**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Connect the Dots
time_limit_ms: 2000
memory_limit_mb: 512
Description: One fine evening, Alice sat down to play the classic game "Connect the Dots", but with a twist.

To play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:

  * She picks three integers $a_i$, $d_i$ ($1 \le d_i \le 10$), and $k_i$.    * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \ldots, a_i+k_i\cdot d_i$ and connects each pair of these points with arcs.

After performing all $m$ operations, she wants to know the number of connected components$^\dagger$ these points form. Please help her find this number.

$^\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.

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 two integers $n$ and $m$ ($1 \le n \le 2 \cdot 10^5$, $1 \le m \le 2 \cdot 10^5$).

The $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \le a_i \le a_i + k_i\cdot d_i \le n$, $1 \le d_i \le 10$, $0 \le k_i \le n$).

It is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \cdot 10^5$.

Output: For each test case, output the number of connected components.

Examples:
input:
3
10 2
1 2 4
2 2 4
100 1
19 2 4
100 3
1 2 5
7 2 6
17 2 31
output:
2
96
61

Note: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\{1, 3, 5, 7, 9\\}$ and $\\{2, 4, 6, 8, 10\\}$.

In the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.

In the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.

Buggy Code:
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

#define print(x) for(auto it:(x)) cout<<it<<" "; cout<<endl;
#define printp(x) for(auto it:(x)) cout<<it.first<<" "<<it.second<<endl;

typedef vector<int> vi;
typedef vector<pair<int,int>> vp;
typedef vector<pair<long long, long long>> vpll;
typedef vector<long long int> vll;
typedef vector<vector<int>> vvi;
typedef long long int ll;
typedef vector<vector<long long int>> vvll;
typedef vector<string> vs;
typedef pair<long long int, long long int> pll;
typedef tree <int ,  null_type ,  less<int> ,  rb_tree_tag ,  tree_order_statistics_node_update> ordered_set; //find_by order(k)->kth element of set(0 counting),  order_of_key(k)-> number of elements strictly smaller than k

ll binpow(ll a, ll b, ll m) { // x^n % m
    a %= m;
    ll res = 1;
    while (b > 0) {
        if (b & 1)
            res = res * a % m;
        a = a * a % m;
        b >>= 1;
    }
    return res;
}

ll n,m;
vll parent(200005,0); // intialise in solve
vll sizen(200005,1);

ll findUPar(ll node){
	if(parent[node] == node) return node;

	return parent[node] = findUPar(parent[node]);
}

void unionBySize(ll a, ll b){
	ll ulp_a = findUPar(a);
	ll ulp_b = findUPar(b);
	if(ulp_a == ulp_b) return;
	if(sizen[ulp_a] > sizen[ulp_b]){
		parent[ulp_b] = ulp_a;
		sizen[ulp_a] += sizen[ulp_b];
	}else{
		parent[ulp_a] = ulp_b;
		sizen[ulp_b] += sizen[ulp_a];
	}
	return;
}

void solve(){
    cin>>n>>m;
    for(ll i=1; i<=n; i++) parent[i] = i, sizen[i] = 1;;
    vector<vpll> darr(11);
    for(ll i=0; i<m; i++){
        ll a,d,k;
        cin>>a>>d>>k;
        darr[d].push_back({a,k});
    }
    for(ll i=0; i<11; i++){
        if(!darr[i].empty()) sort(darr[i].begin(),darr[i].end());
    }
    for(ll i=0; i<11; i++){
        if(darr[i].empty()) continue;
        // d value = i
        map<ll,ll> ub;
        for(auto it: darr[i]){
            ll st = it.first;
            ll en = it.first + it.second*i;
            ll j = max(st, ub[st%i]);
            ub[st%i] = en;
            while(j <= en){
                ub[j] = en;
                unionBySize(st,j);
                j+=i;
            }
        }
    }
    set<ll> ans;
    for(ll i=1; i<=n; i++) ans.insert(parent[i]);
    cout<<ans.size()<<endl;
}

int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t=1;
    cin>>t;
    while(t--){
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios that might cause issues in the implementation. Here are some key points to consider:

1. **Small-scale Test Cases**: The test input should be small enough to be manageable but complex enough to reveal issues.
2. **Edge Cases**: Consider edge cases such as:
   - Single-point components.
   - Overlapping operations.
   - Operations that start at the boundary of the range.
3. **Maximize Interactions**: Create scenarios where multiple operations interact in non-trivial ways.

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

### Test Input
```plaintext
3
10 2
1 2 4
2 2 4
100 1
19 2 4
100 3
1 2 5
7 2 6
17 2 31
```

### Explanation
1. **First Test Case**:
   - **n = 10, m = 2**
   - **Operations**:
     - `1 2 4`: Connects points 1, 3, 5, 7, 9.
     - `2 2 4`: Connects points 2, 4, 6, 8, 10.
   - **Expected Output**: 2 (Two connected components: {1, 3, 5, 7, 9} and {2, 4, 6, 8, 10}).

2. **Second Test Case**:
   - **n = 100, m = 1**
   - **Operation**:
     - `19 2 4`: Connects points 19, 21, 23, 25, 27.
   - **Expected Output**: 96 (One connected component of size 5 and 95 single-point components).

3. **Third Test Case**:
   - **n = 100, m = 3**
   - **Operations**:
     - `1 2 5`: Connects points 1, 3, 5, 7, 9.
     - `7 2 6`: Connects points 7, 9, 11, 13, 15, 17.
     - `17 2 31`: Connects points 17, 19, 21, ..., 79.
   - **Expected Output**: 61 (One connected component of size 40 and 60 single-point components).

This test input is designed to cover various scenarios and edge cases to ensure the code handles them correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Connect the Dots
time_limit_ms: 2000
memory_limit_mb: 512
Description: One fine evening, Alice sat down to play the classic game "Connect the Dots", but with a twist.

To play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:

  * She picks three integers $a_i$, $d_i$ ($1 \le d_i \le 10$), and $k_i$.    * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \ldots, a_i+k_i\cdot d_i$ and connects each pair of these points with arcs.

After performing all $m$ operations, she wants to know the number of connected components$^\dagger$ these points form. Please help her find this number.

$^\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.

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 two integers $n$ and $m$ ($1 \le n \le 2 \cdot 10^5$, $1 \le m \le 2 \cdot 10^5$).

The $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \le a_i \le a_i + k_i\cdot d_i \le n$, $1 \le d_i \le 10$, $0 \le k_i \le n$).

It is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \cdot 10^5$.

Output: For each test case, output the number of connected components.

Examples:
input:
3
10 2
1 2 4
2 2 4
100 1
19 2 4
100 3
1 2 5
7 2 6
17 2 31
output:
2
96
61

Note: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\{1, 3, 5, 7, 9\\}$ and $\\{2, 4, 6, 8, 10\\}$.

In the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.

In the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.

Buggy Code:
import sys

def find(u):
	if un[u] != u:
		un[u] = find(un[u])
	return un[u]

def union(u, v):
	ru = find(u)
	rv = find(v)
	if ru == rv:
		return 0
	un[rv] = ru
	return 1

input = sys.stdin.buffer.readline
for _ in range(int(input())):
	n, m = map(int, input().split())
	grp = [[[] for _ in range(10)] for _ in range(11)]
	un = list(range(n + 1))
	ans = n
	for _ in range(m):
		a, d, k = map(int, input().split())
		grp[d][a % d].append([a, a + d * k])
	for d in range(1, 10):
		for mod in range(d):
			mx = -1
			grp[d][mod].sort()
			for l, r in grp[d][mod]:
				for u in range(max(mx, l), r, d):
					ans -= union(u, u + d)
				mx = max(r, mx)
	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
3
10 2
1 2 4
2 2 4
100 1
19 2 4
100 3
1 2 5
7 2 6
17 2 31
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Connect the Dots
time_limit_ms: 2000
memory_limit_mb: 512
Description: One fine evening, Alice sat down to play the classic game "Connect the Dots", but with a twist.

To play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:

  * She picks three integers $a_i$, $d_i$ ($1 \le d_i \le 10$), and $k_i$.    * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \ldots, a_i+k_i\cdot d_i$ and connects each pair of these points with arcs.

After performing all $m$ operations, she wants to know the number of connected components$^\dagger$ these points form. Please help her find this number.

$^\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.

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 two integers $n$ and $m$ ($1 \le n \le 2 \cdot 10^5$, $1 \le m \le 2 \cdot 10^5$).

The $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \le a_i \le a_i + k_i\cdot d_i \le n$, $1 \le d_i \le 10$, $0 \le k_i \le n$).

It is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \cdot 10^5$.

Output: For each test case, output the number of connected components.

Examples:
input:
3
10 2
1 2 4
2 2 4
100 1
19 2 4
100 3
1 2 5
7 2 6
17 2 31
output:
2
96
61

Note: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\{1, 3, 5, 7, 9\\}$ and $\\{2, 4, 6, 8, 10\\}$.

In the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.

In the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.

Buggy Code:
from types import GeneratorType
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
    return wrappedfunc

def sol():
    n, m = map(int, input().split())

    # Dsu
    fa = [i for i in range(n + 1)]
    def fd(x: int) -> int:
        if fa[x] == x:
            return x
        fa[x] = fd(fa[x])
        return fa[x]
    def merge(x: int, y: int):
        fx = fd(x)
        fy = fd(y)
        if fx == fy:
            return
        fa[fx] = fy

    f = [[0 for _ in range(11)] for _ in range(n + 1)]

    @bootstrap
    def dfs(x: int, k: int, d: int):
        if f[x][d] >= k:
            yield
        lst = f[x][d]
        tmp = d
        now = 1
        while tmp <= 10:
            f[x][tmp] = max(f[x][tmp], k // now)
            now += 1
            tmp += d
        merge(x, x + (lst + 1) * d)
        yield (yield dfs(x + (lst + 1) * d, k - lst - 1, d))

    while m:
        m -= 1
        a, d, k = map(int, input().split())
        dfs(a, k, d)
    cnt = 0
    vis = [False for _ in range(n + 1)]
    for i in range(1, n + 1):
        x = fd(i)
        if not vis[x]:
            vis[x] = True
            cnt += 1
    print(cnt)

t = int(input())
while t:
    t -= 1
    sol()

**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
10 1
1 1 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Connect the Dots
time_limit_ms: 2000
memory_limit_mb: 512
Description: One fine evening, Alice sat down to play the classic game "Connect the Dots", but with a twist.

To play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:

  * She picks three integers $a_i$, $d_i$ ($1 \le d_i \le 10$), and $k_i$.    * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \ldots, a_i+k_i\cdot d_i$ and connects each pair of these points with arcs.

After performing all $m$ operations, she wants to know the number of connected components$^\dagger$ these points form. Please help her find this number.

$^\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.

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 two integers $n$ and $m$ ($1 \le n \le 2 \cdot 10^5$, $1 \le m \le 2 \cdot 10^5$).

The $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \le a_i \le a_i + k_i\cdot d_i \le n$, $1 \le d_i \le 10$, $0 \le k_i \le n$).

It is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \cdot 10^5$.

Output: For each test case, output the number of connected components.

Examples:
input:
3
10 2
1 2 4
2 2 4
100 1
19 2 4
100 3
1 2 5
7 2 6
17 2 31
output:
2
96
61

Note: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\{1, 3, 5, 7, 9\\}$ and $\\{2, 4, 6, 8, 10\\}$.

In the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.

In the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class connectTheDots {
    static class DisJointSet {
        int[] parent;
        int[] rank;
        int N;
        int numOfComp = 0;

        DisJointSet(int n) {
            this.parent = new int[n];
            this.rank = new int[n];
            this.N = n;

            for (int i = 0; i < N; i++) {
                parent[i] = i;
                rank[i] = 1;
            }
        }

        int findParent(int node) {
            if (parent[node] == node) return node;
            return parent[node] = findParent(parent[node]);
        }

        void unionByRank(int u, int v) {
            int PU = findParent(u);
            int PV = findParent(v);

            if (rank[PU] > rank[PV]) {
                parent[PV] = PU;
            } else if (rank[PV] > rank[PU]) {
                parent[PU] = PV;
            } else {
                parent[PU] = PV;
                rank[PV]++;
            }
        }

        int components() {
            for (int i = 0; i < N; i++) {
                if (findParent(i) == i) numOfComp++;
            }
            return numOfComp-1;
        }
    }

    public static void main(String[] args) throws IOException {
        BufferedReader reader= new BufferedReader(new InputStreamReader(System.in));
        int t =Integer.parseInt(reader.readLine());

        while(t-->0){
            String[] inputs= reader.readLine().split(" ");
            int n =Integer.parseInt(inputs[0]);
            int m =Integer.parseInt(inputs[1]);
            DisJointSet ds= new DisJointSet(n+1);
            for(int i=0;i<m;i++){
                String[] inputs2= reader.readLine().split(" ");

                int a =Integer.parseInt(inputs2[0]);
                int d =Integer.parseInt(inputs2[1]);
                int k =Integer.parseInt(inputs2[2]);
                int an=a+(d*k);

                for(int j=an;j>=a;j-=d){
                    if(ds.findParent(a)==ds.findParent(j)){
                        for(int l=a+d;l<j;l+=d){
                            if(ds.findParent(l)==ds.findParent(a)) break;
                            ds.unionByRank(l,a);
                        }
                        break;
                    }
                    else ds.unionByRank(j,a);
                }

            }

            System.out.println(ds.components());

        }

    }
}

**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 2
1 1 3
2 1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Connect the Dots
time_limit_ms: 2000
memory_limit_mb: 512
Description: One fine evening, Alice sat down to play the classic game "Connect the Dots", but with a twist.

To play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:

  * She picks three integers $a_i$, $d_i$ ($1 \le d_i \le 10$), and $k_i$.    * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \ldots, a_i+k_i\cdot d_i$ and connects each pair of these points with arcs.

After performing all $m$ operations, she wants to know the number of connected components$^\dagger$ these points form. Please help her find this number.

$^\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.

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 two integers $n$ and $m$ ($1 \le n \le 2 \cdot 10^5$, $1 \le m \le 2 \cdot 10^5$).

The $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \le a_i \le a_i + k_i\cdot d_i \le n$, $1 \le d_i \le 10$, $0 \le k_i \le n$).

It is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \cdot 10^5$.

Output: For each test case, output the number of connected components.

Examples:
input:
3
10 2
1 2 4
2 2 4
100 1
19 2 4
100 3
1 2 5
7 2 6
17 2 31
output:
2
96
61

Note: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\{1, 3, 5, 7, 9\\}$ and $\\{2, 4, 6, 8, 10\\}$.

In the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.

In the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.

Buggy Code:
/*
https://codeforces.com/contest/2020/problem/D
 */

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.HashSet;
import java.util.Set;
import java.util.StringTokenizer;

public class ConnectTheDots {

    public static void main(String[] args) throws IOException {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        while (t-- > 0) {
            int n = scanner.nextInt();
            int m = scanner.nextInt();
            DisjointSetDataStructure disjointSetDataStructure = new DisjointSetDataStructure(n);
            int count = 1;
            while (m-- > 0) {
                int a = scanner.nextInt();
                int d = scanner.nextInt();
                int k = scanner.nextInt();
                for (int i = 0; i <= k; i++) {
                    int val = a + i * d;
//                    if (!disjointSetDataStructure.union(val, a)) {
//                        count++;
//                    }
                    disjointSetDataStructure.union(val, a);
                }
            }
//            System.out.println(count);
            Set<Integer> set = new HashSet<>();
            for (int i = 1; i < disjointSetDataStructure.parent.length; i++) {
                int parI = disjointSetDataStructure.parent[i];
                int rankI = disjointSetDataStructure.rank[i];
                if(rankI == 1) set.add(parI);
            }
            System.out.println(set.size());
        }
    }

    private static class DisjointSetDataStructure {

        int forests;
        int[] parent;
        int[] rank;

        public DisjointSetDataStructure(int n) {
            this.forests = n + 1;
            this.parent = new int[this.forests];
            this.rank = new int[this.forests];
            for (int i = 0; i < this.forests; i++) {
                parent[i] = i;
                rank[i] = 1;
            }
        }

        public int find(int i) {
            if (this.parent[i] == i) {
                return i;
            }
            int rep = find(parent[i]);
            this.parent[i] = rep;
            return rep;
        }

        public boolean union(int i, int j) {
            int irep = this.find(i);
            int jrep = this.find(j);
            if (irep == jrep) {
                return false;
            }
            int irank = rank[irep];
            int jrank = rank[jrep];
            if (irank > jrank) {
                this.parent[jrep] = irep;
            } else if (irank < jrank) {
                this.parent[irep] = jrep;
            } else {
                this.parent[irep] = jrep;
                rank[jrep]++;
            }
            return true;
        }
    }

    private static class Scanner {

        StringTokenizer st;

        BufferedReader br;

        public Scanner(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(br.readLine());
            }
            return st.nextToken();
        }

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

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

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

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public boolean ready() throws IOException {
            return br.ready() || st.hasMoreTokens();
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases and scenarios that can highlight issues in the union-find implementation. Specifically, the code has a potential issue in how it counts the number of connected components. The main problem lies in the way the `set` is used to track the number of unique parents, which might not correctly account for all connected components.

Here are some scenarios to consider:

1. **Single Point Components**: Ensure that points that are not connected to any other points are correctly counted as single-point components.
2. **Multiple Operations on the Same Points**: Ensure that multiple operations on the same set of points do not incorrectly count the components.
3. **Edge Cases for `d` and `k`**: Ensure that the code handles cases where `d` and `k` are at their limits correctly.
4. **Overlap of Points**: Ensure that points that are connected in multiple operations are correctly counted as part of the same component.

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

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

### Explanation:
1. **First Test Case**:
   - Points: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
   - Operations:
     - `1 2 4`: Connects points 1, 3, 5, 7, 9
     - `2 2 4`: Connects points 2, 4, 6, 8, 10
   - Result: Two connected components: {1, 3, 5, 7, 9} and {2, 4, 6, 8, 10}

2. **Second Test Case**:
   - Points: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
   - Operations:
     - `1 1 9`: Connects points 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
     - `2 1 8`: Connects points 2, 3, 4, 5, 6, 7, 8, 9
   - Result: One connected component: {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

3. **Third Test Case**:
   - Points: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
   - Operations:
     - `1 1 9`: Connects points 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
     - `3 1 7`: Connects points 3, 4, 5, 6, 7, 8, 9, 10 (overlapping with the first operation)
   - Result: One connected component: {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

This test input should help expose the bug in the code by ensuring that the union-find structure correctly handles the union and find operations, and that the count of connected components is accurate.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Connect the Dots
time_limit_ms: 2000
memory_limit_mb: 512
Description: One fine evening, Alice sat down to play the classic game "Connect the Dots", but with a twist.

To play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:

  * She picks three integers $a_i$, $d_i$ ($1 \le d_i \le 10$), and $k_i$.    * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \ldots, a_i+k_i\cdot d_i$ and connects each pair of these points with arcs.

After performing all $m$ operations, she wants to know the number of connected components$^\dagger$ these points form. Please help her find this number.

$^\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.

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 two integers $n$ and $m$ ($1 \le n \le 2 \cdot 10^5$, $1 \le m \le 2 \cdot 10^5$).

The $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \le a_i \le a_i + k_i\cdot d_i \le n$, $1 \le d_i \le 10$, $0 \le k_i \le n$).

It is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \cdot 10^5$.

Output: For each test case, output the number of connected components.

Examples:
input:
3
10 2
1 2 4
2 2 4
100 1
19 2 4
100 3
1 2 5
7 2 6
17 2 31
output:
2
96
61

Note: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\{1, 3, 5, 7, 9\\}$ and $\\{2, 4, 6, 8, 10\\}$.

In the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.

In the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.

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

class UnionFind {
public:
    vector<int> parent;
    vector<int> rank;
    int count;

    UnionFind(int size) {
        parent.resize(size);
        rank.resize(size, 0);
        for (int i = 0; i < size; ++i) {
            parent[i] = i;
        }
        count = size;
    }

    int find(int x) {
        if (parent[x] != x) {
            parent[x] = find(parent[x]); // Path compression
        }
        return parent[x];
    }

    void unionSet(int x, int y) {
        int rootX = find(x);
        int rootY = find(y);
        if (rootX != rootY) {
            if (rank[rootX] > rank[rootY]) {
                parent[rootY] = rootX;
            } else if (rank[rootX] < rank[rootY]) {
                parent[rootX] = rootY;
            } else {
                parent[rootY] = rootX;
                rank[rootX]++;
            }
            count--;
        }
    }
};

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n, m;
        cin >> n >> m;
        UnionFind uf(n);
        for (int i = 0; i < m; ++i) {
            int a, d, k;
            cin >> a >> d >> k;
            for (int j = 0; j <= k; ++j) {
                uf.unionSet(a - 1, a - 1 + j * d);
            }
        }
        cout << uf.count << endl;
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
10 1
1 1 9
```

**Explanation:**
- This test case is designed to expose a potential issue in the `unionSet` function of the Union-Find data structure.
- The input specifies 10 points and 1 operation.
- The operation connects all points from 1 to 10 in a single connected component.
- The expected output is `1` because all points are connected in one component.
- The buggy code in the `unionSet` function might fail to correctly handle the range of indices, especially when `a - 1 + j * d` exceeds the bounds of the `parent` and `rank` vectors. This test case ensures that the function correctly handles the full range of indices from 0 to 9.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Connect the Dots
time_limit_ms: 2000
memory_limit_mb: 512
Description: One fine evening, Alice sat down to play the classic game "Connect the Dots", but with a twist.

To play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:

  * She picks three integers $a_i$, $d_i$ ($1 \le d_i \le 10$), and $k_i$.    * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \ldots, a_i+k_i\cdot d_i$ and connects each pair of these points with arcs.

After performing all $m$ operations, she wants to know the number of connected components$^\dagger$ these points form. Please help her find this number.

$^\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.

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 two integers $n$ and $m$ ($1 \le n \le 2 \cdot 10^5$, $1 \le m \le 2 \cdot 10^5$).

The $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \le a_i \le a_i + k_i\cdot d_i \le n$, $1 \le d_i \le 10$, $0 \le k_i \le n$).

It is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \cdot 10^5$.

Output: For each test case, output the number of connected components.

Examples:
input:
3
10 2
1 2 4
2 2 4
100 1
19 2 4
100 3
1 2 5
7 2 6
17 2 31
output:
2
96
61

Note: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\{1, 3, 5, 7, 9\\}$ and $\\{2, 4, 6, 8, 10\\}$.

In the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.

In the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.

Buggy Code:
#include <iostream>
#include <vector>
#include <unordered_set>

using namespace std;

class DSU {
public:
    vector<int> parent;
    vector<int> rank;

    DSU(int n) {
        parent.resize(n);
        rank.resize(n, 0);
        for (int i = 0; i < n; ++i) {
            parent[i] = i;
        }
    }

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

    void unionSets(int x, int y) {
        int rootX = find(x);
        int rootY = find(y);
        if (rootX != rootY) {
            if (rank[rootX] > rank[rootY]) {
                parent[rootY] = rootX;
            } else if (rank[rootX] < rank[rootY]) {
                parent[rootX] = rootY;
            } else {
                parent[rootY] = rootX;
                rank[rootX]++;
            }
        }
    }
};

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

    int t;
    cin >> t;
    while (t--) {
        int n, m;
        cin >> n >> m;
        DSU dsu(n);

        for (int i = 0; i < m; ++i) {
            int a, d, k;
            cin >> a >> d >> k;
            for (int j = 0; j < k; ++j) {
                dsu.unionSets(a - 1, a - 1 + d * (j + 1));
            }
        }

        unordered_set<int> components;
        for (int i = 0; i < n; ++i) {
            components.insert(dsu.find(i));
        }

        cout << components.size() << '\n';
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
10 1
1 1 9
100 1
1 1 99
10 2
1 1 4
6 1 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Connect the Dots
time_limit_ms: 2000
memory_limit_mb: 512
Description: One fine evening, Alice sat down to play the classic game "Connect the Dots", but with a twist.

To play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:

  * She picks three integers $a_i$, $d_i$ ($1 \le d_i \le 10$), and $k_i$.    * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \ldots, a_i+k_i\cdot d_i$ and connects each pair of these points with arcs.

After performing all $m$ operations, she wants to know the number of connected components$^\dagger$ these points form. Please help her find this number.

$^\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.

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 two integers $n$ and $m$ ($1 \le n \le 2 \cdot 10^5$, $1 \le m \le 2 \cdot 10^5$).

The $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \le a_i \le a_i + k_i\cdot d_i \le n$, $1 \le d_i \le 10$, $0 \le k_i \le n$).

It is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \cdot 10^5$.

Output: For each test case, output the number of connected components.

Examples:
input:
3
10 2
1 2 4
2 2 4
100 1
19 2 4
100 3
1 2 5
7 2 6
17 2 31
output:
2
96
61

Note: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\{1, 3, 5, 7, 9\\}$ and $\\{2, 4, 6, 8, 10\\}$.

In the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.

In the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.

Buggy Code:
class DSU:
    def __init__(self, n):
        self.parents = list(range(n))
        self.sizes = [1] * n

    def find(self, node):
        while self.parents[node] != node:
            node = self.parents[node]
        return node

    def union(self, a, b):
        root_a = self.find(a)
        root_b = self.find(b)
        if root_a != root_b:
            if self.sizes[root_a] > self.sizes[root_b]:
                self.parents[root_b] = root_a
                self.sizes[root_a] += self.sizes[root_b]
            else:
                self.parents[root_a] = root_b
                self.sizes[root_b] += self.sizes[root_a]

def main():
    t = int(input())
    for _ in range(t):
        n, m = map(int, input().split())
        dsu = DSU(n+1)
        for _ in range(m):
            a, d, k = map(int, input().split())
            for i in range(k):
                dsu.union(a+i*d, a+(i+1)*d)
        print(len(set(dsu.find(i) for i in range(1, n+1))))

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 where the `union` operation might not behave as expected. Specifically, the `union` operation in the `DSU` class is designed to merge two sets, but it does not handle the case where the points are already connected in a way that might cause issues in the `find` operation.

### Step-by-Step Analysis:
1. **Edge Case: Overlapping Ranges**:
   - If two operations have overlapping ranges, the `union` operation should correctly merge the sets. However, if the ranges are such that the `find` operation might not correctly identify the root of the set, it can lead to incorrect results.

2. **Small Test Case**:
   - We can create a small test case with a small number of points and operations that overlap in a way that might expose the bug.

### Test Case:
Let's create a test case with 5 points and 2 operations that overlap in a way that might cause issues.

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

### Explanation:
- **Test Case**:
  - **n = 5**: We have 5 points labeled 1 to 5.
  - **m = 2**: We have 2 operations.
  - **First Operation**: `1 1 3` - Connects points 1, 2, and 3.
  - **Second Operation**: `3 1 2` - Connects points 3, 4, and 5.

- **Expected Output**:
  - After the first operation, points 1, 2, and 3 are connected.
  - After the second operation, points 3, 4, and 5 are connected.
  - Since point 3 is common in both operations, all points 1, 2, 3, 4, and 5 should be in the same connected component.
  - Therefore, the expected output is `1`.

### Output:
```plaintext
1
5 2
1 1 3
3 1 2
```

This test case should help expose any issues in the `union` and `find` operations of the `DSU` class.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Connect the Dots
time_limit_ms: 2000
memory_limit_mb: 512
Description: One fine evening, Alice sat down to play the classic game "Connect the Dots", but with a twist.

To play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:

  * She picks three integers $a_i$, $d_i$ ($1 \le d_i \le 10$), and $k_i$.    * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \ldots, a_i+k_i\cdot d_i$ and connects each pair of these points with arcs.

After performing all $m$ operations, she wants to know the number of connected components$^\dagger$ these points form. Please help her find this number.

$^\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.

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 two integers $n$ and $m$ ($1 \le n \le 2 \cdot 10^5$, $1 \le m \le 2 \cdot 10^5$).

The $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \le a_i \le a_i + k_i\cdot d_i \le n$, $1 \le d_i \le 10$, $0 \le k_i \le n$).

It is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \cdot 10^5$.

Output: For each test case, output the number of connected components.

Examples:
input:
3
10 2
1 2 4
2 2 4
100 1
19 2 4
100 3
1 2 5
7 2 6
17 2 31
output:
2
96
61

Note: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\{1, 3, 5, 7, 9\\}$ and $\\{2, 4, 6, 8, 10\\}$.

In the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.

In the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.

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

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

    def union(self, x, y):
        rootX = self.find(x)
        rootY = self.find(y)
        if rootX != rootY:
            # Union by rank
            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 solve():
    t = int(input())  # Number of test cases
    for _ in range(t):
        n, m = map(int, input().split())  # n points, m operations
        uf = UnionFind(n)

        for _ in range(m):
            a, d, k = map(int, input().split())
            # Points a, a+d, a+2d, ..., a+k*d are connected
            last = a - 1  # Convert to 0-indexed
            for i in range(1, k + 1):
                next_point = a + i * d - 1  # 0-indexed
                uf.union(last, next_point)
                last = next_point

        # Count unique connected components
        unique_components = set()
        for i in range(n):
            unique_components.add(uf.find(i))

        print(len(unique_components))

# Reading input and running the solution
solve()

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

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

Think step by step.

Response:
```plaintext
3
10 2
1 2 4
2 2 4
100 1
1 2 50
100 2
1 2 50
2 2 49
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Connect the Dots
time_limit_ms: 2000
memory_limit_mb: 512
Description: One fine evening, Alice sat down to play the classic game "Connect the Dots", but with a twist.

To play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:

  * She picks three integers $a_i$, $d_i$ ($1 \le d_i \le 10$), and $k_i$.    * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \ldots, a_i+k_i\cdot d_i$ and connects each pair of these points with arcs.

After performing all $m$ operations, she wants to know the number of connected components$^\dagger$ these points form. Please help her find this number.

$^\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.

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 two integers $n$ and $m$ ($1 \le n \le 2 \cdot 10^5$, $1 \le m \le 2 \cdot 10^5$).

The $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \le a_i \le a_i + k_i\cdot d_i \le n$, $1 \le d_i \le 10$, $0 \le k_i \le n$).

It is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \cdot 10^5$.

Output: For each test case, output the number of connected components.

Examples:
input:
3
10 2
1 2 4
2 2 4
100 1
19 2 4
100 3
1 2 5
7 2 6
17 2 31
output:
2
96
61

Note: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\{1, 3, 5, 7, 9\\}$ and $\\{2, 4, 6, 8, 10\\}$.

In the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.

In the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.

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

public class D_Connect_the_Dots {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(br.readLine().trim()); // Number of test cases

        while (t-- > 0) {
            StringTokenizer st = new StringTokenizer(br.readLine());
            int n = Integer.parseInt(st.nextToken()); // Number of physical nodes
            int m = Integer.parseInt(st.nextToken()); // Number of connections

            int[] arr = new int[n + 1]; // Stores the connection assigned to physical nodes
            PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] == b[0] ? b[2] - a[2] : a[0] - b[0]);
            boolean[][] vis = new boolean[n + 1][11]; // Visited array for each node and its distance

            // Reading connections and adding them to the priority queue
            for (int i = 0; i < m; i++) {
                st = new StringTokenizer(br.readLine());
                int a = Integer.parseInt(st.nextToken()); // Physical node
                int d = Integer.parseInt(st.nextToken()); // Distance
                int k = Integer.parseInt(st.nextToken()); // Steps
                pq.add(new int[]{a, d, k, i + 1}); // Add to queue with connection ID i+1
            }

            // Initialize graph for m connections
            List<Set<Integer>> graph = new ArrayList<>(m + 1);
            for (int i = 0; i <= m; i++) graph.add(new HashSet<>());

            // Processing priority queue
            while (!pq.isEmpty()) {
                int[] a = pq.poll(); // Get the top element from the priority queue
                int node = a[0], dist = a[1], steps = a[2], connID = a[3];

                // Skip if already visited with this distance
                if (vis[node][dist]) {
                    graph.get(connID).add(arr[node]);
                    graph.get(arr[node]).add(connID);
                    continue;
                }

                // Add connections for all distances already visited for this node
                for (int i = 1; i <= 10; i++) {
                    if (vis[node][i]) {
                        graph.get(connID).add(arr[node]);
                        graph.get(arr[node]).add(connID);
                    }
                }

                // Mark the node with current distance as visited
                vis[node][dist] = true;
                arr[node] = connID; // Assign this connection to the node

                // If steps left, propagate to the next node
                if (steps > 0 && node + dist <= n) {
                    pq.add(new int[]{node + dist, dist, steps - 1, connID});
                }
            }

            // Count connected components in the graph
            int numOfConnectedComponents = 0;
            boolean[] mark = new boolean[m + 1]; // Visited array for DFS on connections

            // Iterative DFS to count connected components
            for (int i = 1; i <= m; i++) {
                if (!mark[i]) {
                    iterativeDFS(graph, i, mark);
                    numOfConnectedComponents++; // Increment component count
                }
            }

            // Count the number of physical nodes connected to some connection
            int cnt = 0;
            for (int i = 1; i <= n; i++) {
                if (arr[i] != 0) cnt++;
            }

            // Output the result
            System.out.println(n - cnt + numOfConnectedComponents);
        }
    }

    // Iterative DFS to explore the connection graph
    public static void iterativeDFS(List<Set<Integer>> graph, int src, boolean[] vis) {
        Stack<Integer> stack = new Stack<>();
        stack.push(src);
        vis[src] = true;

        while (!stack.isEmpty()) {
            int node = stack.pop();
            for (int neighbor : graph.get(node)) {
                if (!vis[neighbor]) {
                    vis[neighbor] = true;
                    stack.push(neighbor);
                }
            }
        }
    }
}

**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
10 2
1 2 4
2 2 4
100 1
19 2 4
100 3
1 2 5
7 2 6
17 2 31
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Connect the Dots
time_limit_ms: 2000
memory_limit_mb: 512
Description: One fine evening, Alice sat down to play the classic game "Connect the Dots", but with a twist.

To play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:

  * She picks three integers $a_i$, $d_i$ ($1 \le d_i \le 10$), and $k_i$.    * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \ldots, a_i+k_i\cdot d_i$ and connects each pair of these points with arcs.

After performing all $m$ operations, she wants to know the number of connected components$^\dagger$ these points form. Please help her find this number.

$^\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.

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 two integers $n$ and $m$ ($1 \le n \le 2 \cdot 10^5$, $1 \le m \le 2 \cdot 10^5$).

The $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \le a_i \le a_i + k_i\cdot d_i \le n$, $1 \le d_i \le 10$, $0 \le k_i \le n$).

It is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \cdot 10^5$.

Output: For each test case, output the number of connected components.

Examples:
input:
3
10 2
1 2 4
2 2 4
100 1
19 2 4
100 3
1 2 5
7 2 6
17 2 31
output:
2
96
61

Note: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\{1, 3, 5, 7, 9\\}$ and $\\{2, 4, 6, 8, 10\\}$.

In the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.

In the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.

Buggy Code:
import java.util.HashMap;
import java.util.Scanner;

public class D_Connect_the_Dots {

    public static class DisjointSet {
        HashMap<Integer, Integer> parent;
        HashMap<Integer, Integer> size;
        int count ;

        DisjointSet(int n) {
            parent = new HashMap<>();
            size = new HashMap<>();
            count = n ;
        }

        int findUlp(int node) {
            if (!parent.containsKey(node)) {
                parent.put(node, node);
                size.put(node, 1);
            }
            if (parent.get(node) == node) {
                return node;
            }
            int ulp = findUlp(parent.get(node));
            parent.put(node, ulp);
            return ulp;
        }

        void unionBySize(int u, int v) {
            int ulp_u = findUlp(u);
            int ulp_v = findUlp(v);

            if (ulp_u != ulp_v) {
                if (size.get(ulp_u) > size.get(ulp_v)) {
                    parent.put(ulp_v, ulp_u);
                    size.put(ulp_u, size.get(ulp_u) + size.get(ulp_v));
                } else {
                    parent.put(ulp_u, ulp_v);
                    size.put(ulp_v, size.get(ulp_u) + size.get(ulp_v));
                }

                count-- ;
            }
        }
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int m = sc.nextInt();

            DisjointSet ds = new DisjointSet(n);

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

                int prev = a;
                ds.findUlp(prev);

                for (int j = 1; j <= k; j++) {
                    int next = a + j * d;
                    if (next > n) break;
                    ds.findUlp(next);
                    ds.unionBySize(prev, next);
                    prev = next;
                }
            }

            System.out.println(ds.count);
        }
    }
}

**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
10 1
1 1 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Elections
time_limit_ms: 2000
memory_limit_mb: 256
Description: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.

The candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.

You found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.

You are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $c$ ($1 \le n \le 2 \cdot 10^5$, $0 \le c \le 10^9$) — the number of candidates in the elections and the number of undecided people.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the number of fans for each candidate.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.

Examples:
input:
5
3 1
2 0 3
2 3
0 10
5 3
5 4 3 2 1
4 5
3 10 7 1
6 0
2 2 2 3 3 3
output:
0 1 2
1 0
0 1 2 3 4
1 0 2 3
1 1 2 0 4 5

Note: In the first test case:

  * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$.   * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$.   * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.

In the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define ppb pop_back
#define f first
#define int long long int
#define sec second
#define MOD 1e9+7
#define vi vector<int>
#define pii pair<int, int>
#define MOD1 1000000009
#define MOD2 998244353
#define PI 3.141592653589793238462
#define set_bits __builtin_popcountll
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, -1, 0, 1};
#define for_(x) for(int i=0;i<x;i++)
const long long inf = 1e18;
typedef long long ll;
typedef unsigned long long ull;
typedef long double lld;
#define all(v) v.begin(), v.end()
#define yes cout << "YES" << endl
#define no cout << "NO" << endl
#define rall(x) (x).rbegin(),(x).rend()
#define fastio()                      \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);
/*---------------------------------------------------------------------------------------------------------------------------*/
ll expo(ll a, ll b, ll mod) {ll res = 1; while (b > 0) {if (b & 1)res = (res * a) % mod; a = (a * a) % mod; b = b >> 1;} return res;}
void extendgcd(ll a, ll b, ll*v) {if (b == 0) {v[0] = 1; v[1] = 0; v[2] = a; return ;} extendgcd(b, a % b, v); ll x = v[1]; v[1] = v[0] - v[1] * (a / b); v[0] = x; return;} //pass an arry of size1 3
ll mminv(ll a, ll b) {ll arr[3]; extendgcd(a, b, arr); return arr[0];} //for non prime b
ll mminvprime(ll a, ll b) {return expo(a, b - 2, b);}
ll combination(ll n, ll r, ll m, ll *fact, ll *ifact) {ll val1 = fact[n]; ll val2 = ifact[n - r]; ll val3 = ifact[r]; return (((val1 * val2) % m) * val3) % m;}
ll phin(ll n) {ll number = n; if (n % 2 == 0) {number /= 2; while (n % 2 == 0) n /= 2;} for (ll i = 3; i <= sqrt(n); i += 2) {if (n % i == 0) {while (n % i == 0)n /= i; number = (number / i * (i - 1));}} if (n > 1)number = (number / n * (n - 1)) ; return number;} //O(sqrt(N))
ll gcd(ll a, ll b) {if (b > a) {return gcd(b, a);} if (b == 0) {return a;} return gcd(b, a % b);}
bool revsort(ll a, ll b) {return a > b;}
void google(int t) {cout << "Case #" << t << ": ";}
vector<ll> sieve(int n) {int*arr = new int[n + 1](); vector<ll> vect; for (int i = 2; i <= n; i++)if (arr[i] == 0) {vect.push_back(i); for (int j = 2 * i; j <= n; j += i)arr[j] = 1;} return vect;}
ll mod_add(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a + b) % m) + m) % m;}
ll mod_mul(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a * b) % m) + m) % m;}
ll mod_div(ll a, ll b, ll m) {a = a % m; b = b % m; return (mod_mul(a, mminvprime(b, m), m) + m) % m;}
ll mod_sub(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a - b) % m) + m) % m;}
vector<int> decimalToBinary(int n) {  vector<int> binary(32, 0); int index = 31;  while (n > 0) {    int remainder = n % 2;    binary[index] = remainder;    n /= 2;    index--; } reverse(all(binary)); return binary;}
ll binpow(long long a, long long b) { if (b == 0)   return 1;  long long res = binpow(a, b / 2);  if (b % 2)return res * res * a; else return res * res;}
ll binaryToDecimal(const std::vector<int>& binary) { int decimal = 0; int power = 1;for (int i = binary.size() - 1; i >= 0; --i) { decimal += binary[i] * power; power *= 2;}return decimal;}
ll binaryToDecimal(string binary) {long long decimal = 0;long long power = 1;for (long long i = binary.size() - 1; i >= 0; --i) {decimal += (binary[i]-'0') * power; power *= 2;}return decimal;}
/*---------------------------------------------------------------------------------------------------------------------------*/

#ifndef ONLINE_JUDGE
#define debug(x)            \
    std::cerr << #x << " "; \
    _print(x);              \
    cerr << endl;
#else
#define debug(x)

#endif
void _print(int t) { std::cerr << t; }
void _print(string t) { std::cerr << t; }
void _print(char t) { std::cerr << t; }
void _print(double t) { std::cerr << t; }
void _print(ull t) { std::cerr << t; }
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.f); cerr << ","; _print(p.sec); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}

/*
try brute force
use clear naming
write clear code
recheck the logic
concentrate
*/
void solve(){
    int n, c; cin >> n >> c;
    vector<int> vec(n);
    for_(n){
        cin >> vec[i];
    }
    int idxMaxi = max_element(all(vec))-vec.begin();
    int sum = accumulate(all(vec), 0ll);
    int curr = 0;
    for_(n){
        if(i == 0){
            if(vec[idxMaxi] > vec[i]+c)  cout << 1 << " ";
            else cout << 0 << " ";
        }
        else if(i == idxMaxi && vec[i] > vec[0]+c){
            cout << 0 << " ";
        }
        else if((vec[i]+curr+c > vec[idxMaxi])||(vec[i]+curr+c == vec[idxMaxi] && i < idxMaxi)){
            cout << i << " ";
        }
        else  cout << i+1 << " ";
        curr+=vec[i];
    }
    cout << endl;
}

signed main(){
	fastio();
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	freopen("error.txt", "w", stderr);
#endif

int n; cin >> n;
while(n--)
    solve();
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases that can cause the algorithm to fail. Here are the steps to think through this:

1. **Identify the Bug**: The buggy code has a logical issue in how it handles the exclusion of candidates and the redistribution of votes. Specifically, it does not correctly handle the scenario where multiple candidates have the same number of votes after some exclusions.

2. **Edge Cases to Consider**:
   - **Equal Votes**: Candidates with equal votes.
   - **Undecided Voters**: A large number of undecided voters.
   - **Single Candidate**: A scenario with a single candidate.
   - **All Candidates with Zero Fans**: All candidates have zero fans.

3. **Constructing the Test Case**:
   - We will create a test case with a small number of candidates to ensure it fits within the output limit.
   - We will use a scenario where candidates have equal votes and a significant number of undecided voters to make the problem more challenging.

Let's construct the test case:

- **Number of Test Cases (t)**: 1
- **Number of Candidates (n)**: 3
- **Number of Undecided Voters (c)**: 5
- **Votes for Each Candidate (a1, a2, a3)**: 2, 2, 2

This test case will challenge the code because all candidates have the same number of fans, and there are a significant number of undecided voters. The code should correctly determine the minimum number of exclusions needed for each candidate to win.

Here is the test input:

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

This input will force the code to handle the scenario where all candidates have equal votes and there are many undecided voters, which can expose the logical issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Elections
time_limit_ms: 2000
memory_limit_mb: 256
Description: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.

The candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.

You found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.

You are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $c$ ($1 \le n \le 2 \cdot 10^5$, $0 \le c \le 10^9$) — the number of candidates in the elections and the number of undecided people.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the number of fans for each candidate.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.

Examples:
input:
5
3 1
2 0 3
2 3
0 10
5 3
5 4 3 2 1
4 5
3 10 7 1
6 0
2 2 2 3 3 3
output:
0 1 2
1 0
0 1 2 3 4
1 0 2 3
1 1 2 0 4 5

Note: In the first test case:

  * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$.   * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$.   * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.

In the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.

Buggy Code:
#include "bits/stdc++.h"
using namespace std;
#define FOR(a, c) for (int(a) = 0; (a) < (c); (a)++)
#define FORL(a, b, c) for (int(a) = (b); (a) <= (c); (a)++)
#define FORR(a, b, c) for (int(a) = (b); (a) >= (c); (a)--)
#define dbg(x) cout << #x << '=' << x << endl;
#define int long long
#define inc(v) sort(v.begin() , v.end())
#define dec(v) sort(v.rbegin() , v.rend())
typedef long double ld;
typedef vector<int> vi;
typedef pair<int, int> pi;
#define F first
#define S second
#define pb push_back
#define pob pop_back
#define mp make_pair

void solve()
{
    int n , k ;
    cin >> n >> k ;
    int x = -1 , j = -1 ;
    vi v(n);
    for(int i=0 ; i<n ; i++){
        cin >> v[i];
        int temp = x ;
        x = max(x,v[i]);
        if(x>temp){
            j = i ;
        }
    }
    vi a;
    v[0] += k ;
    if(v[0]>=x){
        a.pb(0);
    }else{
        a.pb(1);
    }
    int t = 1 ;
    for(int i=1 ; i<n ; i++){
        if(i == j){
            if(a[0] == 1){
                a.pb(0);
            }else{
                a.pb(t);
            }
            t++;
            continue;
        }
        a.pb(t);
        t++;
    }

    for(int i=0 ; i<n ; i++){
        cout << a[i] << " ";
    }
    cout << '\n';
}

int32_t main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int T;
    cin >> T;
    while (T--) {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can highlight the issues in the logic. The main issues in the buggy code include:

1. Incorrect handling of the maximum votes and the candidate with the most votes.
2. Incorrect logic for determining the minimum number of candidates to exclude.

Let's consider the following edge cases:

1. **Equal votes for multiple candidates**: This can expose issues in determining the minimum number of candidates to exclude.
2. **Large number of undecided voters**: This can affect the final vote count and the logic for excluding candidates.
3. **Candidates with zero votes**: This can test the logic for handling candidates with no fans.

Here is a small-scale test input that includes these edge cases:

```plaintext
3
3 1
2 0 3
2 3
0 10
4 5
3 10 7 1
```

Explanation:
1. **Test Case 1**:
   - `n = 3`, `c = 1`
   - `a = [2, 0, 3]`
   - The candidate with the most fans is 3. If candidate 1 is excluded, candidate 2 will win. If candidate 2 is excluded, candidate 1 will win. If candidate 3 is excluded, candidate 1 will win.
   - Expected output: `0 1 2`

2. **Test Case 2**:
   - `n = 2`, `c = 3`
   - `a = [0, 10]`
   - The candidate with the most fans is 2. If candidate 2 is excluded, candidate 1 will win.
   - Expected output: `1 0`

3. **Test Case 3**:
   - `n = 4`, `c = 5`
   - `a = [3, 10, 7, 1]`
   - The candidate with the most fans is 2. If candidate 1 is excluded, candidate 3 will win. If candidate 2 is excluded, candidate 1 will win. If candidate 3 is excluded, candidate 1 will win. If candidate 4 is excluded, candidate 1 will win.
   - Expected output: `1 0 2 3`

This test input should help expose the bugs in the provided buggy code by testing various edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Elections
time_limit_ms: 2000
memory_limit_mb: 256
Description: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.

The candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.

You found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.

You are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $c$ ($1 \le n \le 2 \cdot 10^5$, $0 \le c \le 10^9$) — the number of candidates in the elections and the number of undecided people.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the number of fans for each candidate.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.

Examples:
input:
5
3 1
2 0 3
2 3
0 10
5 3
5 4 3 2 1
4 5
3 10 7 1
6 0
2 2 2 3 3 3
output:
0 1 2
1 0
0 1 2 3 4
1 0 2 3
1 1 2 0 4 5

Note: In the first test case:

  * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$.   * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$.   * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.

In the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.

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

#define fast                          \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);
#define int long long
#define double long double
#define pb push_back
#define pii pair<int, int>
#define vi vector<int>
#define all(x) (x).begin(), (x).end()
#define fr(i, n) for (int i = 1; i <= n; i++)

const int M = 1e9 + 7;

void solve()
{
    // Your code here
    int n, c;
    cin >> n >> c;
    vi nums(n + 1, 0);
    fr(i, n) cin >> nums[i];
    vi pre_sum(n + 1);
    nums[1] += c;
    fr(i, n) pre_sum[i] = pre_sum[i - 1] + nums[i];
    map<int, vi> mp;
    fr(i, n) mp[nums[i]].pb(i);
    int resev = *mp[*max_element(all(nums))].begin();
    int max_ = *max_element(all(nums));
    int a = lower_bound(all(pre_sum), max_) - pre_sum.begin();
    fr(i, n)
    {
        int ans = (i == resev) ? 0 : max(i - 1, a - 1);
        cout << ans << " ";
    }
    cout << endl;
}

signed main()
{
    fast;
    int t;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios where the logic of the code might fail. The code seems to have issues with handling the undecided voters and the correct calculation of the minimum number of exclusions required for each candidate to win.

Here are the key points to consider for a challenging test input:
1. **Undecided Voters**: The code might not handle the addition of undecided voters correctly.
2. **Ties**: The code might not handle ties between candidates correctly, especially when the lowest-numbered candidate should win in a tie.
3. **Exclusions**: The code might not correctly calculate the minimum number of exclusions required for each candidate to win.

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

1. **Small Number of Candidates**: To keep the input small and manageable.
2. **Equal Votes**: To create a tie scenario.
3. **Undecided Voters**: To see how the code handles additional votes from undecided voters.

Here is a test input that meets these criteria:

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

**Explanation:**
- **Number of Test Cases**: 1
- **Number of Candidates (n)**: 3
- **Undecided Voters (c)**: 2
- **Votes for Each Candidate**: 1, 1, 1

**Expected Output:**
- **Candidate 1**: 0 (Candidate 1 will win with 3 votes (1 from fans + 2 from undecided))
- **Candidate 2**: 1 (If Candidate 1 is excluded, Candidate 2 will win with 3 votes (1 from fans + 2 from undecided))
- **Candidate 3**: 2 (If Candidates 1 and 2 are excluded, Candidate 3 will win with 3 votes (1 from fans + 2 from undecided))

This test input should expose any issues in the code's handling of ties and the addition of undecided voters.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Elections
time_limit_ms: 2000
memory_limit_mb: 256
Description: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.

The candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.

You found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.

You are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $c$ ($1 \le n \le 2 \cdot 10^5$, $0 \le c \le 10^9$) — the number of candidates in the elections and the number of undecided people.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the number of fans for each candidate.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.

Examples:
input:
5
3 1
2 0 3
2 3
0 10
5 3
5 4 3 2 1
4 5
3 10 7 1
6 0
2 2 2 3 3 3
output:
0 1 2
1 0
0 1 2 3 4
1 0 2 3
1 1 2 0 4 5

Note: In the first test case:

  * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$.   * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$.   * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.

In the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.

Buggy Code:
def ii():
    return int(input())

def I():
    return input()

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

def mii():
    return map(int, input().split())

def lmsi():
    return list(map(str, input().split()))

# import copy
import math

# M=1e9+7
# import sys
mod = 10**9 + 7
M = 998244353

# input = sys.stdin.readline

# for _ in range(int(input())):
#     n, k = mii()
#     mx = 0
#     for i in range(1, n + 1):
#         mx += abs(n - i * 2 + 1)
#     if k % 2 or k > mx:
#         print("No")
#         continue
#     res = [i for i in range(1, n + 1)]
#     sw = 2 * (n - 1)
#     n_swp = k // 2
#     j = n - 1
#     for i in range(n_swp):
#         if j <= 1:
#             j = n - 1
#         while j > 1 and n_swp > 0:
#             if abs(res[j] - j - 1) + abs(res[j - 1] - j) < abs(
#                 res[j - 1] - j - 1
#             ) + abs(res[j] - j):
#                 n_swp -= 1
#             res[j], res[j - 1] = res[j - 1], res[j]
#             j -= 1
#     print("Yes")
#     print(*res)

for _ in range(ii()):
    n, c = mii()
    a = lmii()
    mx = max(a)
    ind = a.index(max(a))
    if ind == 0 or c + a[0] >= max(a):
        res = [i for i in range(n)]
    else:
        temp = c
        ind1 = -1
        for i in range(ind):
            temp += a[i]
            if temp >= mx:
                ind1 = i
                break
        # print(ind1, "ind")
        res = []
        if ind1 != -1:
            for i in range(ind1):
                res.append(i + 1)
            res.append(res[-1])
            for i in range(ind1 + 1, ind):
                res.append(ind - i + 1)
            res.append(0)
            for i in range(len(res), n):
                res.append(i)
        else:
            for i in range(ind):
                res.append(i + 1)
            res.append(0)
            for i in range(len(res), n):
                res.append(i)
    print(*res)

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

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

Think step by step.

Response:
```plaintext
1
4 5
10 20 30 40
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Elections
time_limit_ms: 2000
memory_limit_mb: 256
Description: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.

The candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.

You found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.

You are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $c$ ($1 \le n \le 2 \cdot 10^5$, $0 \le c \le 10^9$) — the number of candidates in the elections and the number of undecided people.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the number of fans for each candidate.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.

Examples:
input:
5
3 1
2 0 3
2 3
0 10
5 3
5 4 3 2 1
4 5
3 10 7 1
6 0
2 2 2 3 3 3
output:
0 1 2
1 0
0 1 2 3 4
1 0 2 3
1 1 2 0 4 5

Note: In the first test case:

  * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$.   * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$.   * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.

In the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.

Buggy Code:
def solve():
    n, c = map(int, input().split())
    a = list(map(int,input().split()))
    if n == 1:
        print(0)
        return
    ans = []
    b = []
    for i in range(n):
        if a[i] == 0: continue
        b.append(a[i])
    ii = a.index(max(a))
    w = max(a)
    if min(b)+c >= w:
        ii = a.index(min(b))
    # print(w,ii)
    for i in range(n):
        if i == ii:
            ans.append(0)
        elif c + a[i] >= w:
            ans.append(i)
        else:
            ans.append(i+1)
        c+= a[i]
    print(*ans)

for _ in range(int(input())):
    solve()

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

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

Think step by step.

Response:
```plaintext
5
3 0
1 1 1
2 0
1 1
4 0
1 1 1 1
5 0
1 1 1 1 1
6 0
1 1 1 1 1 1
```