**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Billetes MX (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, it is guaranteed that $q \leq 10^5$. 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 $q$ ($2 \le n, m \le 10^5$; $0 \le k, q \leq 10^5$) — 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:
3
3 3 8 1
2 1 6
3 2 12
1 2 6
2 2 0
1 3 10
1 1 0
2 3 12
3 1 10
3 3 1
2 5 2 0
1 1 10
1 2 30
2 5 0 2
1 1 10
1 2 30
output:
1
0
489373567
651321892
769740174
489373567

Note: In the first test case of the example, we initially 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 $0$, so the first answer is $1$. For the second query, the grid does not satisfy the condition, and thus the answer is $0$.

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

using namespace std;

#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")

template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }
template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) { os << '{'; string sep; for (const T &x : v) os << sep << x, sep = ", "; return os << '}'; }
void dbg_out() { cerr << endl; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T) { cerr << ' ' << H; dbg_out(T...); }
#ifdef LOCAL
#define dbg(...) cerr << "(" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__)
#else
#define dbg(...)
#endif

#define ar array
#define ll long long
#define ld long double
#define sze(x) ((int)x.size())
#define all(a) (a).begin(), (a).end()
#define sui cout.tie(NULL); cin.tie(NULL); ios_base::sync_with_stdio(false)
#define mset(a, x) memset(a, x, sizeof(a))
#define md(a) (a%MOD+MOD)%MOD
typedef priority_queue <int, vector<int>, greater<int> > max_heap;
typedef priority_queue <int> min_heap;
const int MAX_N = 1e5 + 5;
const int MOD = 1e9 + 7;
const int INF = 1e9;
const ld EPS = 1e-9;
const int LOG = 30;

int ADD(int a, int b) { return md(a+b); }
int SUB(int a, int b) { return md(a-b); }
int MUL(int a, int b) { return md(1ll*a*b); }
int POW(int a, int b) { int res=1;while(b){if(b&1)res=MUL(res,a);a=MUL(a,a);b>>=1;}return res; }

int col[LOG][2][MAX_N];
vector<pair<int, int>> sz[LOG][2][MAX_N];
pair<int, int> par[LOG][2][MAX_N];
int cnt[LOG];

pair<int, int> getpar(int x, int y, int z) {return (par[x][y][z] == make_pair(y, z) ? make_pair(y, z) : par[x][y][z] = getpar(x, par[x][y][z].first, par[x][y][z].second));}
bool ok = true;
void merge(int x, pair<int, int> u, pair<int, int> v, int dif)
{
    auto uu = u;
    auto vv = v;
    u = getpar(x, u.first, u.second);
    v = getpar(x, v.first, v.second);
    if (sz[x][u.first][u.second].size() < sz[x][v.first][v.second].size()) swap(u, v);
    if (u == v)
    {
        if (col[x][uu.first][uu.second] ^ col[x][vv.first][vv.second] != dif) ok = false;
        return;
    } else
    {
        if (col[x][uu.first][uu.second] ^ col[x][vv.first][vv.second] != dif) for (auto xx : sz[x][v.first][v.second]) col[x][xx.first][xx.second] ^= 1;
        for (auto xx : sz[x][v.first][v.second]) sz[x][u.first][u.second].push_back(xx);
    }
    cnt[x]--;
    par[x][v.first][v.second] = u;
}

void solve() {
    int n, m, k, q;
    cin >> n >> m >> k >> q;
    ok = true;
    for (int i = 0; i < LOG; i++)
    {
        cnt[i] = n + m;
        for (int j = 1; j <= n; j++) col[i][0][j] = 0, sz[i][0][j].clear(), sz[i][0][j].push_back(make_pair(0, j)), par[i][0][j] = make_pair(0, j);
        for (int j = 1; j <= m; j++) col[i][1][j] = 0, sz[i][1][j].clear(), sz[i][1][j].push_back(make_pair(1, j)), par[i][1][j] = make_pair(1, j);
    }
    for (int i = 0; i < k; i++)
    {
        int r, c, v;
        cin >> r >> c >> v;
        for (int i = 0; i < LOG; i++)
        {
            if (v & (1 << i))
            {
                merge(i, make_pair(0, r), make_pair(1, c), 1);
            } else
            {
                merge(i, make_pair(0, r), make_pair(1, c), 0);
            }
        }
    }
    if (!ok) cout << 0 << "\n";
    else cout << POW(2, (cnt[0] - 1) * LOG) << "\n";

    while (q--)
    {
        int r, c, v;
        cin >> r >> c >> v;
        for (int i = 0; i < LOG; i++)
        {
            if (v & (1 << i))
            {
                merge(i, make_pair(0, r), make_pair(1, c), 1);
            } else
            {
                merge(i, make_pair(0, r), make_pair(1, c), 0);
            }
        }
        if (!ok) cout << 0 << "\n";
        else cout << POW(2, (cnt[0] - 1) * LOG) << "\n";
    }
}

int main() {
    sui;
    int tc = 1;
    cin >> tc;
    for (int t = 1; t <= tc; t++) {
        solve();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input for the given problem, we need to focus on edge cases and scenarios that can expose potential bugs in the code. Here are the steps to consider:

1. **Small-Scale Input**: Since the output length is limited, we should use a small grid size.
2. **Initial Fill**: Ensure some cells are initially filled to test the initial state.
3. **Updates**: Include updates that can potentially conflict with the initial filled cells or each other.
4. **Edge Cases**: Consider edge cases such as all cells filled, no cells filled, and cells that form subgrids with conflicting XOR conditions.

Let's construct a test input that meets these criteria:

- **Grid Size**: 3x3
- **Initial Filled Cells**: 2 cells
- **Updates**: 2 updates

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

### Explanation
- **Test Case 1**: We have one test case.
- **Grid Size**: 3x3 grid.
- **Initial Filled Cells**:
  - Cell (1, 1) is filled with value 1.
  - Cell (2, 2) is filled with value 2.
- **Updates**:
  - Update cell (1, 3) with value 3.
  - Update cell (2, 3) with value 4.

### Expected Behavior
- **Initial State**: The grid is:
  ```
  1 0 0
  0 2 0
  0 0 0
  ```
  The number of ways to fill the grid to be beautiful is calculated based on the initial filled cells.

- **After First Update**: The grid becomes:
  ```
  1 0 3
  0 2 0
  0 0 0
  ```
  The number of ways to fill the grid to be beautiful is recalculated.

- **After Second Update**: The grid becomes:
  ```
  1 0 3
  0 2 4
  0 0 0
  ```
  The number of ways to fill the grid to be beautiful is recalculated.

This test input should help expose any issues in the code related to handling initial filled cells and subsequent updates, especially in maintaining 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 (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, it is guaranteed that $q \leq 10^5$. 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 $q$ ($2 \le n, m \le 10^5$; $0 \le k, q \leq 10^5$) — 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:
3
3 3 8 1
2 1 6
3 2 12
1 2 6
2 2 0
1 3 10
1 1 0
2 3 12
3 1 10
3 3 1
2 5 2 0
1 1 10
1 2 30
2 5 0 2
1 1 10
1 2 30
output:
1
0
489373567
651321892
769740174
489373567

Note: In the first test case of the example, we initially 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 $0$, so the first answer is $1$. For the second query, the grid does not satisfy the condition, and thus the answer is $0$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define MAX (int)1e5+50
#define inf (int)1e9
const ll mod = 1000000007;
ll pw[MAX];
int fa[MAX], val[MAX], sz[MAX];
int flag, sum, res;

int find(int x)
{
	while (x != fa[x]) {
		sum ^= val[x];
		x = fa[x];
	}
	return x;
}

void merge(int x, int y)
{
	int fx, fy;
	fx = find(x), fy = find(y);
	if (fx == fy) {
		if (sum)
			flag = 1;
		return;
	}
	if (sz[fx] > sz[fy])
		swap(fx, fy);
	fa[fx] = fy;
	sz[fy] += sz[fx];
	val[fx] = sum;
	--res;
}

int main()
{
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int T;
	cin >> T;
	ll tmp = (1LL << 30);
	pw[0] = 1;
	for (ll i = 1; i <= (ll)1e5; i++) {
		pw[i] = pw[i - 1] * tmp % mod;
	}
	while (T--) {
		int i, j, k, n, m;
		int q;
		cin >> n >> m >> k >> q;
		int r, c, v;
		res = n + m - 1;
		flag = 0;
		for (i = 1; i <= n + m; i++) {
			fa[i] = i;
			val[i] = 0;
			sz[i] = 1;
		}
		for (i = 0; i < k; i++) {
			cin >> r >> c >> v;
			c += n;
			sum = v;
			merge(r, c);
		}
		if (flag)
			cout << "0" << '\n';
		else
			cout << pw[res] << '\n';
		for (i = 0; i < q; i++) {
			cin >> r >> c >> v;
			c += n;
			sum = v;
			merge(r, c);
			if (flag)
				cout << "0" << '\n';
			else
				cout << pw[res] << '\n';
		}
	}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are the steps to create such a test input:

1. **Small Scale Test Case**: Ensure the test case is small enough to be manageable and readable.
2. **Edge Cases**: Include cases where the grid is partially filled and the updates can potentially invalidate the grid.
3. **Corner Cases**: Test with extreme values of \( n \), \( m \), \( k \), and \( q \) within the constraints.
4. **Conflict Scenarios**: Create scenarios where the XOR condition is violated to test the bug in the merging logic.

Given the constraints and the need to expose the bug, let's create a test case:

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

### Explanation:
- **Test Case 1**:
  - **Grid Size**: \( n = 3 \), \( m = 3 \)
  - **Initial Fixed Cells**: \( k = 2 \)
    - \( G_{1,1} = 0 \)
    - \( G_{1,2} = 1 \)
  - **Updates**: \( q = 2 \)
    - \( G_{2,1} = 1 \)
    - \( G_{2,2} = 2 \)

### Expected Output:
- **Initial State**:
  - The grid is:
    ```
    0 1 ?
    1 ? ?
    ? ? ?
    ```
  - The number of ways to fill the grid to make it beautiful is \( 1 \) (since \( G_{2,2} = 0 \) to satisfy the XOR condition).

- **After First Update**:
  - The grid becomes:
    ```
    0 1 ?
    1 1 ?
    ? ? ?
    ```
  - The number of ways to fill the grid to make it beautiful is \( 0 \) (since \( G_{2,2} = 1 \) and \( G_{1,1} \oplus G_{1,2} \oplus G_{2,1} \oplus G_{2,2} \neq 0 \)).

- **After Second Update**:
  - The grid becomes:
    ```
    0 1 ?
    1 2 ?
    ? ? ?
    ```
  - The number of ways to fill the grid to make it beautiful is \( 0 \) (since \( G_{2,2} = 2 \) and \( G_{1,1} \oplus G_{1,2} \oplus G_{2,1} \oplus G_{2,2} \neq 0 \)).

### Final Output:
```plaintext
1
0
0
```

This test case should help expose the bug in the merging logic of the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Billetes MX (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, it is guaranteed that $q \leq 10^5$. 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 $q$ ($2 \le n, m \le 10^5$; $0 \le k, q \leq 10^5$) — 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:
3
3 3 8 1
2 1 6
3 2 12
1 2 6
2 2 0
1 3 10
1 1 0
2 3 12
3 1 10
3 3 1
2 5 2 0
1 1 10
1 2 30
2 5 0 2
1 1 10
1 2 30
output:
1
0
489373567
651321892
769740174
489373567

Note: In the first test case of the example, we initially 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 $0$, so the first answer is $1$. For the second query, the grid does not satisfy the condition, and thus the answer is $0$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int MAXN = 100005, P = 1000000007;
int fa[MAXN], off[MAXN];
int qpow(int a, int b)
{
    int c = 1;
    for (; b; b >>= 1, a = a * a % P)
        if (b & 1)
            c = c * a % P;
    return c;
}
pair<int, int> root(int x)
{
    if (fa[x] == x)
        return {x, 0};
    pair<int, int> p = root(fa[x]);
    fa[x] = p.first;
    off[x] ^= p.second;
    // cout << x << " " << fa[x] << " " << off[x] << endl;
    return {fa[x], off[x]};
}
int merge(int x, int y, int w)
{
    int x0, y0;
    tie(x, x0) = root(x);
    tie(y, y0) = root(y);
    if (x == y)
    {
        // cout << w << " " << x << " " << y << " " << x0 << " " << y0 << endl;
        if ((x0 ^ y0) != w)
            return -1;
        return 0;
    }
    off[y] = x0 ^ y0 ^ w;
    fa[y] = x;
    // cout << y << " " << fa[y] << " " << off[y] << endl;
    return 1;
}
signed main()
{
    int T;
    cin >> T;
    while (T--)
    {
        int n, m, k, q;
        cin >> n >> m >> k >> q;
        for (int i = 0; i < n + m; i++)
            fa[i] = i, off[i] = 0;
        bool dead = 0;
        int cnt = (n + m - 1) * 30;
        for (int i = 0; i < k; i++)
        {
            int r, c, v;
            cin >> r >> c >> v;
            --r, c += n - 1;
            if (!dead)
            {
                int s = merge(r, c, v);
                if (s == -1)
                    dead = 1;
                else
                    cnt -= s * 30;
            }
            // cout << r << " " << c << " " << v << " " << dead << endl;
        }
        cout << (dead ? 0 : qpow(2, cnt)) << endl;
        for (int i = 0; i < q; i++)
        {
            int r, c, v;
            cin >> r >> c >> v;
            --r, c += n - 1;
            if (!dead)
            {
                int s = merge(r, c, v);
                if (s == -1)
                    dead = 1;
                else
                    cnt -= s * 30;
            }
            cout << (dead ? 0 : qpow(2, cnt)) << endl;
        }
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
3 3 4 2
1 1 1
1 2 2
2 1 3
2 2 3
3 3 1 1
1 1 0
1 2 1
2 2 3 2
1 1 0
1 2 0
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 (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, it is guaranteed that $q \leq 10^5$. 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 $q$ ($2 \le n, m \le 10^5$; $0 \le k, q \leq 10^5$) — 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:
3
3 3 8 1
2 1 6
3 2 12
1 2 6
2 2 0
1 3 10
1 1 0
2 3 12
3 1 10
3 3 1
2 5 2 0
1 1 10
1 2 30
2 5 0 2
1 1 10
1 2 30
output:
1
0
489373567
651321892
769740174
489373567

Note: In the first test case of the example, we initially 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 $0$, so the first answer is $1$. For the second query, the grid does not satisfy the condition, and thus the answer is $0$.

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

const long long MOD = 1000000007;
const int MAXN = 200005;
const int MAXP = 20000005;
const int BITS = 30;

long long pow2[MAXP];
int par[BITS][MAXN];
int sizes[BITS][MAXN];

int find(int bit, int node) {
    if (par[bit][node] == node) {
        return node;
    }
    par[bit][node] = find(bit, par[bit][node]);
    return par[bit][node];
}

int merge(int bit, int a, int b) {
    a = find(bit, a);
    b = find(bit, b);
    if (a != b) {
        if (sizes[bit][a] > sizes[bit][b]) {
            swap(a, b);
        }
        par[bit][a] = b;
        sizes[bit][b] += sizes[bit][a];
        return 1;
    }
    return 0;
}

void solve() {
    int n, m, k, q;
    cin >> n >> m >> k >> q;

    for (int i = 0; i < 2 * (n + m); i++) {
        for (int j = 0; j < BITS; j++) {
            par[j][i] = i;
            sizes[j][i] = 1;
        }
    }

    bool possible = true;
    int comps = 30 * 2 * (n + m);
    if (k == 0) {
        cout << pow2[comps / 2 - BITS] << "\n";
    }
    for (int i = 0; i < k + q; i++) {
        int r, c, v;
        cin >> r >> c >> v;
        r--; c--;
        for (int j = 0; j < BITS; j++) {
            if (v & (1 << j)) {
                comps -= merge(j, r, (n + m) + n + c);
                comps -= merge(j, n + c, (n + m) + r);
            } else {
                comps -= merge(j, r, n + c);
                comps -= merge(j, (n + m) + r, (n + m) + n + c);
            }
            if (find(j, r) == find(j, (n + m) + r) || find(j, n + c) == find(j, (n + m) + n + c)) {
                possible = false;
            }
        }
        if (i >= k - 1) {
            if (!possible) {
                cout << "0\n";
            } else {
                assert(comps / 2 - BITS >= 0);
                cout << pow2[comps / 2 - BITS] << "\n";
            }
        }
    }
}

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

    pow2[0] = 1;
    for (int i = 1; i < MAXP; i++) {
        pow2[i] = pow2[i - 1] * 2 % MOD;
    }

    int 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
3 3 2 2
1 1 1
2 2 2
1 3 3
2 3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Billetes MX (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, it is guaranteed that $q \leq 10^5$. 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 $q$ ($2 \le n, m \le 10^5$; $0 \le k, q \leq 10^5$) — 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:
3
3 3 8 1
2 1 6
3 2 12
1 2 6
2 2 0
1 3 10
1 1 0
2 3 12
3 1 10
3 3 1
2 5 2 0
1 1 10
1 2 30
2 5 0 2
1 1 10
1 2 30
output:
1
0
489373567
651321892
769740174
489373567

Note: In the first test case of the example, we initially 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 $0$, so the first answer is $1$. For the second query, the grid does not satisfy the condition, and thus the answer is $0$.

Buggy Code:
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
from collections import defaultdict as ddic, deque
from itertools import *
import random
import math
from bisect import *
import string
from functools import cache

# MOD = 998244353
MOD = 10**9 + 7
inf = float("inf")

class DSU:
    def __init__(self, N):
        self.par = list(range(N))
        self.sz = [1] * N
        self.parity = [0] * N
        self.components = N
        self.bad = False

    def find(self, x):
        if self.par[x] != x:
            p = self.par[x]
            self.par[x] = self.find(p)
            self.parity[x] ^= self.parity[p]
        return self.par[x]

    def union(self, x, y, v):
        xr, yr = self.find(x), self.find(y)
        if xr == yr:
            if (self.parity[x] ^ self.parity[y]) != v:
                self.bad = True
            return False
        if self.sz[xr] < self.sz[yr]:
            xr, yr = yr, xr
        self.par[yr] = xr
        self.parity[yr] = self.parity[x] ^ self.parity[y] ^ v
        self.sz[xr] += self.sz[yr]
        self.components -= 1
        return True

    def size(self, x):
        return self.sz[self.find(x)]

MX = 200001
pow2 = [1] * MX
for i in range(1, MX):
    pow2[i] = pow2[i - 1] * 2 % MOD

def solve():
    R, C, K, Q = rlist()
    updates = [rlist() for _ in range(K)]
    queries = [rlist() for _ in range(Q)]
    ans = [0] * len(queries)

    total = 1
    dsus = [DSU(R + C) for bit in range(30)]
    for bit in range(30):
        for r, c, v in updates:
            r -= 1
            c -= 1
            b = v >> bit & 1
            dsus[bit].union(r, R + c, b)

        if not dsus[bit].bad:
            total *= pow2[dsus[bit].components - 1]
            total %= MOD
        else:
            total = 0
            break

    print(total)
    for r, c, v in queries:
        r -= 1
        c -= 1
        for bit in range(30):
            b = v >> bit & 1
            dsus[bit].union(r, R + c, b)
        if any(dsus[b].bad for b in range(30)):
            print(0)
        else:
            total = 1
            for b in range(30):
                total *= pow2[dsus[b].components - 1]
                total %= MOD
            print(total)

def main():
    T = 1
    T = rint()

    for tc in range(1, 1 + T):
        ans = solve()

        # print(ans)
        # print(*ans)

        # print("Yes" if ans else "No")
        # print("YES" if ans else "NO")
        # print("Alice" if ans else "Bob")
        # print("First" if ans else "Second")
        # print("Case #{}: {}".format(tc, ans))
        # print(len(ans))
        # for row in ans: print(*row)

# region fastio

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
rint = lambda: int(input())

def rlist():
    return list(map(int, input().split()))

def rgrid(n):
    return [rlist() for _ in range(n)]

# endregion

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To create a challenging test input that exposes bugs in the given code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are the steps to create such a test input:

1. **Small Scale**: Ensure the test input is small enough to be manageable but still challenging.
2. **Edge Cases**: Consider cases where the grid is partially filled and the updates can cause conflicts or unexpected behavior.
3. **Valid Input**: Ensure the input adheres to the problem constraints and is complete.

### Step-by-Step Reasoning:

1. **Test Case with Conflicting Updates**:
   - Create a grid where the initial assignments are such that the grid can be beautiful.
   - Add updates that create conflicts, leading to the grid no longer being beautiful.

2. **Test Case with Full Grid**:
   - Create a grid that is fully filled initially and is beautiful.
   - Add updates that do not change the beauty of the grid.

3. **Test Case with Empty Grid**:
   - Create an empty grid and add updates that make the grid beautiful or not.

### Chosen Test Input:
Let's create a test input with the following characteristics:
- A small grid (3x3) with some initial assignments.
- Updates that create conflicts.

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

### Explanation:
1. **Test Case 1**:
   - Initial grid:
     ```
     0 | 1 | ?
     1 | 0 | ?
     ? | ? | ?
     ```
   - Updates:
     - Update (2, 1, 1) -> Grid remains beautiful.
     - Update (2, 2, 0) -> Grid remains beautiful.
   - Expected Output:
     ```
     1
     1
     1
     ```

2. **Test Case 2**:
   - Initial grid:
     ```
     0 | ? | ?
     ? | 0 | ?
     ? | ? | ?
     ```
   - Updates:
     - Update (2, 1, 1) -> Grid is no longer beautiful.
   - Expected Output:
     ```
     1
     0
     ```

3. **Test Case 3**:
   - Initial grid:
     ```
     ? | ? | ?
     ? | ? | ?
     ? | ? | ?
     ```
   - Updates:
     - Update (1, 1, 1) -> Grid is not beautiful.
   - Expected Output:
     ```
     1073741824
     0
     ```

This test input should help expose any bugs in the given code by testing various scenarios including initial and updated conflicts.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is a hard version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le 5 \cdot 10^3$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval. Note that there may exist identical intervals (i.e., there may exist two different indices $i, j$ such that $l_i = l_j$ and $r_i = r_j$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^3$ and the sum of $m$ over all test cases does not exceed $5 \cdot 10^3$.

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
8
2 0
2 1
1 2
5 1
2 4
8 3
1 4
2 5
7 8
7 2
1 4
4 7
7 3
1 2
1 7
3 7
7 4
1 3
4 7
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
18
-1
-1
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 3, 8, 7, 6, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3] = [2, 2, 7]$.

In the fifth and the sixth test case, it can be proven that there is no interesting permutation.

In the seventh test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3, k_4] = [1, 6, 1, 6]$.

In the eighth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int maxn=5e3+100;
int f[maxn][2],g[maxn][2],a[maxn];
void add(int &x,int y) {
	x=max(x,y);
}
int main() {
	ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	memset(f,-63,sizeof(f));
	memset(g,-63,sizeof(g));
	int T;cin>>T;
	while (T--) {
		int n,m,lst=0,ans=0;
		cin>>n>>m;
		f[0][0]=0;f[0][1]=0;
		for (int i=1;i<=m;i++) {
			int l,r;
			cin>>l>>r;
			a[l]++;
			a[r]--;
			if (a[l]<=0||a[r]>=0) {
				ans=-1;
			}
		}
		for (int i=1,sum=0,lst=0,bl=0;i<=n;i++,lst++) {
			for (int x=0;x<=lst;x++) {
				if (a[i]==0&&sum==0) {
					add(g[x+1][0],f[x][0]+lst);
					add(g[x+1][0],f[x][1]+lst);
					add(g[x][1],f[x][0]+lst-x);
					add(g[x][1],f[x][1]+lst-x);
				}
				if (a[i]>0&&sum==0) {
					add(g[x+1][0],f[x][0]+lst);
					add(g[x+1][0],f[x][1]+lst);
				}
				if (a[i]>0&&sum>0) {
					if (bl) ans=-1;
					add(g[x+1][0],f[x][0]+lst);
				}
				if (a[i]==0&&sum>0) {
					add(g[x+1][0],f[x][0]+lst);
					add(g[x][1],f[x][0]+lst-x);
				}
				if (a[i]<0&&sum>0) {
					add(g[x][1],f[x][0]+lst-x);
					add(g[x][1],f[x][1]+lst-x);
				}
			}
			memcpy(f,g,sizeof(f));
			memset(g,-63,sizeof(g));
			sum+=a[i];
			if (a[i]<0) bl=1;
			if (sum==0) bl=0;
		}
		if (ans!=-1) {
			for (int i=0;i<=n;i++) add(ans,f[i][0]),add(ans,f[i][1]);
		}
		cout<<ans<<"\n";
		memset(a,0,sizeof(a));
	}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are the steps to create such a test input:

1. **Check for Overlapping Intervals**: Ensure that the intervals overlap in such a way that the conditions for interesting permutations are not trivially satisfied.
2. **Check for Edge Cases**: Test with the smallest and largest possible values of `n` and `m`.
3. **Check for No Interesting Permutations**: Create a test case where no interesting permutation exists to ensure the code handles this correctly.
4. **Check for Maximum Inversions**: Create a test case where the number of inversions is maximized.

Given these considerations, here is a small-scale test input that should expose the bug:

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

Explanation:
1. **Test Case 1**: `n = 2, m = 0`
   - No intervals, so any permutation of length 2 is interesting.
   - Expected output: 1 (since the only permutation is [2, 1] with 1 inversion).

2. **Test Case 2**: `n = 2, m = 1, [1, 2]`
   - One interval [1, 2], which covers the entire permutation.
   - Expected output: 0 (since the only permutation is [1, 2] with 0 inversions).

3. **Test Case 3**: `n = 5, m = 2, [1, 3], [3, 5]`
   - Two intervals that do not overlap and cover the entire permutation.
   - Expected output: 8 (since the permutation [5, 4, 3, 2, 1] has 10 inversions, but the maximum interesting permutation is [5, 3, 4, 2, 1] with 8 inversions).

4. **Test Case 4**: `n = 8, m = 3, [1, 4], [2, 5], [7, 8]`
   - Three intervals that cover most of the permutation.
   - Expected output: -1 (since it can be proven that there is no interesting permutation for these intervals).

This test input should help identify issues in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is a hard version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le 5 \cdot 10^3$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval. Note that there may exist identical intervals (i.e., there may exist two different indices $i, j$ such that $l_i = l_j$ and $r_i = r_j$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^3$ and the sum of $m$ over all test cases does not exceed $5 \cdot 10^3$.

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
8
2 0
2 1
1 2
5 1
2 4
8 3
1 4
2 5
7 8
7 2
1 4
4 7
7 3
1 2
1 7
3 7
7 4
1 3
4 7
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
18
-1
-1
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 3, 8, 7, 6, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3] = [2, 2, 7]$.

In the fifth and the sixth test case, it can be proven that there is no interesting permutation.

In the seventh test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3, k_4] = [1, 6, 1, 6]$.

In the eighth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int maxn=5e3+9;
int c[maxn],f[maxn][maxn],n,m,T,tg,tp,pre[maxn];
struct node{
    int x,y;
}range[maxn],nrg[maxn];
int cmp(node t1,node t2){
    return t1.x<t2.x;
}
/*
f[i][j]:前i个数，填j个大数的方案数
*/
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
cin>>T;
while(T--){
    tg=0;tp=0;
    cin>>n>>m;
    for(int i=1;i<=m;i++)
        cin>>range[i].x>>range[i].y;
    sort(range+1,range+m+1,cmp);
    for(int i=1;i<=n;i++)
        c[i]=-1,pre[i]=0;//可0可1
    for(int i=1;i<=m;i++){
        int x=range[i].x,y=range[i].y;
        if(c[x]==1)
            tg=1;
        else
            c[x]=0;
        if(c[y]==0)
            tg=0;
        else
            c[y]=1;
    }
    if(tg){
        cout<<"-1\n";
        continue;
    }
    int nl=range[1].x,nr=range[1].y;
    for(int i=2;i<=m;i++) {
       // cout<<nl<<"----------"<<nr<<" "<<min(nr,range[i].y)+1 <<" "<<max(nr,range[i].y)-1<<endl;
        if(nr<range[i].x){
            node tmp;
            tmp.x=nl;tmp.y=nr;
            nrg[++tp]=tmp;
            nl=range[i].x,nr=range[i].y;
            continue;
        }
        for(int j=nl+1;j<range[i].x;j++) {
            if(c[j]==1)
                tg=1;
            else
                c[j]=0;
        }
        for(int j=min(nr,range[i].y)+1;j<=max(nr,range[i].y)-1;j++){
        if(c[j]==0)
            tg=1;
        else
            c[j]=1;
        }
        nl=range[i].x;
        nr=min(range[i].y,nr);
    }
    tp++;
    nrg[tp].x=nl;nrg[tp].y=nr;
    if(tg==1){
        cout<<"-1\n";
        continue;
    }
    for(int i=1;i<=tp;i++)
        pre[nrg[i].y]=nrg[i].x;
    /*
    for(int i=1;i<=n;i++)
        cout<<c[i]<<" ";
    cout<<endl;
    for(int i=1;i<=tp;i++)
        cout<<nrg[i].x<<" "<<nrg[i].y<<endl;*/
        for(int i=0;i<=n;i++)
        for(int j=0;j<=n;j++)
           f[i][j]=-1e8;
    f[0][0]=0;
    for(int i=1;i<=n;i++)
    for(int j=0;j<=i;j++){
        if(!pre[i]){
            if(c[i]==0){
                f[i][j]=f[i-1][j]+j;
            }
            else if(c[i]==1&&j>=1){
                f[i][j]=f[i-1][j-1];
            }
            else{
                if(j>=1)
                f[i][j]=max(f[i-1][j]+j,f[i-1][j-1]);
            else
                f[i][j]=f[i-1][j]+j;
            }
        }
        else{
            int len=(i-pre[i]+1);
            for(int k=1;k<len;k++) {//枚举1的个数
                if(k>j) continue;
                f[i][j]=max(f[i][j],f[pre[i]-1][j-k]+(len-k)*(j-k));
            }
        }
        if(f[i][j]<0)
            f[i][j]=-1e8;
    }
    /*
    for(int i=1;i<=n;i++) {
        for(int j=0;j<=n;j++)
           cout<<max(-1,f[i][j])<<" ";
    cout<<endl;
    }
    cout<<"----------\n";*/
    int res=0;
    for(int i=0;i<=n;i++)
        if(f[n][i]>=0)
        res=max(res,f[n][i]+i*(i-1)/2+(n-i)*(n-i-1)/2);
    cout<<res<<'\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:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases or scenarios that maximize the failure probability. Here are the steps to think through this:

1. **Understanding the Bug**: The provided code has several logical steps, including sorting intervals, marking positions, and dynamic programming to calculate the maximum number of inversions. The bug could be in any of these steps.

2. **Identifying Edge Cases**:
   - **Overlapping Intervals**: Intervals that overlap in complex ways.
   - **Intervals that are very close to each other**: This can cause issues in the marking and dynamic programming steps.
   - **Intervals that span the entire range**: This can cause issues in the boundary conditions.
   - **No interesting permutations**: This can help verify if the code correctly identifies when no interesting permutations exist.

3. **Constructing the Test Case**:
   - We will create a test case with a small number of test cases to keep the input small.
   - Each test case will have a small number of intervals to ensure it fits the output length limit.
   - The intervals will be designed to challenge the code's ability to handle overlapping and close intervals.

Here is a small-scale test input that is complete and valid:

```plaintext
3
5 2
1 3
4 5
5 2
1 2
3 5
5 3
1 3
2 4
4 5
```

### Explanation:
1. **Test Case 1**:
   - Intervals: `[1, 3]` and `[4, 5]`
   - This test case has two intervals that do not overlap but are close to each other. The code should correctly handle this and find an interesting permutation.

2. **Test Case 2**:
   - Intervals: `[1, 2]` and `[3, 5]`
   - This test case has one interval that is very short and another that is longer. The code should correctly handle the short interval and find an interesting permutation.

3. **Test Case 3**:
   - Intervals: `[1, 3]`, `[2, 4]`, and `[4, 5]`
   - This test case has overlapping intervals. The code should correctly handle the overlapping intervals and find an interesting permutation.

These test cases are designed to expose potential issues in the code's handling of interval overlaps and close intervals.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is a hard version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le 5 \cdot 10^3$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval. Note that there may exist identical intervals (i.e., there may exist two different indices $i, j$ such that $l_i = l_j$ and $r_i = r_j$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^3$ and the sum of $m$ over all test cases does not exceed $5 \cdot 10^3$.

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
8
2 0
2 1
1 2
5 1
2 4
8 3
1 4
2 5
7 8
7 2
1 4
4 7
7 3
1 2
1 7
3 7
7 4
1 3
4 7
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
18
-1
-1
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 3, 8, 7, 6, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3] = [2, 2, 7]$.

In the fifth and the sixth test case, it can be proven that there is no interesting permutation.

In the seventh test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3, k_4] = [1, 6, 1, 6]$.

In the eighth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int T,n,m;
long long dp[5005][5005];
pair<int,int> seg[5005];
int ed[5005], constrain[5005];

typedef long long LL;
void solve(){
    for(int i=1;i<=m;i++){
        cin >> seg[i].first >> seg[i].second;
    }
    sort(seg+1, seg+m+1);

    int mnl,mnr,mxl,mxr;
    mnl = mxl = seg[1].first;
    mnr = mxr = seg[1].second;

    vector<pair<int,int>> lr;
    for(int i=2;i<=m;i++){
        if(seg[i].first > mxr){
            lr.push_back({mxl, mnr});
            for(int i=mnl; i<mxl;i++) constrain[i] = 0;
            for(int j=mnr+1;j<=mxr;j++) constrain[j] = 1;
            mnl = mxl = seg[i].first;
            mnr = mxr = seg[i].second;
        }
        else{
            mnl = min(mnl, seg[i].first);
            mnr = min(mnr, seg[i].second);
            mxl = max(mxl, seg[i].first);
            mxr = max(mxr, seg[i].second);
        }

        if(mxl >= mnr){
            cout << -1 << endl;
            return;
        }
    }
    lr.push_back({mxl, mnr});
    for(auto[x,y]: lr){
        // cout << x << " " << y << "!!!!!!\n";
        for(int i=x;i<=y;i++) ed[i] = y;
    }

    for(int i=1;i<=n;i++){
        for(int z=0;z<=i;z++){
            if(constrain[i] == -1){
                if(ed[i]){
                    if(ed[i] == i) continue;
                    // i is last pos fill 0
                    if(z) dp[ed[i]][z] = max(dp[ed[i]][z], dp[i-1][z-1] + (i-z));

                    // or normally fill 0
                    if(z) dp[i][z] = max(dp[i-1][z-1] + (i-z), dp[i][z]);
                }
                else{
                    // fill 1
                    dp[i][z] = max(dp[i-1][z], dp[i][z]);

                    // fill 0
                    if(z) dp[i][z] = max(dp[i-1][z-1] + (i-z), dp[i][z]);
                }
            }
            else{
                if(constrain[i] == 1){
                    dp[i][z] = max(dp[i-1][z], dp[i][z]);
                }
                else{
                    if(z) dp[i][z] = max(dp[i-1][z-1] + (i-z), dp[i][z]);
                }
            }
        }
    }

    // for(int i=0;i<=n;i++){
    //     for(int j=0;j<=i;j++){
    //         cout << dp[i][j] << " ";
    //     }
    //     cout << endl;
    // }
    long long ans = 0;
    for(int i=0;i<=n;i++){
        LL tmp1 = 1LL * i * (i-1) / 2;
        LL tmp2 = 1LL * (n-i) * (n-i-1) / 2;
        ans = max(ans, dp[n][i] + tmp1 + tmp2);
    }
    cout << ans << endl;
}

int main(){
    cin >> T;
    while(T--){
        cin >> n >> m;
        for(int i=0;i<=n;i++){
            for(int j=0;j<=n;j++){
                dp[i][j] = -1e18;
            }
        }
        dp[0][0] = 0;
        for(int i=0;i<=n;i++){
            ed[i] = 0;
            constrain[i] = -1;
        }
        solve();
    }
}

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

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

Think step by step.

Response:
```plaintext
3
2 0
2 1
1 2
5 1
2 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is a hard version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le 5 \cdot 10^3$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval. Note that there may exist identical intervals (i.e., there may exist two different indices $i, j$ such that $l_i = l_j$ and $r_i = r_j$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^3$ and the sum of $m$ over all test cases does not exceed $5 \cdot 10^3$.

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
8
2 0
2 1
1 2
5 1
2 4
8 3
1 4
2 5
7 8
7 2
1 4
4 7
7 3
1 2
1 7
3 7
7 4
1 3
4 7
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
18
-1
-1
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 3, 8, 7, 6, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3] = [2, 2, 7]$.

In the fifth and the sixth test case, it can be proven that there is no interesting permutation.

In the seventh test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3, k_4] = [1, 6, 1, 6]$.

In the eighth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
#include <iostream>
#include <array>
#include <vector>
#include <cmath>
#include <algorithm>
#include <functional>
#include <utility>
#include <map>
#include <queue>
#define pb(x) push_back(x)
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef array<ll,2> ll2;
typedef array<ll,3> ll3;
typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> OS;
void solve(ll T){
  ll n,i,k,j,l,m;
  cin>>n>>m;
  if(T==66)cout<<n<<','<<m<<',';
  vector<ll2>interv(m);
  vector<ll>mars(n+2),in(n+2);
  set<ll>st,dr;
  for(i=0;i<m;i++){
    if(T==66)cout<<interv[i][0]<<','<<interv[i][1]<<',';
    cin>>interv[i][0]>>interv[i][1];
    st.insert(interv[i][0]);
    dr.insert(interv[i][1]);
    mars[interv[i][0]]++;
    mars[interv[i][1]+1]--;
  }
  for(i=1;i<=n;i++)in[i]=in[i-1]+mars[i];
  vector<ll>S(n+2),D(n+2);
  vector<ll>marS(n+2),marD(n+2);
  for(i=0;i<m;i++){
    marS[interv[i][0]]++;
    marD[interv[i][1]+1]--;
    //cout<<interv[i][0]<<' ';
    auto it=st.upper_bound(interv[i][1]);
    if(it!=st.begin()){
      it--;
      marS[(*it)+1]--;
      //cout<<(*it)<<' ';
    }
    else{
      marS[interv[i][0]+1]--;
      //cout<<interv[i][0]<<' ';
    }
    it=dr.lower_bound(interv[i][0]);
    if(it!=dr.end()){
      marD[*it]++;
      //cout<<(*it)<<' ';
    }
    else{
      marD[interv[i][1]]++;
      //cout<<interv[i][1]<<' ';
    }
    //cout<<interv[i][1]<<'\n';
  }
  ll delt=0,mx=0;
  OS ST,DR;
  for(i=1;i<=n;i++){
    S[i]=S[i-1]+marS[i];
    D[i]=D[i-1]+marD[i];
    if(S[i]*D[i]!=0)return cout<<"-1\n",void();
    if(S[i])ST.insert(i),delt+=(ll)DR.size();
    else DR.insert(i);
  }
  mx=delt+(ll)DR.size()*((ll)DR.size()-1)/2+(ll)ST.size()*((ll)ST.size()-1)/2;
  //cout<<ST.size()<<' '<<DR.size()<<' '<<delt<<'\n';
  vector<ll>poz;
  for(i=n;i>=1;i--){
    if(D[i])continue;
    if(in[i]==0) poz.pb(i);
    ll p=i;
    while(in[p]&&S[p-1]&&!D[p])S[p]=1,poz.pb(p),S[p++]=1;
  }
  for(ll x:poz){
    DR.erase(x);
    delt+=(ll)DR.order_of_key(x)-(ll(ST.size())-(ll)ST.order_of_key(x));
    ST.insert(x);
    //cout<<ST.size()<<' '<<DR.size()<<' '<<delt<<' '<<x<<'\n';
    mx=max(mx,delt+(ll)DR.size()*((ll)DR.size()-1)/2+(ll)ST.size()*((ll)ST.size()-1)/2);
  }
  cout<<mx<<'\n';
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    ll T=0;
    ll t;cin>>t;while(t--)
    solve(++T);
    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 2
1 2
2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is a hard version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le 5 \cdot 10^3$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval. Note that there may exist identical intervals (i.e., there may exist two different indices $i, j$ such that $l_i = l_j$ and $r_i = r_j$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^3$ and the sum of $m$ over all test cases does not exceed $5 \cdot 10^3$.

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
8
2 0
2 1
1 2
5 1
2 4
8 3
1 4
2 5
7 8
7 2
1 4
4 7
7 3
1 2
1 7
3 7
7 4
1 3
4 7
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
18
-1
-1
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 3, 8, 7, 6, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3] = [2, 2, 7]$.

In the fifth and the sixth test case, it can be proven that there is no interesting permutation.

In the seventh test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3, k_4] = [1, 6, 1, 6]$.

In the eighth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
#include<cstdio>
#include<algorithm>
#define N 5010
using namespace std;
int read()
{ char ch=getchar();int p=0;
  for(;ch<'0'||'9'<ch;ch=getchar());
  for(;'0'<=ch&&ch<='9';p=(p<<3)+(p<<1)+ch-'0',ch=getchar());
  return p;
}
void print(int x)
{ if(x<10) { putchar(x+'0'); return; }
  print(x/10),putchar(x%10+'0');
}
struct LINE
{ int l,r;
  #define l(x) line[x].l
  #define r(x) line[x].r
  #define L(x) Line[x].l
  #define R(x) Line[x].r
}line[N>>1],Line[N>>1];
bool cmp(const LINE &A,const LINE &B)
{ return A.l==B.r?A.r<B.r:A.l<B.l;
}
int b[2][N];
int pre(int n)
{ int m=read();
  for(int i=1;i<=m;i++) l(i)=read(),r(i)=read();
  sort(line+1,line+m+1,cmp);
  int lnow=l(1),rnow=r(1),e=0;
  for(int i=2;i<=m;i++)
	if(rnow<l(i)) L(++e)=lnow,R(e)=rnow,lnow=l(i),rnow=r(i);
	else
	  { if(rnow==l(i)) return -1;
		b[0][lnow]++,b[0][l(i)]--;
		b[1][min(rnow,r(i))+1]++,b[1][max(rnow,r(i))+1]--;
		lnow=l(i),rnow=min(rnow,r(i));
	  }
  L(++e)=lnow,R(e)=rnow;
  for(int i=1;i<=n+1;i++) b[0][i]+=b[0][i-1],b[1][i]+=b[1][i-1];
  for(int i=1;i<=n;i++)
	if(b[0][i]&&b[1][i])
	  return -1;
  for(int i=1;i<=e;i++)
	if(b[1][L(i)]||b[0][R(i)])
	  return -1;
  return e;
}
int f[N][N];
void clear(int n)
{ for(int i=1;i<=n;i++)
	{ b[0][i]=b[1][i]=0;
	  for(int j=1;j<i;j++)
		f[i][j]=0;
	}
}
int main()
{ int T=read();
  while(T--)
	{ int n=read(),m=pre(n);
	  if(m<0)
		{ putchar('-'),putchar('1'),putchar('\n'),clear(n);
		  continue;
		}
	  int p=1,minn=0,maxn=0;
	  for(int i=1;i<=n;i++)
		if(p>m||i<L(p))
		  { if(!b[1][i])
			  for(int j=max(minn,1);j<=maxn;j++) f[i][j]=f[i-1][j]+j;
			else minn++;
			if(!b[0][i])
			  { maxn++;
				for(int j=max(minn,1);j<=maxn;j++)
				  f[i][j]=max(f[i][j],f[i-1][j-1]);
			  }
		  }
		else
		  { int up=R(p)-L(p)+1;
			for(int k=1;k<up;k++)
			  { int num=up-k;
				for(int j=max(minn,1);j<=maxn;j++)
				  f[R(p)][j+k]=max(f[R(p)][j+k],f[i-1][j]+j*num);
			  }
			minn++,maxn+=up-1,i=R(p++);
		  }
	  int ans=0;
	  for(int i=max(1,minn);i<=maxn;i++)
		ans=max(ans,f[n][i]+i*(i-1)/2+(n-i)*(n-i-1)/2);
	  print(ans),putchar('\n');
	  clear(n);
	}
}

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

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

Think step by step.

Response:
To generate a challenging test input for the given buggy code, we need to focus on edge cases and scenarios that can expose the bugs. Here are some key points to consider:

1. **Edge Cases with Intervals**: Intervals that are very small or overlap in specific ways can cause issues.
2. **Constraints**: Ensure the input respects the constraints but still challenges the algorithm.
3. **Special Conditions**: Intervals that are identical or have specific overlaps can cause unexpected behavior.

Given the constraints and the nature of the problem, let's construct a test case that includes overlapping intervals and small intervals to test the edge cases.

### Test Case:
```plaintext
1
6 4
1 3
2 4
4 5
5 6
```

### Explanation:
1. **Test Case Description**:
   - **Number of Test Cases**: 1
   - **Permutation Length**: 6
   - **Number of Intervals**: 4
   - **Intervals**: [1, 3], [2, 4], [4, 5], [5, 6]

2. **Purpose**:
   - **Overlapping Intervals**: The intervals [1, 3] and [2, 4] overlap, and the intervals [4, 5] and [5, 6] also overlap. This can test how the algorithm handles overlapping intervals.
   - **Small Intervals**: The intervals are relatively small, which can test the algorithm's ability to handle small ranges.
   - **Edge Cases**: The intervals are designed to create a scenario where the algorithm must carefully manage the intervals and their overlaps to determine if there is an interesting permutation and to calculate the maximum number of inversions.

This test case should help expose any issues in the buggy code, especially in handling overlapping intervals and small ranges.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is a hard version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le 5 \cdot 10^3$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval. Note that there may exist identical intervals (i.e., there may exist two different indices $i, j$ such that $l_i = l_j$ and $r_i = r_j$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^3$ and the sum of $m$ over all test cases does not exceed $5 \cdot 10^3$.

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
8
2 0
2 1
1 2
5 1
2 4
8 3
1 4
2 5
7 8
7 2
1 4
4 7
7 3
1 2
1 7
3 7
7 4
1 3
4 7
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
18
-1
-1
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 3, 8, 7, 6, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3] = [2, 2, 7]$.

In the fifth and the sixth test case, it can be proven that there is no interesting permutation.

In the seventh test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3, k_4] = [1, 6, 1, 6]$.

In the eighth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

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

using LL = long long;

void solve() {
    int n, m; cin >> n >> m;
    vector<int> a(m), b(m);
    for (int i = 0; i < m; i ++) cin >> a[i] >> b[i];
    vector<int> L(n + 1, 0);
    vector<int> vis(n + 1, -1);
    vector<int> id(m);
    iota(id.begin(), id.end(), 0);
    sort(id.begin(), id.end(), [&] (int x, int y) {
        return a[x] < a[y];
    });
    if (m) {
        int la = a[id[0]], lb = a[id[0]], ra = b[id[0]], rb = b[id[0]];
        for (int i = 1; i <= m; i ++) {
            if (i == m || a[id[i]] > rb) {
                if (ra <= lb) {
                    cout << "-1\n";
                    return;
                }
                for (int j = la; j < lb; j ++) vis[j] = 0;
                for (int j = ra + 1; j <= rb; j ++) vis[j] = 1;
                L[lb] = ra;
                la = lb = a[id[i]];
                ra = rb = b[id[i]];
            } else {
                la = min(la, a[id[i]]);
                lb = max(lb, a[id[i]]);
                ra = min(ra, b[id[i]]);
                rb = max(rb, b[id[i]]);
            }
        }
    }
    vector<vector<int>> dp(n + 1, vector<int>(n + 1, -1));
    dp[0][0] = 0;
    for (int i = 1; i <= n; i ++) {
        for (int j = 0; j < i; j ++) {
            if (dp[i - 1][j] == -1) continue;
            if (L[i]) {
                int r = L[i], d = r - i + 1;
                for (int k = 1; k < d; k ++) {
                    dp[r][j + d - k] = max(dp[r][j + d - k], dp[i - 1][j] + j * k);
                }
            } else {
                if (~vis[i]) {
                    dp[i][j + vis[i]] = max(dp[i][j + vis[i]], dp[i - 1][j] + j * (1 - vis[i]));
                } else {
                    dp[i][j + 1] = max(dp[i][j + 1], dp[i - 1][j]);
                    dp[i][j] = max(dp[i][j], dp[i - 1][j] + j);
                }
            }
        }
    }
    int rlt = 0;
    for (int i = 0; i <= n; i ++) {
        if (dp[n][i] == -1) continue;
        rlt = max(rlt, dp[n][i] + i * (i - 1) / 2 + (n - i) * (n - i - 1) / 2);
    }
    cout << rlt << '\n';
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(0);
    cin.tie(0);

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

1. **Understand the Bug**: The provided code seems to have issues with handling intervals and ensuring the conditions for interesting permutations are met. Specifically, the logic for merging intervals and checking the validity of the intervals might be flawed.

2. **Identify Edge Cases**:
   - **Single Interval**: Test with a single interval that spans the entire permutation.
   - **Multiple Overlapping Intervals**: Test with multiple intervals that overlap in complex ways.
   - **Non-overlapping Intervals**: Test with intervals that do not overlap but are close to each other.
   - **Empty Intervals**: Test with no intervals.
   - **Invalid Intervals**: Test with intervals that cannot form an interesting permutation.

3. **Construct the Test Input**:
   - **Single Interval Spanning the Entire Permutation**:
     - This should be a valid case where the maximum number of inversions can be calculated.
   - **Multiple Overlapping Intervals**:
     - This should be a complex case where the intervals overlap in a way that makes it difficult to determine the interesting permutations.
   - **Non-overlapping Intervals**:
     - This should be a case where the intervals are disjoint but close to each other.
   - **Empty Intervals**:
     - This should be a simple case where no intervals are provided.
   - **Invalid Intervals**:
     - This should be a case where the intervals cannot form an interesting permutation.

Based on these considerations, here is a small-scale test input that covers these edge cases:

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

### Explanation of the Test Input:
1. **Test Case 1**:
   - **Input**: `5 1 1 5`
   - **Description**: A single interval `[1, 5]` that spans the entire permutation.
   - **Expected Output**: The maximum number of inversions for a permutation of length 5.

2. **Test Case 2**:
   - **Input**: `5 2 1 3 3 5`
   - **Description**: Two overlapping intervals `[1, 3]` and `[3, 5]`.
   - **Expected Output**: The maximum number of inversions for a permutation of length 5.

3. **Test Case 3**:
   - **Input**: `5 3 1 2 2 3 3 5`
   - **Description**: Three non-overlapping intervals `[1, 2]`, `[2, 3]`, and `[3, 5]`.
   - **Expected Output**: The maximum number of inversions for a permutation of length 5.

4. **Test Case 4**:
   - **Input**: `5 0`
   - **Description**: No intervals.
   - **Expected Output**: The maximum number of inversions for a permutation of length 5 without any intervals.

5. **Test Case 5**:
   - **Input**: `5 2 1 3 4 5`
   - **Description**: Two intervals `[1, 3]` and `[4, 5]` that are close but do not overlap.
   - **Expected Output**: The maximum number of inversions for a permutation of length 5.

This test input should help expose any bugs in the provided code by covering a variety of edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is a hard version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le 5 \cdot 10^3$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval. Note that there may exist identical intervals (i.e., there may exist two different indices $i, j$ such that $l_i = l_j$ and $r_i = r_j$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^3$ and the sum of $m$ over all test cases does not exceed $5 \cdot 10^3$.

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
8
2 0
2 1
1 2
5 1
2 4
8 3
1 4
2 5
7 8
7 2
1 4
4 7
7 3
1 2
1 7
3 7
7 4
1 3
4 7
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
18
-1
-1
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 3, 8, 7, 6, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3] = [2, 2, 7]$.

In the fifth and the sixth test case, it can be proven that there is no interesting permutation.

In the seventh test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3, k_4] = [1, 6, 1, 6]$.

In the eighth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
import math
import heapq
import itertools
import bisect
import random
import time
from collections import deque
import sys
from cmath import exp,pi
from functools import cmp_to_key
input=sys.stdin.readline

t=int(input())
for _ in range(t):
    n,m=map(int,input().split())
    arr=[]
    data=[-1]*n
    flag=0
    nodown=[0]*n
    for i in range(m):
        a,b=map(int,input().split())
        a-=1
        b-=1
        if data[a]==1:
            flag=1
        if data[b]==0:
            flag=1
        data[a]=0
        data[b]=1
        nodown[a]+=1
        nodown[b]-=1

    for i in range(1,n):
        nodown[i]+=nodown[i-1]
    for i in range(n):
        if nodown[i]>0:
            nodown[i]=1
    #print(nodown)

    if flag==1:
        print(-1)
    else:
        dp=[[[-10**9]*(2) for _ in range(n+1)] for _ in range(n+1)] #total, down, cur
        dp[0][0][1]=dp[0][0][0]=0
        for i in range(n):
            for j in range(n+1):
                if data[i]!=1: #0
                    #d->d
                    if j>0:
                        dp[i+1][j][0]=max(dp[i+1][j][0],dp[i][j-1][0]+i)
                    #u->d
                    if j>0 and (i==0 or nodown[i-1]==0):
                        dp[i+1][j][0]=max(dp[i+1][j][0],dp[i][j-1][1]+i)

                if data[i]!=0: #1
                    #u->u
                    dp[i+1][j][1]=max(dp[i+1][j][1],dp[i][j][1]+(i-j))
                    #d->u
                    dp[i+1][j][1]=max(dp[i+1][j][1],dp[i][j][0]+(i-j))

        ans=-1
        for i in range(n+1):
            ans=max(ans,max(dp[n][i]))
        if ans<0:
            print(-1)
        else:
            print(ans)

**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 for the given buggy code, let's focus on edge cases that can expose issues in the algorithm. The code seems to have issues with how it handles the intervals and the conditions for forming an interesting permutation. Here are the steps to consider:

1. **Edge Case: No Intervals (m = 0)**:
   - This should be straightforward, but it's a good baseline to check if the code handles it correctly.

2. **Edge Case: Single Interval (m = 1)**:
   - Check if the code can handle a single interval correctly, especially when the interval is very small or very large.

3. **Edge Case: Multiple Overlapping Intervals**:
   - Intervals that overlap or are adjacent can cause issues with the data array and nodown array.

4. **Edge Case: No Interesting Permutation**:
   - Create a set of intervals that make it impossible to form an interesting permutation.

5. **Edge Case: Maximum Inversions**:
   - Create a set of intervals that should result in the maximum number of inversions.

Based on these considerations, here is a small-scale test input that includes these edge cases:

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

This test input includes:
1. A test case with no intervals.
2. A test case with a single interval.
3. A test case with multiple overlapping intervals.
4. A test case where no interesting permutation is possible.
5. A test case that should result in the maximum number of inversions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is a hard version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le 5 \cdot 10^3$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval. Note that there may exist identical intervals (i.e., there may exist two different indices $i, j$ such that $l_i = l_j$ and $r_i = r_j$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^3$ and the sum of $m$ over all test cases does not exceed $5 \cdot 10^3$.

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
8
2 0
2 1
1 2
5 1
2 4
8 3
1 4
2 5
7 8
7 2
1 4
4 7
7 3
1 2
1 7
3 7
7 4
1 3
4 7
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
18
-1
-1
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 3, 8, 7, 6, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3] = [2, 2, 7]$.

In the fifth and the sixth test case, it can be proven that there is no interesting permutation.

In the seventh test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2, k_3, k_4] = [1, 6, 1, 6]$.

In the eighth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
import math
import heapq
import itertools
import bisect
import random
import time
from collections import deque
import sys
from cmath import exp,pi
from functools import cmp_to_key
input=sys.stdin.readline

t=int(input())
for _ in range(t):
    n,m=map(int,input().split())
    arr=[]
    data=[-1]*n
    flag=0
    nodown=[0]*n
    for i in range(m):
        a,b=map(int,input().split())
        a-=1
        b-=1
        if data[a]==1:
            flag=1
        if data[b]==0:
            flag=1
        data[a]=0
        data[b]=1
        for j in range(a,b):
            nodown[j]=1

    if flag==1:
        print(-1)
    else:
        dp=[[[-10**9]*(2) for _ in range(n+1)] for _ in range(n+1)] #total, down, cur
        dp[0][0][1]=dp[0][0][0]=0
        for i in range(n):
            for j in range(n+1):
                if data[i]!=1: #0
                    #d->d
                    if j>0:
                        dp[i+1][j][0]=max(dp[i+1][j][0],dp[i][j-1][0]+i)
                    #u->d
                    if j>0 and (i==0 or nodown[i-1]==0):
                        dp[i+1][j][0]=max(dp[i+1][j][0],dp[i][j-1][1]+i)

                if data[i]!=0: #1
                    #u->u
                    dp[i+1][j][1]=max(dp[i+1][j][1],dp[i][j][1]+(i-j))
                    #d->u
                    dp[i+1][j][1]=max(dp[i+1][j][1],dp[i][j][0]+(i-j))

        ans=-1
        for i in range(n+1):
            ans=max(ans,max(dp[n][i]))
        if ans<0:
            print(-1)
        else:
            print(ans)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. The buggy code has issues with handling overlapping intervals and ensuring the conditions for interesting permutations are met correctly. Here are the steps to create such a test input:

1. **Test Case with Overlapping Intervals**: The code might not handle overlapping intervals correctly, leading to incorrect marking of positions as `nodown`.

2. **Test Case with No Valid Permutation**: The code might not correctly determine when there is no valid permutation that meets the conditions.

3. **Test Case with Maximum Inversions**: The code might not correctly calculate the maximum number of inversions for a valid permutation.

Based on these considerations, here is a small-scale test input that includes these edge cases:

```plaintext
3
3 2
1 2
2 3
4 2
1 2
3 4
5 3
1 2
2 3
4 5
```

### Explanation:
1. **Test Case 1**:
   - Intervals: `[1, 2]` and `[2, 3]`
   - This case has overlapping intervals. The code should correctly handle the `nodown` array and determine if there is a valid permutation.
   - Expected output: `-1` (No valid permutation)

2. **Test Case 2**:
   - Intervals: `[1, 2]` and `[3, 4]`
   - This case has non-overlapping intervals. The code should correctly determine if there is a valid permutation and calculate the maximum number of inversions.
   - Expected output: `2` (Permutation: `[2, 1, 4, 3]` with 2 inversions)

3. **Test Case 3**:
   - Intervals: `[1, 2]`, `[2, 3]`, and `[4, 5]`
   - This case has overlapping and non-overlapping intervals. The code should correctly handle the `nodown` array and determine if there is a valid permutation.
   - Expected output: `4` (Permutation: `[2, 1, 3, 5, 4]` with 4 inversions)

This test input is designed to challenge the buggy code by including scenarios that are likely to expose its issues.