**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Intersected Segments
time_limit_ms: 5000
memory_limit_mb: 512
Description: Turtle just received $n$ segments and a sequence $a_1, a_2, \ldots, a_n$. The $i$-th segment is $[l_i, r_i]$.

Turtle will create an undirected graph $G$. If segment $i$ and segment $j$ intersect, then Turtle will add an undirected edge between $i$ and $j$ with a weight of $|a_i - a_j|$, for every $i \ne j$.

Turtle wants you to calculate the sum of the weights of the edges of the minimum spanning tree of the graph $G$, or report that the graph $G$ has no spanning tree.

We say two segments $[l_1, r_1]$ and $[l_2, r_2]$ intersect if and only if $\max(l_1, l_2) \le \min(r_1, r_2)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^5$) — the number of segments.

The $i$-th of the following $n$ lines contains three integers $l_i, r_i, a_i$ ($1 \le l_i \le r_i \le 10^9, 1 \le a_i \le 10^9$) — the $i$-th segment and the $i$-th element of the sequence.

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

Output: For each test case, output a single integer — the sum of the weights of the edges of the minimum spanning tree of the graph $G$. If the graph $G$ has no spanning tree, output $-1$.

Examples:
input:
4
5
1 7 3
2 4 6
3 5 5
6 7 9
3 4 4
5
2 7 3
1 3 6
4 5 5
6 7 9
1 1 4
4
1 4 3
1 2 1
3 4 5
1 4 4
3
1 3 1
2 3 3
4 5 8
output:
9
13
4
-1

Note: In the first test case, the graph $G$ is as follows:

![](CDN_BASE_URL/7b0a32f8f1d54a5dbbaffee1efd8ece1)

One of the minimum spanning trees of $G$ is as follows:

![](CDN_BASE_URL/7a035af2d5227648781d0c24b81c076e)

The sum of the weights of the edges of the minimum spanning tree is $9$.

In the second test case, the graph $G$ is as follows:

![](CDN_BASE_URL/62be880a08fa85071b491d6badd5f58e)

$G$ is already a tree, and the sum of the weights of the tree is $13$.

In the third test case, the graph $G$ is as follows:

![](CDN_BASE_URL/b667aa771d019e3a8a3a4c4372200f03)

In the fourth test case, the graph $G$ is as follows:

![](CDN_BASE_URL/45bc2d052677dbf032ff37fd1d723d83)

It's easy to see that $G$ is not connected, so $G$ has no spanning tree.

Buggy Code:
class DisjointSet:
    def __init__(self, n):
        self.parent = list(range(n))
        self.size = [1] * n

    def find(self, node):
        if node != self.parent[node]:
            self.parent[node] = self.find(self.parent[node])

        # print(self.parent[node])
        return self.parent[node]

    def union(self, u, v):
        root_u = self.find(u)
        root_v = self.find(v)
        if root_u == root_v:
            return
        if self.size[root_u] < self.size[root_v]:
            self.parent[root_u] = root_v
            self.size[root_v] += self.size[root_u]
        else:
            self.parent[root_v] = root_u
            self.size[root_u] += self.size[root_v]

testcase = int(input())
for _ in range(testcase):
    n = int(input())
    intervals = []
    weights = []
    for i in range(n):
        l, r, w = map(int, input().split())
        weights.append(w)
        intervals.append((l, 1, i))
        intervals.append((r + 1, -1, i))
    intervals.sort()
    edges = []
    active = set()
    for _, type, i in intervals:
        if type == -1:
            active.remove((weights[i], i))
        else:
            num = weights[i]
            it = sorted(active, key=lambda x: abs(x[0] - num))
            if it:
                val, id = it[0]
                edges.append((abs(num - val), id, i))
            if len(it) > 1:
                val, id = it[1]
                edges.append((abs(num - val), id, i))
            active.add((weights[i], i))

        # print(active, edges)
    edges.sort()
    # print(edges)
    ds = DisjointSet(n)
    ans = 0
    for wt, node1, node2 in edges:
        if ds.find(node1) != ds.find(node2):
            ans += wt
            ds.union(node1, node2)

    # print(ds.parent, ds.size)
    par = ds.find(0)

    for i in range(1, n):
        if ds.find(i) != par:
            ans = -1
    # print(ds.parent, ds.size)

    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:
```plaintext
1
5
1 2 1
2 3 2
3 4 3
4 5 4
5 6 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Intersected Segments
time_limit_ms: 5000
memory_limit_mb: 512
Description: Turtle just received $n$ segments and a sequence $a_1, a_2, \ldots, a_n$. The $i$-th segment is $[l_i, r_i]$.

Turtle will create an undirected graph $G$. If segment $i$ and segment $j$ intersect, then Turtle will add an undirected edge between $i$ and $j$ with a weight of $|a_i - a_j|$, for every $i \ne j$.

Turtle wants you to calculate the sum of the weights of the edges of the minimum spanning tree of the graph $G$, or report that the graph $G$ has no spanning tree.

We say two segments $[l_1, r_1]$ and $[l_2, r_2]$ intersect if and only if $\max(l_1, l_2) \le \min(r_1, r_2)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^5$) — the number of segments.

The $i$-th of the following $n$ lines contains three integers $l_i, r_i, a_i$ ($1 \le l_i \le r_i \le 10^9, 1 \le a_i \le 10^9$) — the $i$-th segment and the $i$-th element of the sequence.

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

Output: For each test case, output a single integer — the sum of the weights of the edges of the minimum spanning tree of the graph $G$. If the graph $G$ has no spanning tree, output $-1$.

Examples:
input:
4
5
1 7 3
2 4 6
3 5 5
6 7 9
3 4 4
5
2 7 3
1 3 6
4 5 5
6 7 9
1 1 4
4
1 4 3
1 2 1
3 4 5
1 4 4
3
1 3 1
2 3 3
4 5 8
output:
9
13
4
-1

Note: In the first test case, the graph $G$ is as follows:

![](CDN_BASE_URL/7b0a32f8f1d54a5dbbaffee1efd8ece1)

One of the minimum spanning trees of $G$ is as follows:

![](CDN_BASE_URL/7a035af2d5227648781d0c24b81c076e)

The sum of the weights of the edges of the minimum spanning tree is $9$.

In the second test case, the graph $G$ is as follows:

![](CDN_BASE_URL/62be880a08fa85071b491d6badd5f58e)

$G$ is already a tree, and the sum of the weights of the tree is $13$.

In the third test case, the graph $G$ is as follows:

![](CDN_BASE_URL/b667aa771d019e3a8a3a4c4372200f03)

In the fourth test case, the graph $G$ is as follows:

![](CDN_BASE_URL/45bc2d052677dbf032ff37fd1d723d83)

It's easy to see that $G$ is not connected, so $G$ has no spanning tree.

Buggy Code:
import java.util.*;

public class TurtleSegmentsMST {

    static class Edge implements Comparable<Edge> {
        int u, v, weight;

        Edge(int u, int v, int weight) {
            this.u = u;
            this.v = v;
            this.weight = weight;
        }

        @Override
        public int compareTo(Edge other) {
            return this.weight - other.weight;
        }
    }

    static class UnionFind {
        int[] parent, rank;

        UnionFind(int n) {
            parent = new int[n];
            rank = new int[n];
            for (int i = 0; i < n; i++) {
                parent[i] = i;
                rank[i] = 0;
            }
        }

        int find(int u) {
            if (parent[u] != u) {
                parent[u] = find(parent[u]);
            }
            return parent[u];
        }

        boolean union(int u, int v) {
            int rootU = find(u);
            int rootV = find(v);
            if (rootU != rootV) {
                if (rank[rootU] > rank[rootV]) {
                    parent[rootV] = rootU;
                } else if (rank[rootU] < rank[rootV]) {
                    parent[rootU] = rootV;
                } else {
                    parent[rootV] = rootU;
                    rank[rootU]++;
                }
                return true;
            }
            return false;
        }
    }

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

        while (t-- > 0) {
            int n = scanner.nextInt();
            int[][] segments = new int[n][3];
            for (int i = 0; i < n; i++) {
                segments[i][0] = scanner.nextInt(); // li
                segments[i][1] = scanner.nextInt(); // ri
                segments[i][2] = scanner.nextInt(); // ai
            }

            List<Edge> edges = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                for (int j = i + 1; j < n; j++) {
                    if (intersects(segments[i][0], segments[i][1], segments[j][0], segments[j][1])) {
                        int weight = Math.abs(segments[i][2] - segments[j][2]);
                        edges.add(new Edge(i, j, weight));
                    }
                }
            }

            Collections.sort(edges);
            UnionFind uf = new UnionFind(n);
            int mstWeight = 0;
            int edgesUsed = 0;

            for (Edge edge : edges) {
                if (uf.union(edge.u, edge.v)) {
                    mstWeight += edge.weight;
                    edgesUsed++;
                    if (edgesUsed == n - 1) {
                        break;
                    }
                }
            }

            if (edgesUsed == n - 1) {
                System.out.println(mstWeight);
            } else {
                System.out.println(-1);
            }
        }

        scanner.close();
    }

    private static boolean intersects(int l1, int r1, int l2, int r2) {
        return Math.max(l1, l2) <= Math.min(r1, r2);
    }
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases that could cause the code to fail. The main issues to focus on are:

1. **Disjoint Segments**: Segments that do not intersect, leading to a disconnected graph.
2. **Overlapping Segments**: Segments that intersect in complex ways, leading to a large number of edges.
3. **Equal Values**: Segments with the same `a_i` values, which might cause the weight of edges to be zero.
4. **Boundary Conditions**: Segments that intersect at the boundaries.

Let's create a small-scale test input that includes these edge cases:

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

### Explanation:

1. **Test Case 1**:
   - Segments: `[1, 2, 1]`, `[3, 4, 2]`, `[5, 6, 3]`
   - These segments are disjoint, so the graph will be disconnected.
   - Expected output: `-1`

2. **Test Case 2**:
   - Segments: `[1, 4, 3]`, `[1, 2, 1]`, `[3, 4, 5]`, `[1, 4, 4]`
   - These segments intersect in various ways, and there are multiple edges with the same weight.
   - Expected output: `4`

3. **Test Case 3**:
   - Segments: `[1, 3, 1]`, `[2, 4, 3]`, `[3, 5, 5]`, `[4, 6, 7]`, `[5, 7, 8]`
   - These segments intersect in a chain, creating a more complex graph.
   - Expected output: `12`

This test input should help expose any issues in the buggy code, particularly related to handling disjoint segments and complex intersections.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Intersected Segments
time_limit_ms: 5000
memory_limit_mb: 512
Description: Turtle just received $n$ segments and a sequence $a_1, a_2, \ldots, a_n$. The $i$-th segment is $[l_i, r_i]$.

Turtle will create an undirected graph $G$. If segment $i$ and segment $j$ intersect, then Turtle will add an undirected edge between $i$ and $j$ with a weight of $|a_i - a_j|$, for every $i \ne j$.

Turtle wants you to calculate the sum of the weights of the edges of the minimum spanning tree of the graph $G$, or report that the graph $G$ has no spanning tree.

We say two segments $[l_1, r_1]$ and $[l_2, r_2]$ intersect if and only if $\max(l_1, l_2) \le \min(r_1, r_2)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^5$) — the number of segments.

The $i$-th of the following $n$ lines contains three integers $l_i, r_i, a_i$ ($1 \le l_i \le r_i \le 10^9, 1 \le a_i \le 10^9$) — the $i$-th segment and the $i$-th element of the sequence.

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

Output: For each test case, output a single integer — the sum of the weights of the edges of the minimum spanning tree of the graph $G$. If the graph $G$ has no spanning tree, output $-1$.

Examples:
input:
4
5
1 7 3
2 4 6
3 5 5
6 7 9
3 4 4
5
2 7 3
1 3 6
4 5 5
6 7 9
1 1 4
4
1 4 3
1 2 1
3 4 5
1 4 4
3
1 3 1
2 3 3
4 5 8
output:
9
13
4
-1

Note: In the first test case, the graph $G$ is as follows:

![](CDN_BASE_URL/7b0a32f8f1d54a5dbbaffee1efd8ece1)

One of the minimum spanning trees of $G$ is as follows:

![](CDN_BASE_URL/7a035af2d5227648781d0c24b81c076e)

The sum of the weights of the edges of the minimum spanning tree is $9$.

In the second test case, the graph $G$ is as follows:

![](CDN_BASE_URL/62be880a08fa85071b491d6badd5f58e)

$G$ is already a tree, and the sum of the weights of the tree is $13$.

In the third test case, the graph $G$ is as follows:

![](CDN_BASE_URL/b667aa771d019e3a8a3a4c4372200f03)

In the fourth test case, the graph $G$ is as follows:

![](CDN_BASE_URL/45bc2d052677dbf032ff37fd1d723d83)

It's easy to see that $G$ is not connected, so $G$ has no spanning tree.

Buggy Code:
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        StringBuilder result = new StringBuilder();

        while (t-- > 0) {
            int n = sc.nextInt();
            Segment[] segments = new Segment[n];
            for (int i = 0; i < n; i++) {
                int l = sc.nextInt();
                int r = sc.nextInt();
                int a = sc.nextInt();
                segments[i] = new Segment(l, r, a, i);
            }

            List<Edge> edges = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                for (int j = i + 1; j < n; j++) {
                    if (segments[i].intersects(segments[j])) {
                        int weight = Math.abs(segments[i].a - segments[j].a);
                        edges.add(new Edge(i, j, weight));
                    }
                }
            }

            int mstWeight = kruskal(n, edges);
            result.append(mstWeight).append("\n");
        }

        System.out.print(result.toString());
        sc.close();
    }

    static class Segment {
        int l, r, a, index;

        Segment(int l, int r, int a, int index) {
            this.l = l;
            this.r = r;
            this.a = a;
            this.index = index;
        }

        boolean intersects(Segment other) {
            return Math.max(this.l, other.l) <= Math.min(this.r, other.r);
        }
    }

    static class Edge implements Comparable<Edge> {
        int u, v, weight;

        Edge(int u, int v, int weight) {
            this.u = u;
            this.v = v;
            this.weight = weight;
        }

        @Override
        public int compareTo(Edge other) {
            return Integer.compare(this.weight, other.weight);
        }
    }

    static class UnionFind {
        int[] parent, rank;

        UnionFind(int n) {
            parent = new int[n];
            rank = new int[n];
            for (int i = 0; i < n; i++) {
                parent[i] = i;
            }
        }

        int find(int u) {
            if (parent[u] != u) {
                parent[u] = find(parent[u]);
            }
            return parent[u];
        }

        boolean union(int u, int v) {
            int rootU = find(u);
            int rootV = find(v);
            if (rootU != rootV) {
                if (rank[rootU] > rank[rootV]) {
                    parent[rootV] = rootU;
                } else if (rank[rootU] < rank[rootV]) {
                    parent[rootU] = rootV;
                } else {
                    parent[rootV] = rootU;
                    rank[rootU]++;
                }
                return true;
            }
            return false;
        }
    }

    static int kruskal(int n, List<Edge> edges) {
        Collections.sort(edges);
        UnionFind uf = new UnionFind(n);
        int mstWeight = 0;
        int edgesUsed = 0;

        for (Edge edge : edges) {
            if (uf.union(edge.u, edge.v)) {
                mstWeight += edge.weight;
                edgesUsed++;
                if (edgesUsed == n - 1) {
                    break;
                }
            }
        }

        return (edgesUsed == n - 1) ? mstWeight : -1;
    }
}

**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 consider edge cases and scenarios that might cause the algorithm to fail. Here are some potential scenarios to consider:

1. **Disjoint Segments**: If the segments are disjoint, the graph will have no spanning tree, and the output should be `-1`.
2. **Segments with Overlaps but Not Connected**: If the segments overlap but do not form a connected graph, the output should also be `-1`.
3. **Segments with Multiple Overlaps**: If the segments have multiple overlaps, the algorithm should correctly compute the minimum spanning tree.
4. **Segments with Equal Values**: If the segments have the same `a` values, the weights of the edges will be zero, which might cause issues if the algorithm does not handle this correctly.

Let's create a test input that covers these scenarios:

1. **Disjoint Segments**:
   - Segments: `[1, 2, 3]`, `[4, 5, 6]`
   - Expected Output: `-1`

2. **Segments with Overlaps but Not Connected**:
   - Segments: `[1, 2, 3]`, `[2, 3, 4]`, `[4, 5, 5]`
   - Expected Output: `-1`

3. **Segments with Multiple Overlaps**:
   - Segments: `[1, 4, 3]`, `[1, 2, 1]`, `[3, 4, 5]`, `[1, 4, 4]`
   - Expected Output: `4`

4. **Segments with Equal Values**:
   - Segments: `[1, 3, 1]`, `[2, 3, 1]`, `[4, 5, 1]`
   - Expected Output: `0`

Here is the test input:

```plaintext
4
2
1 2 3
4 5 6
3
1 2 3
2 3 4
4 5 5
4
1 4 3
1 2 1
3 4 5
1 4 4
3
1 3 1
2 3 1
4 5 1
```

This input will help expose any issues in the given code, particularly in handling disjoint segments and segments with equal values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Intersected Segments
time_limit_ms: 5000
memory_limit_mb: 512
Description: Turtle just received $n$ segments and a sequence $a_1, a_2, \ldots, a_n$. The $i$-th segment is $[l_i, r_i]$.

Turtle will create an undirected graph $G$. If segment $i$ and segment $j$ intersect, then Turtle will add an undirected edge between $i$ and $j$ with a weight of $|a_i - a_j|$, for every $i \ne j$.

Turtle wants you to calculate the sum of the weights of the edges of the minimum spanning tree of the graph $G$, or report that the graph $G$ has no spanning tree.

We say two segments $[l_1, r_1]$ and $[l_2, r_2]$ intersect if and only if $\max(l_1, l_2) \le \min(r_1, r_2)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^5$) — the number of segments.

The $i$-th of the following $n$ lines contains three integers $l_i, r_i, a_i$ ($1 \le l_i \le r_i \le 10^9, 1 \le a_i \le 10^9$) — the $i$-th segment and the $i$-th element of the sequence.

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

Output: For each test case, output a single integer — the sum of the weights of the edges of the minimum spanning tree of the graph $G$. If the graph $G$ has no spanning tree, output $-1$.

Examples:
input:
4
5
1 7 3
2 4 6
3 5 5
6 7 9
3 4 4
5
2 7 3
1 3 6
4 5 5
6 7 9
1 1 4
4
1 4 3
1 2 1
3 4 5
1 4 4
3
1 3 1
2 3 3
4 5 8
output:
9
13
4
-1

Note: In the first test case, the graph $G$ is as follows:

![](CDN_BASE_URL/7b0a32f8f1d54a5dbbaffee1efd8ece1)

One of the minimum spanning trees of $G$ is as follows:

![](CDN_BASE_URL/7a035af2d5227648781d0c24b81c076e)

The sum of the weights of the edges of the minimum spanning tree is $9$.

In the second test case, the graph $G$ is as follows:

![](CDN_BASE_URL/62be880a08fa85071b491d6badd5f58e)

$G$ is already a tree, and the sum of the weights of the tree is $13$.

In the third test case, the graph $G$ is as follows:

![](CDN_BASE_URL/b667aa771d019e3a8a3a4c4372200f03)

In the fourth test case, the graph $G$ is as follows:

![](CDN_BASE_URL/45bc2d052677dbf032ff37fd1d723d83)

It's easy to see that $G$ is not connected, so $G$ has no spanning tree.

Buggy Code:
# import sys
# sys.stdout = open('output.txt', 'w')
# sys.stdin = open('input.txt', 'r')

import sys
from heapq import heappush, heappop
from collections import defaultdict

input = sys.stdin.read

def find(parent, x):
    if parent[x] != x:
        parent[x] = find(parent, parent[x])
    return parent[x]

def union(parent, rank, x, y):
    rootX = find(parent, x)
    rootY = find(parent, y)

    if rootX != rootY:
        if rank[rootX] > rank[rootY]:
            parent[rootY] = rootX
        elif rank[rootX] < rank[rootY]:
            parent[rootX] = rootY
        else:
            parent[rootY] = rootX
            rank[rootX] += 1

def minimum_spanning_tree(n, edges):
    parent = list(range(n))
    rank = [0] * n
    mst_weight = 0
    edge_count = 0

    for weight, u, v in sorted(edges):
        if find(parent, u) != find(parent, v):
            union(parent, rank, u, v)
            mst_weight += weight
            edge_count += 1
            if edge_count == n - 1:
                break

    if edge_count == n - 1:
        return mst_weight
    else:
        return -1

data = input().split()

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

for _ in range(t):
    n = int(data[index])
    index += 1
    segments = []
    events = []
    for i in range(n):
        l = int(data[index])
        r = int(data[index+1])
        a = int(data[index+2])
        segments.append((l, r, a, i))
        events.append((l, 0, i)) # Segment start
        events.append((r, 1, i)) # Segment end
        index += 3

    # Sort events by position
    events.sort()
    active_segments = set()
    edges = []

    for position, event_type, idx in events:
        if event_type == 0: # Start of segment
            for active in active_segments:
                li, ri, ai, _ = segments[active]
                lj, rj, aj, _ = segments[idx]
                if max(li, lj) <= min(ri, rj):  # They intersect
                    edges.append((abs(ai - aj), active, idx))
            active_segments.add(idx)
        else: # End of segment
            active_segments.remove(idx)

    if not edges and n > 1:
        results.append(-1)
    else:
        mst_weight = minimum_spanning_tree(n, edges)
        results.append(mst_weight)

sys.stdout.write("\n".join(map(str, results)) + "\n")

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Intersected Segments
time_limit_ms: 5000
memory_limit_mb: 512
Description: Turtle just received $n$ segments and a sequence $a_1, a_2, \ldots, a_n$. The $i$-th segment is $[l_i, r_i]$.

Turtle will create an undirected graph $G$. If segment $i$ and segment $j$ intersect, then Turtle will add an undirected edge between $i$ and $j$ with a weight of $|a_i - a_j|$, for every $i \ne j$.

Turtle wants you to calculate the sum of the weights of the edges of the minimum spanning tree of the graph $G$, or report that the graph $G$ has no spanning tree.

We say two segments $[l_1, r_1]$ and $[l_2, r_2]$ intersect if and only if $\max(l_1, l_2) \le \min(r_1, r_2)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^5$) — the number of segments.

The $i$-th of the following $n$ lines contains three integers $l_i, r_i, a_i$ ($1 \le l_i \le r_i \le 10^9, 1 \le a_i \le 10^9$) — the $i$-th segment and the $i$-th element of the sequence.

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

Output: For each test case, output a single integer — the sum of the weights of the edges of the minimum spanning tree of the graph $G$. If the graph $G$ has no spanning tree, output $-1$.

Examples:
input:
4
5
1 7 3
2 4 6
3 5 5
6 7 9
3 4 4
5
2 7 3
1 3 6
4 5 5
6 7 9
1 1 4
4
1 4 3
1 2 1
3 4 5
1 4 4
3
1 3 1
2 3 3
4 5 8
output:
9
13
4
-1

Note: In the first test case, the graph $G$ is as follows:

![](CDN_BASE_URL/7b0a32f8f1d54a5dbbaffee1efd8ece1)

One of the minimum spanning trees of $G$ is as follows:

![](CDN_BASE_URL/7a035af2d5227648781d0c24b81c076e)

The sum of the weights of the edges of the minimum spanning tree is $9$.

In the second test case, the graph $G$ is as follows:

![](CDN_BASE_URL/62be880a08fa85071b491d6badd5f58e)

$G$ is already a tree, and the sum of the weights of the tree is $13$.

In the third test case, the graph $G$ is as follows:

![](CDN_BASE_URL/b667aa771d019e3a8a3a4c4372200f03)

In the fourth test case, the graph $G$ is as follows:

![](CDN_BASE_URL/45bc2d052677dbf032ff37fd1d723d83)

It's easy to see that $G$ is not connected, so $G$ has no spanning tree.

Buggy Code:
class DisjointSet:
    def __init__(self, size):
        self.parent = list(range(size))
        self.rank = [0] * size

    def find(self, u):
        if self.parent[u] != u:
            self.parent[u] = self.find(self.parent[u])
        return self.parent[u]

    def union(self, u, v):
        rootU = self.find(u)
        rootV = self.find(v)

        if rootU != rootV:
            if self.rank[rootU] > self.rank[rootV]:
                self.parent[rootV] = rootU
            elif self.rank[rootU] < self.rank[rootV]:
                self.parent[rootU] = rootV
            else:
                self.parent[rootV] = rootU
                self.rank[rootU] += 1
            return True
        return False

def kruskal(edges, n):
    ds = DisjointSet(n)
    mst_weight = 0
    edge_count = 0

    for w, u, v in edges:
        if ds.union(u, v):
            mst_weight += w
            edge_count += 1
            if edge_count == n - 1:
                break

    if edge_count == n - 1:
        return mst_weight
    return -1

import sys
input = sys.stdin.read
data = input().split()

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

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

    for i in range(n):
        l = int(data[index])
        r = int(data[index + 1])
        a = int(data[index + 2])
        segments.append((l, r, a, i))
        index += 3

    edges = []

    for i in range(n):
        for j in range(i + 1, n):
            l1, r1, a1, idx1 = segments[i]
            l2, r2, a2, idx2 = segments[j]

            if max(l1, l2) <= min(r1, r2):
                weight = abs(a1 - a2)
                edges.append((weight, idx1, idx2))

    edges.sort()

    result = kruskal(edges, n)
    results.append(result)

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

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases that can potentially cause issues. Here are the steps to generate such a test input:

1. **Non-Intersecting Segments**: Ensure that there are cases where segments do not intersect, leading to a graph that is not connected.
2. **Overlapping Segments**: Include cases with overlapping segments to test the correctness of the edge weight calculation.
3. **Identical Segments**: Test cases with identical segments to ensure the union-find logic handles them correctly.
4. **Large Range of Values**: Use a wide range of values for segment endpoints and sequence values to test the robustness of the algorithm.

Given these considerations, let's create a small-scale test input:

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

### Explanation:

1. **Test Case 1**:
   - Segments: `[1, 3]`, `[4, 6]`, `[7, 9]`, `[2, 5]`, `[8, 10]`
   - Sequence: `1, 2, 3, 4, 5`
   - This test case includes non-intersecting segments (`[1, 3]` and `[4, 6]`, `[7, 9]` and `[8, 10]`) and intersecting segments (`[2, 5]` and `[1, 3]`, `[2, 5]` and `[4, 6]`, `[8, 10]` and `[7, 9]`).

2. **Test Case 2**:
   - Segments: `[1, 2]`, `[3, 4]`, `[5, 6]`, `[7, 8]`
   - Sequence: `1, 2, 3, 4`
   - This test case includes segments that are non-intersecting, ensuring that the graph is not connected.

3. **Test Case 3**:
   - Segments: `[1, 3]`, `[2, 4]`, `[3, 5]`
   - Sequence: `1, 2, 3`
   - This test case includes segments that are overlapping, testing the correctness of the edge weight calculation and the union-find logic.

This test input should help expose any issues in the buggy code, particularly in handling non-intersecting segments and ensuring the graph is connected.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Intersected Segments
time_limit_ms: 5000
memory_limit_mb: 512
Description: Turtle just received $n$ segments and a sequence $a_1, a_2, \ldots, a_n$. The $i$-th segment is $[l_i, r_i]$.

Turtle will create an undirected graph $G$. If segment $i$ and segment $j$ intersect, then Turtle will add an undirected edge between $i$ and $j$ with a weight of $|a_i - a_j|$, for every $i \ne j$.

Turtle wants you to calculate the sum of the weights of the edges of the minimum spanning tree of the graph $G$, or report that the graph $G$ has no spanning tree.

We say two segments $[l_1, r_1]$ and $[l_2, r_2]$ intersect if and only if $\max(l_1, l_2) \le \min(r_1, r_2)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^5$) — the number of segments.

The $i$-th of the following $n$ lines contains three integers $l_i, r_i, a_i$ ($1 \le l_i \le r_i \le 10^9, 1 \le a_i \le 10^9$) — the $i$-th segment and the $i$-th element of the sequence.

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

Output: For each test case, output a single integer — the sum of the weights of the edges of the minimum spanning tree of the graph $G$. If the graph $G$ has no spanning tree, output $-1$.

Examples:
input:
4
5
1 7 3
2 4 6
3 5 5
6 7 9
3 4 4
5
2 7 3
1 3 6
4 5 5
6 7 9
1 1 4
4
1 4 3
1 2 1
3 4 5
1 4 4
3
1 3 1
2 3 3
4 5 8
output:
9
13
4
-1

Note: In the first test case, the graph $G$ is as follows:

![](CDN_BASE_URL/7b0a32f8f1d54a5dbbaffee1efd8ece1)

One of the minimum spanning trees of $G$ is as follows:

![](CDN_BASE_URL/7a035af2d5227648781d0c24b81c076e)

The sum of the weights of the edges of the minimum spanning tree is $9$.

In the second test case, the graph $G$ is as follows:

![](CDN_BASE_URL/62be880a08fa85071b491d6badd5f58e)

$G$ is already a tree, and the sum of the weights of the tree is $13$.

In the third test case, the graph $G$ is as follows:

![](CDN_BASE_URL/b667aa771d019e3a8a3a4c4372200f03)

In the fourth test case, the graph $G$ is as follows:

![](CDN_BASE_URL/45bc2d052677dbf032ff37fd1d723d83)

It's easy to see that $G$ is not connected, so $G$ has no spanning tree.

Buggy Code:
class UnionFind:
    def __init__(self, size):
        self.parent = list(range(size))
        self.rank = [1] * size

    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, x, y):
        rootX = self.find(x)
        rootY = self.find(y)

        if rootX != rootY:
            if self.rank[rootX] > self.rank[rootY]:
                self.parent[rootY] = rootX
            elif self.rank[rootX] < self.rank[rootY]:
                self.parent[rootX] = rootY
            else:
                self.parent[rootY] = rootX
                self.rank[rootX] += 1

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

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

    for _ in range(t):
        n = int(data[index])
        index += 1
        segments = []
        for i in range(n):
            li = int(data[index])
            ri = int(data[index + 1])
            ai = int(data[index + 2])
            index += 3
            segments.append((li, ri, ai))

        # Sort segments by the starting point
        segments.sort(key=lambda x: x[0])

        uf = UnionFind(n)
        total_weight = 0
        edges = []

        for i in range(n):
            for j in range(i + 1, n):
                if max(segments[i][0], segments[j][0]) <= min(segments[i][1], segments[j][1]):
                    edges.append((abs(segments[i][2] - segments[j][2]), i, j))

        edges.sort()
        num_edges = 0
        for weight, u, v in edges:
            if uf.find(u) != uf.find(v):
                uf.union(u, v)
                total_weight += weight
                num_edges += 1

        if num_edges == n - 1:
            results.append(total_weight)
        else:
            results.append(-1)

    for result in results:
        print(result)

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 expose the bug in the given code, we need to consider scenarios where the union-find structure might not behave as expected, especially in edge cases. The primary issues to focus on are:

1. **Non-overlapping segments**: The graph might not be connected, which should result in `-1`.
2. **Overlapping segments with equal weights**: This can cause issues in the union-find structure if not handled correctly.
3. **Segments with very large or very small ranges**: This can cause issues in the sorting and comparison steps.
4. **Edge cases with minimal and maximal segment overlaps**: These can reveal issues in the union-find structure and the edge selection process.

Let's construct a small-scale test input that covers these scenarios:

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

Explanation:
1. **Test Case 1**: All segments are non-overlapping, which should result in `-1` since the graph is not connected.
2. **Test Case 2**: All segments overlap minimally with the last segment not overlapping with the others, which should result in `-1` since the graph is not connected.
3. **Test Case 3**: All segments are non-overlapping and have the same length, which should result in `-1` since the graph is not connected.
4. **Test Case 4**: All segments overlap, but the segments are arranged in a way that the union-find structure must handle multiple edges correctly to form a minimum spanning tree.

This input should help identify issues in the union-find structure and the edge selection process.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Intersected Segments
time_limit_ms: 5000
memory_limit_mb: 512
Description: Turtle just received $n$ segments and a sequence $a_1, a_2, \ldots, a_n$. The $i$-th segment is $[l_i, r_i]$.

Turtle will create an undirected graph $G$. If segment $i$ and segment $j$ intersect, then Turtle will add an undirected edge between $i$ and $j$ with a weight of $|a_i - a_j|$, for every $i \ne j$.

Turtle wants you to calculate the sum of the weights of the edges of the minimum spanning tree of the graph $G$, or report that the graph $G$ has no spanning tree.

We say two segments $[l_1, r_1]$ and $[l_2, r_2]$ intersect if and only if $\max(l_1, l_2) \le \min(r_1, r_2)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^5$) — the number of segments.

The $i$-th of the following $n$ lines contains three integers $l_i, r_i, a_i$ ($1 \le l_i \le r_i \le 10^9, 1 \le a_i \le 10^9$) — the $i$-th segment and the $i$-th element of the sequence.

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

Output: For each test case, output a single integer — the sum of the weights of the edges of the minimum spanning tree of the graph $G$. If the graph $G$ has no spanning tree, output $-1$.

Examples:
input:
4
5
1 7 3
2 4 6
3 5 5
6 7 9
3 4 4
5
2 7 3
1 3 6
4 5 5
6 7 9
1 1 4
4
1 4 3
1 2 1
3 4 5
1 4 4
3
1 3 1
2 3 3
4 5 8
output:
9
13
4
-1

Note: In the first test case, the graph $G$ is as follows:

![](CDN_BASE_URL/7b0a32f8f1d54a5dbbaffee1efd8ece1)

One of the minimum spanning trees of $G$ is as follows:

![](CDN_BASE_URL/7a035af2d5227648781d0c24b81c076e)

The sum of the weights of the edges of the minimum spanning tree is $9$.

In the second test case, the graph $G$ is as follows:

![](CDN_BASE_URL/62be880a08fa85071b491d6badd5f58e)

$G$ is already a tree, and the sum of the weights of the tree is $13$.

In the third test case, the graph $G$ is as follows:

![](CDN_BASE_URL/b667aa771d019e3a8a3a4c4372200f03)

In the fourth test case, the graph $G$ is as follows:

![](CDN_BASE_URL/45bc2d052677dbf032ff37fd1d723d83)

It's easy to see that $G$ is not connected, so $G$ has no spanning tree.

Buggy Code:
class UnionFind:
    def __init__(self, size):
        self.parent = list(range(size))
        self.rank = [0] * size

    def find(self, p):
        if self.parent[p] != p:
            self.parent[p] = self.find(self.parent[p])
        return self.parent[p]

    def union(self, p, q):
        rootP = self.find(p)
        rootQ = self.find(q)
        if rootP != rootQ:
            if self.rank[rootP] > self.rank[rootQ]:
                self.parent[rootQ] = rootP
            elif self.rank[rootP] < self.rank[rootQ]:
                self.parent[rootP] = rootQ
            else:
                self.parent[rootQ] = rootP
                self.rank[rootP] += 1
            return True
        return False

def min_spanning_tree(segments, n):
    edges = []
    for i in range(n):
        for j in range(i + 1, n):
            if max(segments[i][0], segments[j][0]) <= min(segments[i][1], segments[j][1]):
                edges.append((abs(segments[i][2] - segments[j][2]), i, j))

    edges.sort()
    uf = UnionFind(n)
    mst_weight = 0
    count = 0

    for weight, u, v in edges:
        if uf.union(u, v):
            mst_weight += weight
            count += 1
            if count == n - 1:
                break

    return mst_weight if count == n - 1 else -1

import sys
input = sys.stdin.read
data = input().split()

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

for _ in range(t):
    n = int(data[index])
    index += 1
    segments = []
    for _ in range(n):
        l = int(data[index])
        r = int(data[index + 1])
        a = int(data[index + 2])
        segments.append((l, r, a))
        index += 3
    results.append(min_spanning_tree(segments, n))

for result in results:
    print(result)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given algorithm, we need to consider edge cases that can cause issues in the implementation. The key points to focus on are:

1. **Disconnected Segments**: Ensure that the segments are such that the graph is not connected, which should result in the output `-1`.
2. **Overlapping Segments**: Create segments that overlap in a way that might cause the union-find structure to fail in detecting the correct connected components.
3. **Equal Weight Edges**: Create segments where the weights of the edges are equal, which can cause issues in the sorting and selection of edges for the minimum spanning tree.
4. **Large Number of Segments**: While the test input should be small, it should still be large enough to challenge the algorithm.

Given these considerations, let's construct a small-scale test input:

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

### Explanation:

1. **First Test Case**:
   - Segments: `[1, 2, 1]`, `[3, 4, 5]`, `[5, 6, 3]`, `[7, 8, 7]`
   - These segments are completely disjoint, so the graph will not be connected, and the output should be `-1`.

2. **Second Test Case**:
   - Segments: `[1, 5, 1]`, `[2, 6, 2]`, `[3, 7, 3]`, `[4, 8, 4]`, `[5, 9, 5]`
   - These segments overlap in a way that forms a connected graph. The minimum spanning tree should be calculated correctly, and the output should be the sum of the weights of the edges in the MST.

3. **Third Test Case**:
   - Segments: `[1, 4, 1]`, `[1, 4, 2]`, `[1, 4, 3]`, `[1, 4, 4]`
   - These segments are all overlapping and have equal weights, which can challenge the sorting and selection of edges in the minimum spanning tree. The output should be the sum of the weights of the edges in the MST.

This test input should expose any issues in the union-find structure or the edge selection process in the given algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Intersected Segments
time_limit_ms: 5000
memory_limit_mb: 512
Description: Turtle just received $n$ segments and a sequence $a_1, a_2, \ldots, a_n$. The $i$-th segment is $[l_i, r_i]$.

Turtle will create an undirected graph $G$. If segment $i$ and segment $j$ intersect, then Turtle will add an undirected edge between $i$ and $j$ with a weight of $|a_i - a_j|$, for every $i \ne j$.

Turtle wants you to calculate the sum of the weights of the edges of the minimum spanning tree of the graph $G$, or report that the graph $G$ has no spanning tree.

We say two segments $[l_1, r_1]$ and $[l_2, r_2]$ intersect if and only if $\max(l_1, l_2) \le \min(r_1, r_2)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^5$) — the number of segments.

The $i$-th of the following $n$ lines contains three integers $l_i, r_i, a_i$ ($1 \le l_i \le r_i \le 10^9, 1 \le a_i \le 10^9$) — the $i$-th segment and the $i$-th element of the sequence.

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

Output: For each test case, output a single integer — the sum of the weights of the edges of the minimum spanning tree of the graph $G$. If the graph $G$ has no spanning tree, output $-1$.

Examples:
input:
4
5
1 7 3
2 4 6
3 5 5
6 7 9
3 4 4
5
2 7 3
1 3 6
4 5 5
6 7 9
1 1 4
4
1 4 3
1 2 1
3 4 5
1 4 4
3
1 3 1
2 3 3
4 5 8
output:
9
13
4
-1

Note: In the first test case, the graph $G$ is as follows:

![](CDN_BASE_URL/7b0a32f8f1d54a5dbbaffee1efd8ece1)

One of the minimum spanning trees of $G$ is as follows:

![](CDN_BASE_URL/7a035af2d5227648781d0c24b81c076e)

The sum of the weights of the edges of the minimum spanning tree is $9$.

In the second test case, the graph $G$ is as follows:

![](CDN_BASE_URL/62be880a08fa85071b491d6badd5f58e)

$G$ is already a tree, and the sum of the weights of the tree is $13$.

In the third test case, the graph $G$ is as follows:

![](CDN_BASE_URL/b667aa771d019e3a8a3a4c4372200f03)

In the fourth test case, the graph $G$ is as follows:

![](CDN_BASE_URL/45bc2d052677dbf032ff37fd1d723d83)

It's easy to see that $G$ is not connected, so $G$ has no spanning tree.

Buggy Code:
def kruskal(n, edges):
    parent = list(range(n))

    def find(x):
        if parent[x] != x:
            parent[x] = find(parent[x])
        return parent[x]

    def union(x, y):
        rootX = find(x)
        rootY = find(y)
        if rootX != rootY:
            parent[rootX] = rootY
            return True
        return False

    # Sort edges by weight
    edges.sort(key=lambda x: x[2])
    mst_weight = 0
    num_edges_used = 0

    for u, v, w in edges:
        if union(u, v):
            mst_weight += w
            num_edges_used += 1
            if num_edges_used == n - 1:
                break

    # Check if all nodes are in the same set
    root = find(0)
    for i in range(1, n):
        if find(i) != root:
            return -1

    return mst_weight

import sys
input = sys.stdin.read
data = input().split()

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

results = []

for _ in range(t):
    n = int(data[index])
    index += 1
    l, r, a = [], [], []
    for i in range(n):
        li, ri, ai = map(int, data[index:index+3])
        index += 3
        l.append(li)
        r.append(ri)
        a.append(ai)

    events = []
    for i in range(n):
        events.append((l[i], 'start', i))
        events.append((r[i]+1, 'end', i))

    events.sort()
    active_segments = {}
    edge_list = []

    for position, event_type, i in events:
        if event_type == 'start':
            for j in active_segments.values():
                edge_list.append((i, j, abs(a[i] - a[j])))
            active_segments[i] = i
        else:
            del active_segments[i]

    results.append(kruskal(n, edge_list))

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

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

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

Think step by step.

Response:
```plaintext
2
3
1 3 1
2 3 3
4 5 8
4
1 4 3
1 2 1
3 4 5
1 4 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Intersected Segments
time_limit_ms: 5000
memory_limit_mb: 512
Description: Turtle just received $n$ segments and a sequence $a_1, a_2, \ldots, a_n$. The $i$-th segment is $[l_i, r_i]$.

Turtle will create an undirected graph $G$. If segment $i$ and segment $j$ intersect, then Turtle will add an undirected edge between $i$ and $j$ with a weight of $|a_i - a_j|$, for every $i \ne j$.

Turtle wants you to calculate the sum of the weights of the edges of the minimum spanning tree of the graph $G$, or report that the graph $G$ has no spanning tree.

We say two segments $[l_1, r_1]$ and $[l_2, r_2]$ intersect if and only if $\max(l_1, l_2) \le \min(r_1, r_2)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^5$) — the number of segments.

The $i$-th of the following $n$ lines contains three integers $l_i, r_i, a_i$ ($1 \le l_i \le r_i \le 10^9, 1 \le a_i \le 10^9$) — the $i$-th segment and the $i$-th element of the sequence.

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

Output: For each test case, output a single integer — the sum of the weights of the edges of the minimum spanning tree of the graph $G$. If the graph $G$ has no spanning tree, output $-1$.

Examples:
input:
4
5
1 7 3
2 4 6
3 5 5
6 7 9
3 4 4
5
2 7 3
1 3 6
4 5 5
6 7 9
1 1 4
4
1 4 3
1 2 1
3 4 5
1 4 4
3
1 3 1
2 3 3
4 5 8
output:
9
13
4
-1

Note: In the first test case, the graph $G$ is as follows:

![](CDN_BASE_URL/7b0a32f8f1d54a5dbbaffee1efd8ece1)

One of the minimum spanning trees of $G$ is as follows:

![](CDN_BASE_URL/7a035af2d5227648781d0c24b81c076e)

The sum of the weights of the edges of the minimum spanning tree is $9$.

In the second test case, the graph $G$ is as follows:

![](CDN_BASE_URL/62be880a08fa85071b491d6badd5f58e)

$G$ is already a tree, and the sum of the weights of the tree is $13$.

In the third test case, the graph $G$ is as follows:

![](CDN_BASE_URL/b667aa771d019e3a8a3a4c4372200f03)

In the fourth test case, the graph $G$ is as follows:

![](CDN_BASE_URL/45bc2d052677dbf032ff37fd1d723d83)

It's easy to see that $G$ is not connected, so $G$ has no spanning tree.

Buggy Code:
def kruskal(edges, num_segments):
    # Sort edges by weight
    edges.sort(key=lambda x: x[2])

    # Union-Find data structure
    parent = list(range(num_segments))
    rank = [0] * num_segments

    def find(x):
        if parent[x] != x:
            parent[x] = find(parent[x])
        return parent[x]

    def union(x, y):
        rootX = find(x)
        rootY = find(y)
        if rootX != rootY:
            if rank[rootX] > rank[rootY]:
                parent[rootY] = rootX
            elif rank[rootX] < rank[rootY]:
                parent[rootX] = rootY
            else:
                parent[rootY] = rootX
                rank[rootX] += 1
            return True
        return False

    mst_weight = 0
    edges_used = 0

    for u, v, weight in edges:
        if union(u, v):
            mst_weight += weight
            edges_used += 1
            if edges_used == num_segments - 1:
                break

    return mst_weight if edges_used == num_segments - 1 else -1

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

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

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

        segments = []
        for i in range(n):
            li = int(data[index])
            ri = int(data[index + 1])
            ai = int(data[index + 2])
            index += 3
            segments.append((li, ri, ai))

        edges = []
        for i in range(n):
            for j in range(i + 1, n):
                li, ri, ai = segments[i]
                lj, rj, aj = segments[j]
                if max(li, lj) <= min(ri, rj):
                    weight = abs(ai - aj)
                    edges.append((i, j, weight))

        if len(edges) == 0:
            results.append(-1)
            continue

        mst_weight = kruskal(edges, n)
        results.append(mst_weight)

    for result in results:
        print(result)

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 expose the bug in the provided code, we need to consider scenarios that might cause the Kruskal's algorithm to fail. The primary issues to consider are:

1. **Disconnected Graphs**: The graph might be disconnected, which would cause the algorithm to return -1.
2. **Edge Cases**: Segments that barely intersect or do not intersect at all.
3. **Large Weights**: Segments with large weights that might cause overflow or incorrect calculations.
4. **Edge Cases with Minimum Spanning Tree**: Scenarios where the minimum spanning tree is not straightforward to compute.

Let's construct a small-scale test input that includes these elements:

1. **Disconnected Graph**: Two sets of segments that do not intersect with each other.
2. **Barely Intersecting Segments**: Segments that intersect just at the boundaries.
3. **Large Weights**: Segments with large values for `a_i` to test the weight calculations.

Here is the test input:

```plaintext
3
4
1 2 1000000000
3 4 1000000000
2 3 1000000000
4 5 1000000000
2
1 2 1
3 4 2
3
1 3 1
2 3 3
4 5 8
```

### Explanation:

1. **First Test Case (Disconnected Graph)**:
   - Segments: `[1, 2]`, `[3, 4]`, `[2, 3]`, `[4, 5]`
   - The segments `[1, 2]` and `[3, 4]` do not intersect with `[4, 5]`.
   - The segments `[2, 3]` and `[4, 5]` do not intersect with `[1, 2]`.
   - This graph is disconnected, so the expected output is `-1`.

2. **Second Test Case (Barely Intersecting Segments)**:
   - Segments: `[1, 2]`, `[3, 4]`
   - These segments do not intersect, so the graph is disconnected.
   - The expected output is `-1`.

3. **Third Test Case (Large Weights)**:
   - Segments: `[1, 3]`, `[2, 3]`, `[4, 5]`
   - The segments `[1, 3]` and `[2, 3]` intersect, and the segment `[4, 5]` does not intersect with the others.
   - The weights are `|1 - 3| = 2`, `|1 - 8| = 7`, and `|3 - 8| = 5`.
   - The minimum spanning tree includes the edges with weights 2 and 5, so the expected output is `7`.

This test input should effectively expose the bug in the provided code by testing various edge cases and ensuring that the algorithm handles disconnected graphs and large weights correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Intersected Segments
time_limit_ms: 5000
memory_limit_mb: 512
Description: Turtle just received $n$ segments and a sequence $a_1, a_2, \ldots, a_n$. The $i$-th segment is $[l_i, r_i]$.

Turtle will create an undirected graph $G$. If segment $i$ and segment $j$ intersect, then Turtle will add an undirected edge between $i$ and $j$ with a weight of $|a_i - a_j|$, for every $i \ne j$.

Turtle wants you to calculate the sum of the weights of the edges of the minimum spanning tree of the graph $G$, or report that the graph $G$ has no spanning tree.

We say two segments $[l_1, r_1]$ and $[l_2, r_2]$ intersect if and only if $\max(l_1, l_2) \le \min(r_1, r_2)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^5$) — the number of segments.

The $i$-th of the following $n$ lines contains three integers $l_i, r_i, a_i$ ($1 \le l_i \le r_i \le 10^9, 1 \le a_i \le 10^9$) — the $i$-th segment and the $i$-th element of the sequence.

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

Output: For each test case, output a single integer — the sum of the weights of the edges of the minimum spanning tree of the graph $G$. If the graph $G$ has no spanning tree, output $-1$.

Examples:
input:
4
5
1 7 3
2 4 6
3 5 5
6 7 9
3 4 4
5
2 7 3
1 3 6
4 5 5
6 7 9
1 1 4
4
1 4 3
1 2 1
3 4 5
1 4 4
3
1 3 1
2 3 3
4 5 8
output:
9
13
4
-1

Note: In the first test case, the graph $G$ is as follows:

![](CDN_BASE_URL/7b0a32f8f1d54a5dbbaffee1efd8ece1)

One of the minimum spanning trees of $G$ is as follows:

![](CDN_BASE_URL/7a035af2d5227648781d0c24b81c076e)

The sum of the weights of the edges of the minimum spanning tree is $9$.

In the second test case, the graph $G$ is as follows:

![](CDN_BASE_URL/62be880a08fa85071b491d6badd5f58e)

$G$ is already a tree, and the sum of the weights of the tree is $13$.

In the third test case, the graph $G$ is as follows:

![](CDN_BASE_URL/b667aa771d019e3a8a3a4c4372200f03)

In the fourth test case, the graph $G$ is as follows:

![](CDN_BASE_URL/45bc2d052677dbf032ff37fd1d723d83)

It's easy to see that $G$ is not connected, so $G$ has no spanning tree.

Buggy Code:
import sys
from collections import defaultdict

# Function to implement disjoint set union (DSU)
class DSU:
    def __init__(self, N):
        self.parent = list(range(N))
        self.rank = [0] * N

    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, x, y):
        xroot = self.find(x)
        yroot = self.find(y)
        if xroot == yroot:
            return False
        if self.rank[xroot] < self.rank[yroot]:
            self.parent[xroot] = yroot
        elif self.rank[xroot] > self.rank[yroot]:
            self.parent[yroot] = xroot
        else:
            self.parent[yroot] = xroot
            self.rank[xroot] += 1
        return True

def kruskals(edges, n):
    dsu = DSU(n)
    mst_weight = 0
    for u, v, w in sorted(edges, key=lambda x: x[2]):
        if dsu.union(u, v):
            mst_weight += w
    # Check if there is more than one component
    components = set()
    for i in range(n):
        components.add(dsu.find(i))
    if len(components) > 1:
        return -1
    return mst_weight

input_lines = sys.stdin.read().strip().split('\n')
test_cases = int(input_lines[0])
line_index = 1
for _ in range(test_cases):
    n = int(input_lines[line_index])
    line_index += 1
    segments = []
    for i in range(n):
        l, r, a = map(int, input_lines[line_index].split())
        line_index += 1
        segments.append((l, r, a, i))
    edges = []
    for i in range(n):
        for j in range(i + 1, n):
            li, ri, ai, ui = segments[i]
            lj, rj, aj, uj = segments[j]
            if max(li, lj) <= min(ri, rj):
                edges.append(((ui, uj), abs(ai - aj)))
    edges = [(u, v, w) for ((u, v), w) in edges]
    result = kruskals(edges, n)
    print(result)

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given code, we need to focus on scenarios that might cause issues in the implementation. Here are some key considerations:

1. **Edge Cases for Segment Intersections**:
   - Segments that just touch each other (e.g., [1, 2] and [2, 3]).
   - Segments that are completely nested (e.g., [1, 5] and [2, 4]).
   - Segments that do not intersect at all (e.g., [1, 2] and [3, 4]).

2. **Multiple Test Cases**:
   - Test cases with different numbers of segments to ensure the code handles varying input sizes.
   - Test cases with and without a spanning tree to verify the correctness of the spanning tree detection.

3. **Special Values**:
   - Segments with the same start and end points (e.g., [1, 1]).
   - Segments with the same sequence values (e.g., a_i = a_j).

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

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

### Explanation:
1. **Test Case 1**:
   - Segments: [1, 2], [2, 3], [3, 4]
   - These segments form a chain where each segment intersects with the next one. The expected output should be the sum of the weights of the edges in the minimum spanning tree.

2. **Test Case 2**:
   - Segments: [1, 1], [2, 2], [3, 3], [4, 4]
   - These segments do not intersect at all, so the graph will have no spanning tree. The expected output is -1.

3. **Test Case 3**:
   - Segments: [1, 5], [2, 4], [3, 3], [6, 6], [7, 7]
   - The first three segments intersect, but the last two do not intersect with any other segments. This test case checks if the code correctly identifies that the graph has multiple components. The expected output is -1.

4. **Test Case 4**:
   - Segments: [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7]
   - These segments form a chain similar to Test Case 1 but with more segments. The expected output should be the sum of the weights of the edges in the minimum spanning tree.

This test input should help in identifying any issues in the implementation of the minimum spanning tree algorithm and the union-find data structure.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Intersected Segments
time_limit_ms: 5000
memory_limit_mb: 512
Description: Turtle just received $n$ segments and a sequence $a_1, a_2, \ldots, a_n$. The $i$-th segment is $[l_i, r_i]$.

Turtle will create an undirected graph $G$. If segment $i$ and segment $j$ intersect, then Turtle will add an undirected edge between $i$ and $j$ with a weight of $|a_i - a_j|$, for every $i \ne j$.

Turtle wants you to calculate the sum of the weights of the edges of the minimum spanning tree of the graph $G$, or report that the graph $G$ has no spanning tree.

We say two segments $[l_1, r_1]$ and $[l_2, r_2]$ intersect if and only if $\max(l_1, l_2) \le \min(r_1, r_2)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^5$) — the number of segments.

The $i$-th of the following $n$ lines contains three integers $l_i, r_i, a_i$ ($1 \le l_i \le r_i \le 10^9, 1 \le a_i \le 10^9$) — the $i$-th segment and the $i$-th element of the sequence.

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

Output: For each test case, output a single integer — the sum of the weights of the edges of the minimum spanning tree of the graph $G$. If the graph $G$ has no spanning tree, output $-1$.

Examples:
input:
4
5
1 7 3
2 4 6
3 5 5
6 7 9
3 4 4
5
2 7 3
1 3 6
4 5 5
6 7 9
1 1 4
4
1 4 3
1 2 1
3 4 5
1 4 4
3
1 3 1
2 3 3
4 5 8
output:
9
13
4
-1

Note: In the first test case, the graph $G$ is as follows:

![](CDN_BASE_URL/7b0a32f8f1d54a5dbbaffee1efd8ece1)

One of the minimum spanning trees of $G$ is as follows:

![](CDN_BASE_URL/7a035af2d5227648781d0c24b81c076e)

The sum of the weights of the edges of the minimum spanning tree is $9$.

In the second test case, the graph $G$ is as follows:

![](CDN_BASE_URL/62be880a08fa85071b491d6badd5f58e)

$G$ is already a tree, and the sum of the weights of the tree is $13$.

In the third test case, the graph $G$ is as follows:

![](CDN_BASE_URL/b667aa771d019e3a8a3a4c4372200f03)

In the fourth test case, the graph $G$ is as follows:

![](CDN_BASE_URL/45bc2d052677dbf032ff37fd1d723d83)

It's easy to see that $G$ is not connected, so $G$ has no spanning tree.

Buggy Code:
class UnionFind:
    def __init__(self, n):
        self.parent = list(range(n))
        self.rank = [1] * n

    def find(self, u):
        if self.parent[u] != u:
            self.parent[u] = self.find(self.parent[u])
        return self.parent[u]

    def union(self, u, v):
        root_u = self.find(u)
        root_v = self.find(v)
        if root_u != root_v:
            if self.rank[root_u] > self.rank[root_v]:
                self.parent[root_v] = root_u
            elif self.rank[root_u] < self.rank[root_v]:
                self.parent[root_u] = root_v
            else:
                self.parent[root_v] = root_u
                self.rank[root_u] += 1
            return True
        return False

def solve():
    import heapq

    t = int(input())
    results = []

    for _ in range(t):
        n = int(input())
        segments = []

        for _ in range(n):
            l, r, a = map(int, input().split())
            segments.append((l, r, a))

        edges = []

        # 构建所有交叉的边
        for i in range(n):
            for j in range(i + 1, n):
                if max(segments[i][0], segments[j][0]) <= min(segments[i][1], segments[j][1]):
                    weight = abs(segments[i][2] - segments[j][2])
                    edges.append((weight, i, j))

        # 使用Kruskal算法构建最小生成树
        edges.sort()
        uf = UnionFind(n)
        mst_weight = 0
        edge_count = 0

        for weight, u, v in edges:
            if uf.union(u, v):
                mst_weight += weight
                edge_count += 1
                if edge_count == n - 1:
                    break

        if edge_count == n - 1:
            results.append(mst_weight)
        else:
            results.append(-1)

    for result in results:
        print(result)

if __name__ == "__main__":
    solve()

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

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

Think step by step.

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

1. **Non-Intersecting Segments**: Ensure that some segments do not intersect to test the case where the graph is not connected.
2. **Identical Segments**: Test with identical segments to ensure the union-find structure handles them correctly.
3. **Edge Cases with Small and Large Values**: Test with segments that have very small and very large values to ensure the code handles the range correctly.
4. **Overlap and Non-Overlap**: Test with segments that have different overlaps to ensure the edge weight calculation is correct.

Here is a small-scale test input that covers these scenarios:

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

This test input includes:
1. A case where the graph is not connected (first test case).
2. A case with identical segments and overlapping segments (second test case).
3. A case with a fully connected graph (third test case).
4. A case with segments that have different overlaps and a disconnected segment (fourth test case).

These scenarios should help expose any bugs in the given code, particularly in the union-find structure and the edge weight calculation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Intersected Segments
time_limit_ms: 5000
memory_limit_mb: 512
Description: Turtle just received $n$ segments and a sequence $a_1, a_2, \ldots, a_n$. The $i$-th segment is $[l_i, r_i]$.

Turtle will create an undirected graph $G$. If segment $i$ and segment $j$ intersect, then Turtle will add an undirected edge between $i$ and $j$ with a weight of $|a_i - a_j|$, for every $i \ne j$.

Turtle wants you to calculate the sum of the weights of the edges of the minimum spanning tree of the graph $G$, or report that the graph $G$ has no spanning tree.

We say two segments $[l_1, r_1]$ and $[l_2, r_2]$ intersect if and only if $\max(l_1, l_2) \le \min(r_1, r_2)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^5$) — the number of segments.

The $i$-th of the following $n$ lines contains three integers $l_i, r_i, a_i$ ($1 \le l_i \le r_i \le 10^9, 1 \le a_i \le 10^9$) — the $i$-th segment and the $i$-th element of the sequence.

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

Output: For each test case, output a single integer — the sum of the weights of the edges of the minimum spanning tree of the graph $G$. If the graph $G$ has no spanning tree, output $-1$.

Examples:
input:
4
5
1 7 3
2 4 6
3 5 5
6 7 9
3 4 4
5
2 7 3
1 3 6
4 5 5
6 7 9
1 1 4
4
1 4 3
1 2 1
3 4 5
1 4 4
3
1 3 1
2 3 3
4 5 8
output:
9
13
4
-1

Note: In the first test case, the graph $G$ is as follows:

![](CDN_BASE_URL/7b0a32f8f1d54a5dbbaffee1efd8ece1)

One of the minimum spanning trees of $G$ is as follows:

![](CDN_BASE_URL/7a035af2d5227648781d0c24b81c076e)

The sum of the weights of the edges of the minimum spanning tree is $9$.

In the second test case, the graph $G$ is as follows:

![](CDN_BASE_URL/62be880a08fa85071b491d6badd5f58e)

$G$ is already a tree, and the sum of the weights of the tree is $13$.

In the third test case, the graph $G$ is as follows:

![](CDN_BASE_URL/b667aa771d019e3a8a3a4c4372200f03)

In the fourth test case, the graph $G$ is as follows:

![](CDN_BASE_URL/45bc2d052677dbf032ff37fd1d723d83)

It's easy to see that $G$ is not connected, so $G$ has no spanning tree.

Buggy Code:
def find(parent, i):
    if parent[i] == i:
        return i
    return find(parent, parent[i])

def union(parent, rank, x, y):
    rootX = find(parent, x)
    rootY = find(parent, y)
    if rootX != rootY:
        if rank[rootX] > rank[rootY]:
            parent[rootY] = rootX
        elif rank[rootX] < rank[rootY]:
            parent[rootX] = rootY
        else:
            parent[rootY] = rootX
            rank[rootX] += 1

def kruskal(edges, n):
    parent = list(range(n))
    rank = [0] * n
    mst_weight = 0

    # Sort edges by weight
    edges.sort(key=lambda x: x[2])

    for u, v, w in edges:
        if find(parent, u) != find(parent, v):
            union(parent, rank, u, v)
            mst_weight += w

    # Check if all nodes are connected
    if len(set(find(parent, i) for i in range(n))) != 1:
        return -1

    return mst_weight

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

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

    results = []
    for _ in range(t):
        n = int(data[index])
        index += 1
        segs = [(int(data[index + 3*i]), int(data[index + 3*i + 1]), int(data[index + 3*i + 2])) for i in range(n)]
        index += 3*n

        # Sort segments by left endpoint
        segs.sort()

        edges = []
        for i in range(n):
            li, ri, ai = segs[i]
            for j in range(i+1, n):
                lj, rj, aj = segs[j]
                if max(li, lj) <= min(ri, rj):  # Intersection condition
                    edges.append((i, j, abs(ai - aj)))

        res = kruskal(edges, n)
        results.append(res)

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

if __name__ == "__main__":
    main()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Intersected Segments
time_limit_ms: 5000
memory_limit_mb: 512
Description: Turtle just received $n$ segments and a sequence $a_1, a_2, \ldots, a_n$. The $i$-th segment is $[l_i, r_i]$.

Turtle will create an undirected graph $G$. If segment $i$ and segment $j$ intersect, then Turtle will add an undirected edge between $i$ and $j$ with a weight of $|a_i - a_j|$, for every $i \ne j$.

Turtle wants you to calculate the sum of the weights of the edges of the minimum spanning tree of the graph $G$, or report that the graph $G$ has no spanning tree.

We say two segments $[l_1, r_1]$ and $[l_2, r_2]$ intersect if and only if $\max(l_1, l_2) \le \min(r_1, r_2)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^5$) — the number of segments.

The $i$-th of the following $n$ lines contains three integers $l_i, r_i, a_i$ ($1 \le l_i \le r_i \le 10^9, 1 \le a_i \le 10^9$) — the $i$-th segment and the $i$-th element of the sequence.

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

Output: For each test case, output a single integer — the sum of the weights of the edges of the minimum spanning tree of the graph $G$. If the graph $G$ has no spanning tree, output $-1$.

Examples:
input:
4
5
1 7 3
2 4 6
3 5 5
6 7 9
3 4 4
5
2 7 3
1 3 6
4 5 5
6 7 9
1 1 4
4
1 4 3
1 2 1
3 4 5
1 4 4
3
1 3 1
2 3 3
4 5 8
output:
9
13
4
-1

Note: In the first test case, the graph $G$ is as follows:

![](CDN_BASE_URL/7b0a32f8f1d54a5dbbaffee1efd8ece1)

One of the minimum spanning trees of $G$ is as follows:

![](CDN_BASE_URL/7a035af2d5227648781d0c24b81c076e)

The sum of the weights of the edges of the minimum spanning tree is $9$.

In the second test case, the graph $G$ is as follows:

![](CDN_BASE_URL/62be880a08fa85071b491d6badd5f58e)

$G$ is already a tree, and the sum of the weights of the tree is $13$.

In the third test case, the graph $G$ is as follows:

![](CDN_BASE_URL/b667aa771d019e3a8a3a4c4372200f03)

In the fourth test case, the graph $G$ is as follows:

![](CDN_BASE_URL/45bc2d052677dbf032ff37fd1d723d83)

It's easy to see that $G$ is not connected, so $G$ has no spanning tree.

Buggy Code:
import sys
from collections import defaultdict
sys.setrecursionlimit(1 << 25)

class UnionFind:
    def __init__(self, size):
        self.parent = list(range(size))
        self.rank = [1] * size

    def find(self, u):
        if self.parent[u] != u:
            self.parent[u] = self.find(self.parent[u])
        return self.parent[u]

    def union(self, u, v):
        root_u = self.find(u)
        root_v = self.find(v)
        if root_u == root_v:
            return False
        if self.rank[root_u] > self.rank[root_v]:
            self.parent[root_v] = root_u
        elif self.rank[root_u] < self.rank[root_v]:
            self.parent[root_u] = root_v
        else:
            self.parent[root_v] = root_u
            self.rank[root_u] += 1
        return True

def kruskal(n, edges):
    uf = UnionFind(n)
    edges.sort(key=lambda x: x[2])  # Sort edges by weight
    mst_weight = 0
    count_edges = 0
    for u, v, w in edges:
        if uf.union(u, v):
            mst_weight += w
            count_edges += 1
            if count_edges == n - 1:
                break
    return mst_weight if count_edges == n - 1 else -1

def main():
    import sys
    input = sys.stdin.read().split()
    idx = 0
    t = int(input[idx])
    idx += 1
    results = []
    while idx < len(input):
        n = int(input[idx])
        idx += 1
        segments = []
        values = []
        for _ in range(n):
            l, r, a = map(int, input[idx:idx+3])
            segments.append((l, r))
            values.append(a)
            idx += 3

        edges = []
        for i in range(n):
            for j in range(i + 1, n):
                l1, r1 = segments[i]
                l2, r2 = segments[j]
                if max(l1, l2) <= min(r1, r2):
                    weight = abs(values[i] - values[j])
                    edges.append((i, j, weight))

        result = kruskal(n, edges)
        results.append(result)

    for res in results:
        print(res)

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider scenarios where the code might fail. The code is expected to handle the following:

1. **Disjoint Segments**: Ensure that the graph is not connected if the segments are disjoint.
2. **Overlapping Segments**: Ensure that the graph is connected if the segments overlap.
3. **Edge Cases**: Handle cases where segments are just touching (e.g., [1, 2] and [2, 3]).

Given these considerations, let's create a small-scale test input that includes a mix of disjoint and overlapping segments.

### Test Input
```plaintext
3
3
1 2 1
3 4 2
5 6 3
4
1 3 1
2 4 2
3 5 3
4 6 4
5
1 3 1
2 4 2
3 5 3
4 6 4
5 7 5
```

### Explanation
1. **First Test Case**:
   - Segments: [1, 2], [3, 4], [5, 6]
   - These segments are disjoint, so the graph should have no spanning tree. The expected output is `-1`.

2. **Second Test Case**:
   - Segments: [1, 3], [2, 4], [3, 5], [4, 6]
   - These segments overlap, forming a connected graph. The expected output is the sum of the weights of the minimum spanning tree. The edges and their weights are:
     - (0, 1) with weight 1
     - (1, 2) with weight 1
     - (2, 3) with weight 1
   - The sum of the weights is `3`.

3. **Third Test Case**:
   - Segments: [1, 3], [2, 4], [3, 5], [4, 6], [5, 7]
   - These segments also overlap, forming a connected graph. The expected output is the sum of the weights of the minimum spanning tree. The edges and their weights are:
     - (0, 1) with weight 1
     - (1, 2) with weight 1
     - (2, 3) with weight 1
     - (3, 4) with weight 1
   - The sum of the weights is `4`.

This test input should help expose any issues in the code, particularly in handling disjoint segments and ensuring the graph is connected when segments overlap.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Intersected Segments
time_limit_ms: 5000
memory_limit_mb: 512
Description: Turtle just received $n$ segments and a sequence $a_1, a_2, \ldots, a_n$. The $i$-th segment is $[l_i, r_i]$.

Turtle will create an undirected graph $G$. If segment $i$ and segment $j$ intersect, then Turtle will add an undirected edge between $i$ and $j$ with a weight of $|a_i - a_j|$, for every $i \ne j$.

Turtle wants you to calculate the sum of the weights of the edges of the minimum spanning tree of the graph $G$, or report that the graph $G$ has no spanning tree.

We say two segments $[l_1, r_1]$ and $[l_2, r_2]$ intersect if and only if $\max(l_1, l_2) \le \min(r_1, r_2)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^5$) — the number of segments.

The $i$-th of the following $n$ lines contains three integers $l_i, r_i, a_i$ ($1 \le l_i \le r_i \le 10^9, 1 \le a_i \le 10^9$) — the $i$-th segment and the $i$-th element of the sequence.

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

Output: For each test case, output a single integer — the sum of the weights of the edges of the minimum spanning tree of the graph $G$. If the graph $G$ has no spanning tree, output $-1$.

Examples:
input:
4
5
1 7 3
2 4 6
3 5 5
6 7 9
3 4 4
5
2 7 3
1 3 6
4 5 5
6 7 9
1 1 4
4
1 4 3
1 2 1
3 4 5
1 4 4
3
1 3 1
2 3 3
4 5 8
output:
9
13
4
-1

Note: In the first test case, the graph $G$ is as follows:

![](CDN_BASE_URL/7b0a32f8f1d54a5dbbaffee1efd8ece1)

One of the minimum spanning trees of $G$ is as follows:

![](CDN_BASE_URL/7a035af2d5227648781d0c24b81c076e)

The sum of the weights of the edges of the minimum spanning tree is $9$.

In the second test case, the graph $G$ is as follows:

![](CDN_BASE_URL/62be880a08fa85071b491d6badd5f58e)

$G$ is already a tree, and the sum of the weights of the tree is $13$.

In the third test case, the graph $G$ is as follows:

![](CDN_BASE_URL/b667aa771d019e3a8a3a4c4372200f03)

In the fourth test case, the graph $G$ is as follows:

![](CDN_BASE_URL/45bc2d052677dbf032ff37fd1d723d83)

It's easy to see that $G$ is not connected, so $G$ has no spanning tree.

Buggy Code:
#include<bits/stdc++.h>
#define Pr pair<int,int>
#define fi first
#define nd second
using namespace std;
typedef long long LL;
inline int read(){
	char ch=getchar(); while(!isdigit(ch) && ch!='-') ch=getchar();
	int x=0,ff=1; if(ch=='-') ff=-1,ch=getchar();
	while(isdigit(ch)) x=(x<<3) + (x<<1) + (ch^48),ch=getchar();
	return x*ff;
}
const int N=1e6+5,inf=1e9;
int n,gfa[N],b[N],m,p[N]; bool vis[N];
int find(int x) {return gfa[x]==x?x:gfa[x]=find(gfa[x]);}
struct dt{
	int l,r,w;
} a[N]; Pr z[N];
struct node{
	Pr x,y;
	void adt(Pr z) { if(!z.nd) return ;
		if(x.nd==z.nd) x=max(x,z);
		else if(y.nd==z.nd) y=max(y,z);
		else if(!x.nd) x=z;
		else if(!y.nd) y=z;
		else if(x.fi<y.fi) x=max(x,z);
		else y=max(y,z);
	}
} tr[N<<2],tg[N<<2],bk;
#define lc(p) p<<1
#define rc(p) p<<1|1
void build(int p,int L,int R){
	tr[p].x=tr[p].y=tg[p].x=tg[p].y={-inf,0}; if(L==R) return ;
	int mid=(L+R)/2; build(lc(p),L,mid); build(rc(p),mid+1,R);
}
void pushdn(int p){
	if(tg[p].x.nd){
		tr[lc(p)].adt(tg[p].x); tr[rc(p)].adt(tg[p].x);
		tg[lc(p)].adt(tg[p].x); tg[rc(p)].adt(tg[p].x);
		tg[p].x={-inf,0};
	}
	if(tg[p].y.nd){
		tr[lc(p)].adt(tg[p].y); tr[rc(p)].adt(tg[p].y);
		tg[lc(p)].adt(tg[p].y); tg[rc(p)].adt(tg[p].y);
		tg[p].y={-inf,0};
	}
}
void pushup(int p){
	tr[p]=tr[lc(p)];
	if(tr[rc(p)].x.nd) tr[p].adt(tr[rc(p)].x);
	if(tr[rc(p)].y.nd) tr[p].adt(tr[rc(p)].y);
}
void Mdf(int p,int L,int R,int lt,int rt,Pr w){
	if(L>=lt && R<=rt) {bk.adt(tr[p].x); bk.adt(tr[p].y); tr[p].adt(w); tg[p].adt(w); return ;}
	int mid=(L+R)/2; pushdn(p);
	if(lt<=mid) Mdf(lc(p),L,mid,lt,rt,w);
	if(rt>mid) Mdf(rc(p),mid+1,R,lt,rt,w);
	pushup(p);
}
void vmain(){
	n=read();
	for(int i=1;i<=n;i++) a[i]={read(),read(),read()},gfa[i]=i;
	sort(a+1,a+n+1,[&](dt x,dt y){return x.w<y.w;});
	int S=n; LL ans=0; m=0;
	for(int i=1;i<=n;i++) b[++m]=a[i].l,b[++m]=a[i].r;
	sort(b+1,b+m+1); m=unique(b+1,b+m+1)-b-1;
	for(int i=1;i<=n;i++) a[i].l=lower_bound(b+1,b+m+1,a[i].l)-b;
	for(int i=1;i<=n;i++) a[i].r=lower_bound(b+1,b+m+1,a[i].r)-b;
	while(S>1){
		build(1,1,m);
		for(int i=1;i<=n;i++) z[i]={inf,0};
		for(int i=1;i<=n;i++){
			int c=find(i); bk.x=bk.y={-inf,0};
			Mdf(1,1,m,a[i].l,a[i].r,{a[i].w,c});
			if(bk.x.nd && bk.x.nd!=c && a[i].w-bk.x.fi<z[c].fi) z[c]={a[i].w-bk.x.fi,bk.x.nd};
			if(bk.y.nd && bk.y.nd!=c && a[i].w-bk.y.fi<z[c].fi) z[c]={a[i].w-bk.y.fi,bk.y.nd};
		} build(1,1,m);
		for(int i=n;i>=1;i--){
			int c=find(i); bk.x=bk.y={-inf,0}; p[i]=i;
			Mdf(1,1,m,a[i].l,a[i].r,{-a[i].w,c});
			if(bk.x.nd && bk.x.nd!=c && -bk.x.fi-a[i].w<z[c].fi) z[c]={-bk.x.fi-a[i].w,bk.x.nd};
			if(bk.y.nd && bk.y.nd!=c && -bk.y.fi-a[i].w<z[c].fi) z[c]={-bk.y.fi-a[i].w,bk.y.nd};
		}
		sort(p+1,p+n+1,[&](int x,int y){return z[x]<z[y];});
		bool rs=0;
		for(int i=1;i<=n;i++){
			int u=find(p[i]),v=find(z[p[i]].nd),w=z[p[i]].fi; if(u==v || !z[p[i]].nd) continue;
			ans+=w; S--; gfa[u]=v; rs=1;
		} if(!rs) {puts("-1"); return ;}
	} printf("%lld\n",ans);
}
int main(){
	// freopen("1.in","r",stdin);
	// freopen("s2.out","w",stdout);
	int T=read(); while(T--) vmain();
	return 0;
}
/*
g++ 2.cpp -o 2 -Wall -std=c++20 -O2 -Wl,-stack=999999999 -static
*/

**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 maximize the failure probability. The code is designed to handle a graph of intersecting segments and calculate the minimum spanning tree (MST) of the graph. Here are some key points to consider:

1. **Disjoint Segments**: Ensure that some segments do not intersect, which can lead to a disconnected graph.
2. **Overlapping Segments**: Create segments that overlap in complex ways to test the intersection logic.
3. **Edge Cases**: Test with the minimum and maximum allowed values for segment endpoints and weights.
4. **Large Number of Segments**: Although we are limited to small-scale tests, we can still use a moderate number of segments to test the algorithm's robustness.

Given these considerations, let's create a test case that includes both intersecting and non-intersecting segments, and segments with a range of weights.

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

### Explanation:
1. **First Test Case**:
   - Segments: [1, 2], [3, 4], [5, 6], [7, 8], [9, 10]
   - Weights: 1, 2, 3, 4, 5
   - These segments do not intersect, so the graph is not connected, and the output should be -1.

2. **Second Test Case**:
   - Segments: [1, 3], [2, 4], [3, 5], [4, 6], [5, 7]
   - Weights: 1, 2, 3, 4, 5
   - These segments overlap, forming a connected graph. The MST should be calculated based on the weights.

3. **Third Test Case**:
   - Segments: [1, 4], [2, 3], [3, 4], [4, 5]
   - Weights: 1, 2, 3, 4
   - These segments overlap in a more complex way, testing the intersection logic.

4. **Fourth Test Case**:
   - Segments: [1, 2], [2, 3], [3, 4]
   - Weights: 1, 2, 3
   - These segments overlap in a simple way, forming a connected graph.

This test input covers a variety of scenarios, including disjoint segments, overlapping segments, and edge cases, which should help expose any bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Intersected Segments
time_limit_ms: 5000
memory_limit_mb: 512
Description: Turtle just received $n$ segments and a sequence $a_1, a_2, \ldots, a_n$. The $i$-th segment is $[l_i, r_i]$.

Turtle will create an undirected graph $G$. If segment $i$ and segment $j$ intersect, then Turtle will add an undirected edge between $i$ and $j$ with a weight of $|a_i - a_j|$, for every $i \ne j$.

Turtle wants you to calculate the sum of the weights of the edges of the minimum spanning tree of the graph $G$, or report that the graph $G$ has no spanning tree.

We say two segments $[l_1, r_1]$ and $[l_2, r_2]$ intersect if and only if $\max(l_1, l_2) \le \min(r_1, r_2)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^5$) — the number of segments.

The $i$-th of the following $n$ lines contains three integers $l_i, r_i, a_i$ ($1 \le l_i \le r_i \le 10^9, 1 \le a_i \le 10^9$) — the $i$-th segment and the $i$-th element of the sequence.

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

Output: For each test case, output a single integer — the sum of the weights of the edges of the minimum spanning tree of the graph $G$. If the graph $G$ has no spanning tree, output $-1$.

Examples:
input:
4
5
1 7 3
2 4 6
3 5 5
6 7 9
3 4 4
5
2 7 3
1 3 6
4 5 5
6 7 9
1 1 4
4
1 4 3
1 2 1
3 4 5
1 4 4
3
1 3 1
2 3 3
4 5 8
output:
9
13
4
-1

Note: In the first test case, the graph $G$ is as follows:

![](CDN_BASE_URL/7b0a32f8f1d54a5dbbaffee1efd8ece1)

One of the minimum spanning trees of $G$ is as follows:

![](CDN_BASE_URL/7a035af2d5227648781d0c24b81c076e)

The sum of the weights of the edges of the minimum spanning tree is $9$.

In the second test case, the graph $G$ is as follows:

![](CDN_BASE_URL/62be880a08fa85071b491d6badd5f58e)

$G$ is already a tree, and the sum of the weights of the tree is $13$.

In the third test case, the graph $G$ is as follows:

![](CDN_BASE_URL/b667aa771d019e3a8a3a4c4372200f03)

In the fourth test case, the graph $G$ is as follows:

![](CDN_BASE_URL/45bc2d052677dbf032ff37fd1d723d83)

It's easy to see that $G$ is not connected, so $G$ has no spanning tree.

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define size(a) (int)a.size()
#ifdef ONPC
#include "mydbg.h"
#define debug(...) cerr << "~~ [ " << #__VA_ARGS__ << " ]: "; _print(", ", __VA_ARGS__)
#define debugv(arr) {cerr << "~~ [ " << #arr << " ]: {"; int f = 0; for (auto &_i : arr) cerr << ((f++) ? ", " : " ") << _i; cerr << " }\n"; }
#define debugg(arr) cerr << "~~ [ " << #arr << " ]:\n"; for (auto &_i : arr) {for (auto &_j : _i) cerr << _j << ' '; cerr << '\n';} ; cerr << "\n";
#define debugif(con, ...) if (con == true) { debug(__VA_ARGS__); }
#define alert(...) _print(" ", __VA_ARGS__)
#else
#define debug(...)
#define debugv(arr)
#define debugg(arr)
#define debugif(con, ...)
#define alert(...)
#endif

#ifndef ONPC
//#pragma GCC optimize("O3")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt,fma")
#pragma GCC optimize("unroll-loops")
#endif

#define all(a) a.begin(), a.end()
#define rall(x) (x).rbegin(), (x).rend()
#define clr(arr) memset(arr, 0, sizeof(arr))
#define int long long

using namespace std;
using namespace __gnu_pbds;
using ll = long long;
using ld = long double;
using pi = pair<int, int>;

template<class T> inline bool ckmin(T& a, const T& b) { return b < a && (a = b, true); }
template<class T> inline bool ckmax(T& a, const T& b) { return a < b && (a = b, true); }
template<class T, class F> inline bool ckpass(T& a, const T &b, const F &cmp) { return cmp && (a = b, true); }
template<class T, class F> inline bool ckpass(T& a, const T &b, const T&c, const F &cmp) { return cmp ? (a = b, true) : (a = c, false); }
template<class T = int> inline T scn() {T x; cin >> x; return x;}

const int INF = 1e17;

map<int, vector<pi>> g;
map<int, vector<int>> id;
bool intersected(pi a, pi b) {
    return max(a.first, b.first) <= min(a.second, b.second);
}

int cur_id = 0;

void make_disjoint_segments(vector<pi> &s) {
    sort(all(s));
    vector<pi> res;
    for (auto &cur : s) {
        pi to_add = cur;
        while (!res.empty() && intersected(res.back(), cur)) {
            ckmin(to_add.first, res.back().first);
            ckmax(to_add.second, res.back().second);
            res.pop_back();
        }
        res.push_back(to_add);
    }
    s = res;
}

int ans = 0;

struct DSU {
    int n;
    vector<int> p, sz;
    DSU(int _n) : n(_n) {
        p.resize(n); iota(all(p), 0);
        sz.resize(n, 1);
    }

    int prev(int v) {return v == p[v] ? v : p[v] = prev(p[v]); }

    void unite(int u, int v, int c) {
        int a = prev(u), b = prev(v);
        if (a == b) return;
        ans += c;
        if (sz[b] > sz[a]) swap(a, b);
        p[b] = a;
        sz[a] += sz[b];
    }
};

inline void solution() {
    int n = scn();
    ans = 0;
    vector<array<int, 4>> line;
    for (int i = 0; i < n; ++i) {
        int l = scn(), r = scn(), c = scn();
        line.push_back({l, 0, c, i});
        line.push_back({r, 1, c, i});
    }
    sort(all(line));
    set<pi> cur;
    DSU dsu(n);
    vector<array<int, 3>> edges;
    for (auto &[x, t, a, i] : line) {
        if (t == 0) {
            cur.emplace(a, i);
        } else {
            auto it = cur.begin();
            if (it != cur.end()) {
                auto prev = it;
                it++;
                while (it != cur.end()) {
                    edges.push_back({it->first - prev->first, prev->second, it->second});
                    prev++, it++;
                }
            }
            cur.erase(cur.find({a, i}));
        }
    }
    alert(edges);
    sort(all(edges));
    for (auto &[c, u, v] : edges) {
        dsu.unite(u, v, c);
    }
    if (dsu.sz[dsu.prev(0)] != n) cout << -1 << '\n';
    else cout << ans << '\n';
}

inline void precalc() {}

signed main() {
    cin.tie(nullptr)->sync_with_stdio(false);
#ifdef ONPC
    freopen("inp.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    freopen("err.txt", "w", stderr);
#endif
    precalc();
    int t = scn();
    while (t--) {
        solution();
    }
#ifdef ONPC
    alert("\nfinished in", clock() * 1.0 / CLOCKS_PER_SEC, "sec\n");
#endif
}

/*
 *    ~ ~~ ^ ~~ ~~ ^ ~~ ~~ ^ ~~ ~~ ^ ~~ ~
 *   ~~ ^ ~~ ~~ ^ ~~  __ ^ ~~ ~~ ^ ~~ ^ ~~
 *   ~ ~~ ^ ~~ ~~ __/ +  \__ ^ ~~ ~~ ^ ~~ ~
 *    ~~ ^ ~~ ~~ |BY PUSHAK!\~ ~~ ^ ~~ ^ ~~
 *   ~ ~~ ^ ~~ ~/ +  <  ^ \  \~ ~~ ^ ~~ ~
 *   ~~ ^ ~~ ~~ ^- ~~ - ~~ ^ ~~ ~~ ^ ~~ ^ ~~
 *   ~ ~~ ^ ~~ ~~ ^ ~~ ~~ ^ ~~ ~~ ^ ~~ ~
 *   ~~ ^ ~~ ~~ ^ ~~ ~~ ^ ~~ ~~ ^ ~~ ^ ~~
*/

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on edge cases where the code might fail. The primary issues in the code are related to how it handles the intersection of segments and the construction of edges for the minimum spanning tree (MST).

### Key Points to Consider:
1. **Edge Cases with Non-Intersecting Segments:**
   - The code should correctly identify when the graph is not connected and output `-1`.

2. **Edge Cases with Overlapping Segments:**
   - The code should correctly handle multiple overlapping segments and ensure that the MST is constructed correctly.

3. **Edge Cases with Identical Segments:**
   - The code should handle identical segments and ensure that they are treated correctly.

### Test Input Generation:
Let's create a small-scale test input that covers these edge cases:

1. **Test Case 1: Non-Intersecting Segments**
   - This case should ensure that the graph is not connected and the output is `-1`.

2. **Test Case 2: Overlapping Segments**
   - This case should ensure that the code correctly handles multiple overlapping segments and constructs the MST.

3. **Test Case 3: Identical Segments**
   - This case should ensure that the code correctly handles identical segments.

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

### Explanation:
1. **Test Case 1:**
   - Segments: `[1, 2], [3, 4], [5, 6]`
   - These segments do not intersect, so the graph is not connected, and the output should be `-1`.

2. **Test Case 2:**
   - Segments: `[1, 4], [2, 3], [3, 5], [4, 6]`
   - These segments overlap, and the code should correctly construct the MST and output the sum of the weights of the edges.

3. **Test Case 3:**
   - Segments: `[1, 3], [1, 3]`
   - These segments are identical, and the code should handle them correctly and output the sum of the weights of the edges.

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