.

An integer grid $A$ with $p$ rows and $q$ columns is called beautiful if:

  * All elements of the grid are integers between $0$ and $2^{30}-1$, and    * For any subgrid, the XOR of the values at the corners is equal to $0$. Formally, for any four integers $i_1$, $i_2$, $j_1$, $j_2$ ($1 \le i_1 < i_2 \le p$; $1 \le j_1 < j_2 \le q$), $A_{i_1, j_1} \oplus A_{i_1, j_2} \oplus A_{i_2, j_1} \oplus A_{i_2, j_2} = 0$, where $\oplus$ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

There is a partially filled integer grid $G$ with $n$ rows and $m$ columns where only $k$ cells are filled. Polycarp wants to know how many ways he can assign integers to the unfilled cells so that the grid is beautiful.

However, Monocarp thinks that this problem is too easy. Therefore, he will perform $q$ updates on the grid. In each update, he will choose an unfilled cell and assign an integer to it. Note that these updates are persistent. That is, changes made to the grid will apply when processing future updates.

For each of the $q + 1$ states of the grid, the initial state and after each of the $q$ queries, determine the number of ways Polycarp can assign integers to the unfilled cells so that the grid is beautiful. Since this number can be very large, you are only required to output their values modulo $10^9+7$.

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

The first line of each test case contains four integers $n$, $m$, $k$ and $q$ ($2 \le n, m \le 10^5$; $0 \le k \le 10^5$; $q = 0$) — the number of rows, the number of columns, the number of fixed cells, and the number of updates.

The following $k$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

The following $q$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

It is guaranteed that the pairs $(r,c)$ over all assignments are distinct.

It is guaranteed that the sum of $n$, $m$, $k$ and $q$ over all test cases does not exceed $10^5$ respectively.

Output: For each test case, output $q + 1$ lines. The $i$-th line of output should contain the answer of the $i$-th state of the grid modulo $10^9 + 7$.

Examples:
input:
2
3 3 8 0
2 1 6
3 2 12
1 2 6
2 2 0
1 3 10
1 1 0
2 3 12
3 1 10
2 5 2 0
1 1 10
1 2 30
output:
1
489373567

Note: In the first test case of the example, we have the following grid:

$0$| $6$| $10$   ---|---|---   $6$| $0$| $12$   $10$| $12$| $?$      It can be proven that the only valid value for tile $(3, 3)$ is $0$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int N=2e6+6;
int n,m,k,q,x,y,z;
#define pb push_back
vector<int>e[N];
int mi;
const int mod=1e9+7;
typedef long long LL;
LL ksm(LL x,int y)
{
    LL ret=1;
    while(y)
    {
        if(y&1)
        {
            ret=ret*x%mod;
        }
        x=x*x%mod;
        y>>=1;
    }
    return ret;
}
int pd,b[N],vis[N];
void dfs(int u)
{
    vis[u]=1;
    pd+=b[u];
    for(int i=0;i<e[u].size();i++)
    {
        int v=e[u][i];
        if(vis[v]==0)
        {
            dfs(v);
        }
    }
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        cin>>n>>m>>k>>q;
        bool fl=0;
        mi=0;
        for(int i=1;i<=k;i++)
        {
            cin>>x>>y>>z;
            if(x>1&&y>1)
            {
                e[x].pb(y+n-1);
                e[y+n-1].pb(x);
            }else if(x==1&&y==1)
            {
                fl=1;
            }else if(x==1)
            {
                b[y+n-1]=1;
            }else if(y==1)
            {
                b[x]=1;
            }
        }
        if(fl)
        {
            for(int i=2;i<=n+m-1;i++)
            {
                if(vis[i]==0)
                {
                    pd=0;
                    dfs(i);
                    if(pd==0)
                    {
                        mi+=30;
                    }
                }
            }
        }else{
            bool yx=1;
            for(int i=2;i<=n+m-1;i++)
            {
                if(vis[i]==0)
                {
                    pd=0;
                    dfs(i);
                    if(pd==0)
                    {
                        mi+=30;
                    }else if(pd>=2)
                    {
                        yx=0;
                    }
                }
            }
            mi+=yx*30;
        }
        cout<<ksm(2,mi)<<'\n';
        for(int i=1;i<=n+m-1;i++)
        {
            vis[i]=0;b[i]=0;
            e[i].clear();
        }
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
3 3 2 0
1 1 0
1 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Billetes MX (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, it is guaranteed that $q = 0$. You can make hacks only if both versions of the problem are solved.

An integer grid $A$ with $p$ rows and $q$ columns is called beautiful if:

  * All elements of the grid are integers between $0$ and $2^{30}-1$, and    * For any subgrid, the XOR of the values at the corners is equal to $0$. Formally, for any four integers $i_1$, $i_2$, $j_1$, $j_2$ ($1 \le i_1 < i_2 \le p$; $1 \le j_1 < j_2 \le q$), $A_{i_1, j_1} \oplus A_{i_1, j_2} \oplus A_{i_2, j_1} \oplus A_{i_2, j_2} = 0$, where $\oplus$ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

There is a partially filled integer grid $G$ with $n$ rows and $m$ columns where only $k$ cells are filled. Polycarp wants to know how many ways he can assign integers to the unfilled cells so that the grid is beautiful.

However, Monocarp thinks that this problem is too easy. Therefore, he will perform $q$ updates on the grid. In each update, he will choose an unfilled cell and assign an integer to it. Note that these updates are persistent. That is, changes made to the grid will apply when processing future updates.

For each of the $q + 1$ states of the grid, the initial state and after each of the $q$ queries, determine the number of ways Polycarp can assign integers to the unfilled cells so that the grid is beautiful. Since this number can be very large, you are only required to output their values modulo $10^9+7$.

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

The first line of each test case contains four integers $n$, $m$, $k$ and $q$ ($2 \le n, m \le 10^5$; $0 \le k \le 10^5$; $q = 0$) — the number of rows, the number of columns, the number of fixed cells, and the number of updates.

The following $k$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

The following $q$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

It is guaranteed that the pairs $(r,c)$ over all assignments are distinct.

It is guaranteed that the sum of $n$, $m$, $k$ and $q$ over all test cases does not exceed $10^5$ respectively.

Output: For each test case, output $q + 1$ lines. The $i$-th line of output should contain the answer of the $i$-th state of the grid modulo $10^9 + 7$.

Examples:
input:
2
3 3 8 0
2 1 6
3 2 12
1 2 6
2 2 0
1 3 10
1 1 0
2 3 12
3 1 10
2 5 2 0
1 1 10
1 2 30
output:
1
489373567

Note: In the first test case of the example, we have the following grid:

$0$| $6$| $10$   ---|---|---   $6$| $0$| $12$   $10$| $12$| $?$      It can be proven that the only valid value for tile $(3, 3)$ is $0$.

Buggy Code:
M=10**9 + 7
for _ in range(int(input())):
    n,m,k,q = map(int,input().split())
    ax = [1 for i in range(n)]
    bx = [1 for i in range(m)]
    bb = 1
    for i in range(k):
        a,b,c = map(int,input().split())
        a-=1
        b-=1
        if (ax[a]==0):bb=0
        if (bx[b]==0):bb=0
        ax[a] = 0
        bx[b] = 0

    print(pow(2**30,sum(ax)+sum(bx),M))

**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 code, we need to focus on the logic that handles the filled cells and the constraints of the problem. The code initializes two arrays `ax` and `bx` to track the rows and columns that are already filled. If a row or column is filled, it sets the corresponding element in `ax` or `bx` to 0. The final result is calculated based on the sum of the elements in `ax` and `bx`.

The bug in the code is that it incorrectly handles the XOR constraint for the grid. The code does not ensure that the XOR of the values at the corners of any subgrid is equal to 0. Instead, it only checks if a row or column is filled and then calculates the number of ways to fill the remaining cells.

To expose this bug, we need to create a test case where the XOR constraint is violated but the code still outputs a result. Here's a small-scale test input that should expose the bug:

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

Explanation:
1. The first test case has a 3x3 grid with 2 filled cells:
   - Cell (1, 1) is 0.
   - Cell (2, 2) is 0.
   The grid looks like this:
   ```
   0 | ? | ?
   ? | 0 | ?
   ? | ? | ?
   ```
   The only valid way to fill the remaining cells while maintaining the XOR constraint is to set all cells to 0. The expected output is 1.

2. The second test case has a 3x3 grid with 2 filled cells:
   - Cell (1, 1) is 0.
   - Cell (1, 2) is 1.
   The grid looks like this:
   ```
   0 | 1 | ?
   ? | ? | ?
   ? | ? | ?
   ```
   This grid cannot be made beautiful because the XOR of the corners of the subgrid (1,1), (1,2), (2,1), (2,2) will not be 0 for any valid values. The expected output should be 0, but the buggy code will still output a non-zero value.

This test input should help expose the bug in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Billetes MX (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, it is guaranteed that $q = 0$. You can make hacks only if both versions of the problem are solved.

An integer grid $A$ with $p$ rows and $q$ columns is called beautiful if:

  * All elements of the grid are integers between $0$ and $2^{30}-1$, and    * For any subgrid, the XOR of the values at the corners is equal to $0$. Formally, for any four integers $i_1$, $i_2$, $j_1$, $j_2$ ($1 \le i_1 < i_2 \le p$; $1 \le j_1 < j_2 \le q$), $A_{i_1, j_1} \oplus A_{i_1, j_2} \oplus A_{i_2, j_1} \oplus A_{i_2, j_2} = 0$, where $\oplus$ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

There is a partially filled integer grid $G$ with $n$ rows and $m$ columns where only $k$ cells are filled. Polycarp wants to know how many ways he can assign integers to the unfilled cells so that the grid is beautiful.

However, Monocarp thinks that this problem is too easy. Therefore, he will perform $q$ updates on the grid. In each update, he will choose an unfilled cell and assign an integer to it. Note that these updates are persistent. That is, changes made to the grid will apply when processing future updates.

For each of the $q + 1$ states of the grid, the initial state and after each of the $q$ queries, determine the number of ways Polycarp can assign integers to the unfilled cells so that the grid is beautiful. Since this number can be very large, you are only required to output their values modulo $10^9+7$.

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

The first line of each test case contains four integers $n$, $m$, $k$ and $q$ ($2 \le n, m \le 10^5$; $0 \le k \le 10^5$; $q = 0$) — the number of rows, the number of columns, the number of fixed cells, and the number of updates.

The following $k$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

The following $q$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) — indicating that $G_{r, c}$ is assigned the integer $v$.

It is guaranteed that the pairs $(r,c)$ over all assignments are distinct.

It is guaranteed that the sum of $n$, $m$, $k$ and $q$ over all test cases does not exceed $10^5$ respectively.

Output: For each test case, output $q + 1$ lines. The $i$-th line of output should contain the answer of the $i$-th state of the grid modulo $10^9 + 7$.

Examples:
input:
2
3 3 8 0
2 1 6
3 2 12
1 2 6
2 2 0
1 3 10
1 1 0
2 3 12
3 1 10
2 5 2 0
1 1 10
1 2 30
output:
1
489373567

Note: In the first test case of the example, we have the following grid:

$0$| $6$| $10$   ---|---|---   $6$| $0$| $12$   $10$| $12$| $?$      It can be proven that the only valid value for tile $(3, 3)$ is $0$.

Buggy Code:
import sys
import threading

def main():
    import sys
    import math
    sys.setrecursionlimit(1 << 25)
    MOD = 10**9 +7

    from collections import defaultdict

    # Precompute powers of 2 up to 6e6
    MAX_E = 6 * 10**6 +1
    pow2 = [1] * (MAX_E)
    for i in range(1, MAX_E):
        pow2[i] = (pow2[i-1] *2) % MOD

    class GaussBit:
        def __init__(self):
            self.basis = dict()  # var index -> (list of vars, c)
            self.inconsistent = False

        def add_eq(self, vars_set, c):
            vars_list = sorted(vars_set)
            while vars_list:
                var = vars_list[0]
                if var in self.basis:
                    basis_vars, basis_c = self.basis[var]
                    # XOR the basis equation with the current equation
                    new_vars_set = set(vars_list).symmetric_difference(set(basis_vars))
                    new_c = c ^ basis_c
                    vars_list = sorted(new_vars_set)
                    c = new_c
                else:
                    # Set this variable as pivot
                    self.basis[var] = (vars_list.copy(), c)
                    return
            # After elimination, check if c ==0
            if c !=0:
                self.inconsistent = True

        def rank(self):
            return len(self.basis)

    input = sys.stdin.read
    data = input().split()
    ptr =0
    t = int(data[ptr]); ptr +=1
    for _ in range(t):
        n = int(data[ptr]); ptr +=1
        m = int(data[ptr]); ptr +=1
        k = int(data[ptr]); ptr +=1
        q = int(data[ptr]); ptr +=1
        pre_filled = []
        for _ in range(k):
            r = int(data[ptr]); ptr +=1
            c = int(data[ptr]); ptr +=1
            v = int(data[ptr]); ptr +=1
            pre_filled.append( (r,c,v))
        updates = []
        for _ in range(q):
            r = int(data[ptr]); ptr +=1
            c = int(data[ptr]); ptr +=1
            v = int(data[ptr]); ptr +=1
            updates.append( (r,c,v))
        # Define S_set
        S_set = set()
        for c in range(1, m+1):
            S_set.add( (1,c))
        for r in range(1, n+1):
            S_set.add( (r,1))
        # Separate pre-filled
        P_S = dict()
        P_non_S = dict()
        for (r,c,v) in pre_filled:
            if (r,c) in S_set:
                P_S[(r,c)] = v
            else:
                P_non_S[(r,c)] = v
        # S_free
        S_free = list(S_set - set(P_S.keys()))
        x = len(S_free)
        # Assign indices
        cell_to_idx = {cell:i for i, cell in enumerate(S_free)}
        # Initialize GaussBit for each bit
        gauss = [GaussBit() for _ in range(30)]
        # Process pre-filled non_S cells
        for (r,c), v in P_non_S.items():
            for b in range(30):
                v_b = (v >>b)&1
                # G[r,c} = G[r-1,c} ^ G[r,c-1} ^ G[r-1,c-1} = v_b
                cells = [ (r-1,c), (r,c-1), (r-1,c-1) ]
                vars_set = set()
                c_sum = v_b
                for (ri,ci) in cells:
                    if (ri,ci) in P_S:
                        fixed_v = (P_S[(ri,ci)] >>b)&1
                        c_sum ^= fixed_v
                    elif (ri,ci) in S_set:
                        vars_set.add( cell_to_idx[(ri,ci)] )
                    else:
                        # Cells not in S_set and not pre-filled, undefined
                        # According to problem statement, all (r,c) in assignments are unique
                        # So, these cells are not pre-filled and not in S_set, hence their values are undefined
                        # But since constraints involve only S_set cells, it's fine
                        pass
                gauss[b].add_eq(vars_set, c_sum)
        # Check consistency and sum ranks
        consistent = True
        sum_rank_b =0
        for b in range(30):
            if gauss[b].inconsistent:
                consistent = False
                break
            sum_rank_b += gauss[b].rank()
        if consistent:
            if x *30 < sum_rank_b:
                # Should not happen
                answer =0
            else:
                if x *30 - sum_rank_b < MAX_E:
                    answer = pow2[x *30 - sum_rank_b]
                else:
                    # Compute pow(2, e, MOD)
                    answer = pow(2, x *30 - sum_rank_b, MOD)
        else:
            answer =0
        answers = [answer]
        # Process updates
        for (r,c,v) in updates:
            if (r,c) in S_set and (r,c) in cell_to_idx:
                # Assign to a S_set cell that was free
                idx = cell_to_idx[(r,c)]
                # Now, fix this variable to v_b
                # x decreases by1
                x -=1
                # Remove this cell from S_free
                # Assign the variable to v_b by adding an equation: idx = v_b
                for b in range(30):
                    v_b = (v >>b)&1
                    gauss[b].add_eq( {idx}, v_b )
                # Remove the cell from cell_to_idx to prevent future assignments
                del cell_to_idx[(r,c)]
            else:
                # Assign to a non-S_set cell
                # Add a new constraint: G[r-1,c} ^ G[r,c-1} ^ G[r-1,c-1} = v_b
                for b in range(30):
                    v_b = (v >>b)&1
                    cells = [ (r-1,c), (r,c-1), (r-1,c-1) ]
                    vars_set = set()
                    c_sum = v_b
                    for (ri,ci) in cells:
                        if (ri,ci) in P_S:
                            fixed_v = (P_S[(ri,ci)] >>b)&1
                            c_sum ^= fixed_v
                        elif (ri,ci) in S_set:
                            if (ri,ci) in cell_to_idx:
                                vars_set.add( cell_to_idx[(ri,ci)] )
                        else:
                            # Cells not in S_set and not pre-filled, undefined
                            pass
                    gauss[b].add_eq(vars_set, c_sum)
            # After the update, check consistency and sum ranks
            consistent = True
            sum_rank_b =0
            for b in range(30):
                if gauss[b].inconsistent:
                    consistent = False
                    break
                sum_rank_b += gauss[b].rank()
            if consistent:
                if x *30 < sum_rank_b:
                    # Should not happen
                    answer =0
                else:
                    if x *30 - sum_rank_b < MAX_E:
                        answer = pow2[x *30 - sum_rank_b]
                    else:
                        # Compute pow(2, e, MOD)
                        answer = pow(2, x *30 - sum_rank_b, MOD)
            else:
                answer =0
            # Append the answer
            answers.append(answer)
        # Output the answers
        for ans in answers:
            print(ans)

threading.Thread(target=main).start()

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

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

Think step by step.

Response:
```plaintext
1
3 3 8 0
1 1 0
1 2 6
1 3 10
2 1 6
2 2 0
2 3 12
3 1 10
3 2 12
```