**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Iris's Full Binary Tree
time_limit_ms: 4000
memory_limit_mb: 1024
Description: Iris likes full binary trees.

Let's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.

Iris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.

Since performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.

Iris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \leq i \leq n$), she'll give you an integer $p_i$ ($1 \leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.

Iris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \le i \le n$. Can you tell her the answer?

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 5 \cdot 10^5$) — the final size of the tree.

The second line of each test case contains $n - 1$ integers $p_2, p_3, \ldots, p_n$ ($1 \leq p_i < i$) — descriptions of all edges of the tree.

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

Output: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.

Examples:
input:
7
3
1 1
6
1 2 3 4 5
7
1 1 3 2 5 1
10
1 1 2 1 4 2 4 5 8
10
1 1 3 1 3 2 2 2 6
20
1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12
25
1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23
output:
1 2 2
1 2 2 3 3 4
1 2 2 3 3 4 4
1 2 2 3 3 3 4 4 5 5
1 2 2 3 3 4 4 4 -1 -1
1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7
1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8

Note: In the first test case, the final tree is shown below:

![](CDN_BASE_URL/5045337022386beaeb8585d242e8d14a)

  * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$).    * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$).    * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).

In the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):

![](CDN_BASE_URL/47a803b975dced95812a171507c56a8b)

The depth of the formed full binary tree is $4$.

In the fifth test case, the final tree is shown below:

![](CDN_BASE_URL/ffe91eb3e0db4e64598df8a805ba1e4a)

It can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
using i64=long long;
const int N=5e5+5;
i64 n;
vector<i64>v[N];
i64 fa[N],an[N][21],dep[N],dfn[N],rev[N],tot,deg[N],ls[N];
struct STree{
	i64 mn[N<<2],lazy[N<<2];
	void init(i64 l,i64 r,i64 p){
		mn[p]=(l==0? 0:1e18),lazy[p]=0;
		if(l==r)return;
		i64 m=(l+r)>>1;
		init(l,m,p*2+1);init(m+1,r,p*2+2);
	}
	void pushdown(i64 p){
		if(!lazy[p])return;
		lazy[p*2+1]+=lazy[p];lazy[p*2+2]+=lazy[p];
		mn[p*2+1]+=lazy[p];mn[p*2+2]+=lazy[p];
		lazy[p]=0;
	}
	void change(i64 l,i64 r,i64 s,i64 t,i64 p,i64 val){
		if(l<=s&&t<=r){
			mn[p]+=val;lazy[p]+=val;
			return;
		}
		i64 m=(s+t)>>1;pushdown(p);
		if(m>=l)change(l,r,s,m,p*2+1,val);
		if(m<r)change(l,r,m+1,t,p*2+2,val);
		mn[p]=min(mn[p*2+1],mn[p*2+2]);
	}
	void change2(i64 pos,i64 l,i64 r,i64 p,i64 val){
		if(l==r){
			mn[p]=val;
			return;
		}
		i64 m=(l+r)>>1;pushdown(p);
		if(m>=pos)change2(pos,l,m,p*2+1,val);
		else change2(pos,m+1,r,p*2+2,val);
		mn[p]=min(mn[p*2+1],mn[p*2+2]);
	}
}seg;
void dfs(i64 x,i64 pre,i64 d){
	dep[x]=d;fa[x]=pre;an[x][0]=fa[x];ls[x]=1;
	for(int i=0;i<__lg(n+1);i++)	if(an[x][i]==-1)an[x][i+1]=-1;else an[x][i+1]=an[an[x][i]][i];
	rev[tot]=x;dfn[x]=tot++;
	for(auto i:v[x])dfs(i,x,d+1),ls[x]+=ls[i];
}
i64 getlca(i64 x,i64 y){
	if(dep[x]<dep[y])swap(x,y);
	i64 d=dep[x]-dep[y];
	for(i64 i=__lg(d);i>=0;--i)if((d>>i)&1)x=an[x][i];
	if(x==y)return x;
	d=dep[x];
	for(i64 i=__lg(d);i>=0;--i)if((1<<i)<=dep[x]&&an[x][i]!=an[y][i])x=an[x][i],y=an[y][i];
	return fa[x];
}
i64 getan(i64 x,i64 y){
	if(dfn[y]>=dfn[x]&&dfn[y]<=ls[x]){
		i64 d=dep[y]-dep[x]-1;
		for(i64 i=0;(1<<i)<=d;++i)if((d>>i)&1)y=an[y][i];
		return y;
	}else return fa[x];
}
void updside(i64 x,i64 y){
	if(dfn[y]>=dfn[x]&&dfn[y]<=ls[x]){
		seg.change(0,n-1,0,n-1,0,1);
		x=getan(x,y);
		seg.change(dfn[x],ls[x],0,n-1,0,-1);
	}else seg.change(dfn[x],ls[x],0,n-1,0,1);
}
i64 getdist(i64 x,i64 y){return dep[x]+dep[y]-2*dep[getlca(x,y)];}
struct Node{
	i64 x,y,len;
	i64 change2(i64 z){
		if(x==y){
			i64 d=getdist(x,z);
			if(d<=len)return d+len;
			updside(y,z);
			y=getan(y,z);
			return d+len;
		}else{
			i64 d1=getdist(x,z),d2=getdist(y,z),d3=min(d1,d2);
			if(d3<=len)return d3+len+1;
			if(d1>d2)swap(x,y);
			updside(y,z);
			y=x;++len;return len*2;
		}
	}
	i64 query(i64 z){
		if(x==y)return len+getdist(x,z);
		else return len+min(getdist(x,z),getdist(y,z))+1;
	}
};
int main(){
	i64 T;
	cin>>T;
	while(T--){
		cin>>n;
		for(int i=0;i<n;i++)	v[i].clear();
		for(int i=1;i<n;i++){
			cin>>fa[i];--fa[i];
			v[fa[i]].push_back(i);
		}
		tot=0;dfs(0,-1,0);seg.init(0,n-1,0);
		for(int i=0;i<n;i++)	deg[i]=0,ls[i]+=dfn[i]-1;
		Node d={0,0,0};
		cout<<"1 ";
		for(int i=1;i<n;i++){
			++deg[fa[i]];++deg[i];
			if(deg[fa[i]]==4){
				for(int j=i;j<n;j++)	cout<<-1<<' ';
				break;
			}
			if(deg[fa[i]]==3)seg.change2(dfn[fa[i]],0,n-1,0,1e18);
			seg.change2(dfn[i],0,n-1,0,d.change2(i));
			cout<<seg.mn[0]+1<<' ';
		}
		cout<<endl;
	}
	return 0;
}

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

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

Think step by step.

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

### Explanation:
1. **Test Case 1 (n=4):**
   - The tree is formed by adding vertices as follows:
     - Vertex 2 connects to 1.
     - Vertex 3 connects to 1.
     - Vertex 4 connects to 2.
   - This tree forms a simple structure with a depth of 2. The binary depth of the tree formed by the first 4 vertices is 2.

2. **Test Case 2 (n=6):**
   - The tree is formed by adding vertices as follows:
     - Vertex 2 connects to 1.
     - Vertex 3 connects to 2.
     - Vertex 4 connects to 3.
     - Vertex 5 connects to 4.
     - Vertex 6 connects to 5.
   - This tree forms a linear structure with a depth of 5. The binary depth of the tree formed by the first 6 vertices is 5.

3. **Test Case 3 (n=7):**
   - The tree is formed by adding vertices as follows:
     - Vertex 2 connects to 1.
     - Vertex 3 connects to 1.
     - Vertex 4 connects to 1.
     - Vertex 5 connects to 3.
     - Vertex 6 connects to 3.
     - Vertex 7 connects to 3.
   - This tree forms a structure where vertex 1 has 3 children, and vertex 3 has 3 children. This structure cannot form a full binary tree by adding vertices, so the binary depth is -1.

These test cases are designed to cover various edge cases, including simple structures, linear structures, and structures that cannot form a full binary tree. They are small enough to be manageable and valid according to the problem constraints.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Iris's Full Binary Tree
time_limit_ms: 4000
memory_limit_mb: 1024
Description: Iris likes full binary trees.

Let's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.

Iris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.

Since performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.

Iris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \leq i \leq n$), she'll give you an integer $p_i$ ($1 \leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.

Iris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \le i \le n$. Can you tell her the answer?

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 5 \cdot 10^5$) — the final size of the tree.

The second line of each test case contains $n - 1$ integers $p_2, p_3, \ldots, p_n$ ($1 \leq p_i < i$) — descriptions of all edges of the tree.

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

Output: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.

Examples:
input:
7
3
1 1
6
1 2 3 4 5
7
1 1 3 2 5 1
10
1 1 2 1 4 2 4 5 8
10
1 1 3 1 3 2 2 2 6
20
1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12
25
1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23
output:
1 2 2
1 2 2 3 3 4
1 2 2 3 3 4 4
1 2 2 3 3 3 4 4 5 5
1 2 2 3 3 4 4 4 -1 -1
1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7
1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8

Note: In the first test case, the final tree is shown below:

![](CDN_BASE_URL/5045337022386beaeb8585d242e8d14a)

  * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$).    * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$).    * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).

In the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):

![](CDN_BASE_URL/47a803b975dced95812a171507c56a8b)

The depth of the formed full binary tree is $4$.

In the fifth test case, the final tree is shown below:

![](CDN_BASE_URL/ffe91eb3e0db4e64598df8a805ba1e4a)

It can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.

Buggy Code:
import heapq

par = []
depth = []
def build(u, p):
    par[u][0] = p
    i = 0
    while par[u][i]:
        par[u][i + 1] = par[par[u][i]][i]
        i += 1
    depth[u] = depth[p] + 1

def lca(u, v):
    d = depth[u] + depth[v]
    if depth[v] > depth[u]:
        u, v = v, u
    for i in range(19, -1, -1):
        if depth[par[u][i]] >= depth[v]:
            u = par[u][i]
    for i in range(19, -1, -1):
        if par[u][i] != par[v][i]:
            u, v = par[u][i], par[v][i]
    p = u if u == v else par[u][0]
    return p, d - 2 * depth[p]

def solve():
    n = int(input())

    global par, depth
    par = [[0] * 20 for _ in range(n + 1)]
    depth = [0] * (n + 1)
    depth[1] = 1

    u, v = 1, 1
    dist = 0
    heap = [(0, 1)]
    deg = [0] * (n + 1)
    max_deg = 1

    print(1, end = ' ')
    i = 1
    for p in list(map(int, input().split())):
        i += 1
        build(i, p)
        deg[i] += 1
        deg[p] += 1
        max_deg = max(max_deg, deg[p])
        if max_deg > 3:
            print(-1, end = ' ')
            continue

        dist1 = lca(i, u)[1]
        dist2 = lca(i, v)[1]
        if dist1 > dist2:
            if dist1 > dist:
                v, dist = i, dist1
        else:
            if dist2 > dist:
                u, dist = i, dist2

        heapq.heappush(heap, (max(lca(i, u)[1], lca(i, v)[1]), i))
        while True:
            d, t = heap[0]
            if deg[t] > 2:
                heapq.heappop(heap)
                continue
            mx = max(lca(t, u)[1], lca(t, v)[1])
            if mx != d:
                heapq.heappop(heap)
                heapq.heappush(heap, (mx, t))
                continue
            print(d + 1, end = ' ')
            break
    print()

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

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Iris's Full Binary Tree
time_limit_ms: 4000
memory_limit_mb: 1024
Description: Iris likes full binary trees.

Let's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.

Iris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.

Since performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.

Iris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \leq i \leq n$), she'll give you an integer $p_i$ ($1 \leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.

Iris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \le i \le n$. Can you tell her the answer?

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 5 \cdot 10^5$) — the final size of the tree.

The second line of each test case contains $n - 1$ integers $p_2, p_3, \ldots, p_n$ ($1 \leq p_i < i$) — descriptions of all edges of the tree.

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

Output: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.

Examples:
input:
7
3
1 1
6
1 2 3 4 5
7
1 1 3 2 5 1
10
1 1 2 1 4 2 4 5 8
10
1 1 3 1 3 2 2 2 6
20
1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12
25
1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23
output:
1 2 2
1 2 2 3 3 4
1 2 2 3 3 4 4
1 2 2 3 3 3 4 4 5 5
1 2 2 3 3 4 4 4 -1 -1
1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7
1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8

Note: In the first test case, the final tree is shown below:

![](CDN_BASE_URL/5045337022386beaeb8585d242e8d14a)

  * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$).    * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$).    * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).

In the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):

![](CDN_BASE_URL/47a803b975dced95812a171507c56a8b)

The depth of the formed full binary tree is $4$.

In the fifth test case, the final tree is shown below:

![](CDN_BASE_URL/ffe91eb3e0db4e64598df8a805ba1e4a)

It can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.

Buggy Code:
#include "bits/stdc++.h"
using namespace std;
#define all(x) x.begin(),x.end()
template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << p.first << " " << p.second; }
template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) { string sep; for (const T &x : v) os << sep << x, sep = " "; return os; }
#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) 42
#define ASSERT(...) 42
#endif
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pi;
const int oo = 1e9;
struct node {
    int mn;
    node() {
        // *this = node(0);
        mn=oo;
    }
    node(int val) {
        mn=val;
    }
    void merge(const node& o) {
        mn=min(mn,o.mn);
    }
    friend node merge(node a, const node& b) {
        a.merge(b);
        return a;
    }
};
struct segtree {
    int ptwo;
    vector<node> seg;
    segtree(){}
    node& operator[](int i) {
        return seg[i+ptwo];
    }
    segtree(int nn) {
        ptwo=nn;
        seg.assign(ptwo*2,{});
    }
    auto query(int l, int r) {
        assert(l>=0 and l<ptwo and r >=0 and r<ptwo);
        l+=ptwo; r+=ptwo;
        node resl, resr;
        while(l and l<=r) {
            if(l%2==1) resl = merge(resl,seg[l++]);
            if(r%2==0) resr = merge(seg[r--],resr);
            l/=2; r/=2;
        }
        return merge(resl,resr);
    }
    void update(int i, int val) {
        assert(i>=0 and i<ptwo);
        i+=ptwo;
        seg[i] = node(val);
        for(i/=2;i>=1;i/=2) {
            seg[i] = merge(seg[2*i],seg[2*i+1]);
        }
    }
    void build() {
        for(int i=ptwo-1;i>=1;--i) {
            seg[i] = merge(seg[2*i],seg[2*i+1]);
        }
    }
};
segtree seg;
struct LCA {
    vi par,d,jmp,in,out;
    vector<basic_string<int>> adj;
    int root;
    vector<set<int>> h;
    LCA(int n,int root=0) : par(n,root),d(n),jmp(n,root),in(n),out(n),adj(n),root(root),h(n) {}

    void add(int i) {
        int p = par[i];
        d[i]=1+d[p];
        if(d[p] - d[jmp[p]] == d[jmp[p]] - d[jmp[jmp[p]]]) jmp[i] = jmp[jmp[p]];
        else jmp[i] = p;
    }
    int jump(int a, int k) {
        int D = max(0,d[a]-k);
        while(d[a]>D) {
            if(d[jmp[a]]>=D) a = jmp[a];
            else a = par[a];
        }
        return a;
    }
    template<typename F> pi find(int a, F f) {
        int steps=d[a];
        while(!f(a)) {
            if(!f(jmp[a])) a = jmp[a];
            else a = par[a];
        }
        steps-=d[a]-1;
        return {a,steps};
    }
    int lca(int a, int b) {
        if(d[a]<d[b]) swap(a,b);
        a = jump(a,d[a]-d[b]);
        while(a!=b) {
            if(jmp[a]!=jmp[b]) a=jmp[a],b=jmp[b];
            else a=par[a],b=par[b];
        }
        return a;
    }
    int dist(int u,int v) {
        return d[u]+d[v]-2*d[lca(u,v)];
    }
    void addE(int u, int v) {
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    int cnt=0;
    void dfs(int at) {
        in[at]=cnt++;
        for(int to: adj[at]) if(to!=par[at])  {
            par[to]=at;
            add(to);
            dfs(to);
        }
        out[at]=cnt;
    }
    void init() {
        dfs(root);
    }
    const int B = 21;

    int get(int a, int b, int w, int di) {
        int l = lca(a,b);
        if(d[a]-d[l]>=w) {
            return jump(a,w);
        } else {
            return jump(b,di-w);
        }
    }
    int a,b;
    bool insub(int u, int v) {
        return in[u]<=in[v] and in[v]<out[u];
    }
    int midpoint(int u, int v) {
        a=u,b=v;
        int di = dist(a,b);
        int at = get(a,b,di/2,di);
        // get your answer immediately
        int ans=oo;
        for(int v=at,rep=0;rep<B;v=par[v],rep++) {
            // do range min query
            int cur = max(dist(v,a),dist(v,b));
            ans=min(ans, cur + seg.query(in[v],out[v]-1).mn - d[v]);
        }
        for(int rep2=0;rep2<2;++rep2) {
            for(int v=at,rep=0;rep<B;rep++) {
                int cur = max(dist(v,a),dist(v,b));
                ans=min(ans, cur + seg.query(in[v],out[v]-1).mn - d[v]);
                int nxt=-1;
                for(int to : adj[at]) {
                    if(to!=par[at] and insub(to,a)) {
                        nxt=to;
                    }
                }
                if(nxt==-1) break;
                v=nxt;
            }
            swap(a,b);
        }
        return ans;

    }
};
void solve() {
    int n; cin >> n;
    LCA lca(n);
    vi par(n);
    for(int i=1;i<n;++i) {
        int p; cin >> p;
        --p;
        par[i]=p;;
        lca.addE(p,i);
    }
    lca.init();
    int a=0,b=0;
    vi deg(n);
    bool bad=0;
    seg = segtree(n);
    seg.build();
    for(int i=0;i<n;++i) {
        int diam = lca.dist(a,b);
        if(lca.dist(a,i)>diam) {
            b=i;
        } else if(lca.dist(i,b)>diam) {
            a=i;
        }

        if(i) {
            deg[par[i]]++,deg[i]++;
            if(deg[par[i]]==3) {
                seg.update(lca.in[par[i]],oo);
            }

        }
        seg.update(lca.in[i],lca.d[i]);
        if(deg[i]>=4) bad=1;
        if(deg[par[i]]>=4) bad=1;
        int ans = lca.midpoint(a,b);
        if(bad) ans=-2;
        cout << ans+1 << ' ';

    }
    cout << '\n';

}
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t; cin >> t;
    while(t--) solve();
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. The code involves several complex operations, including segment trees and LCA (Lowest Common Ancestor) calculations. Let's consider the following aspects:

1. **Tree Structure**: The tree should have a structure that maximizes the chances of incorrect LCA or diameter calculations.
2. **Node Degrees**: The code checks for nodes with degrees greater than or equal to 4, which could lead to incorrect results.
3. **Edge Cases**: Small trees with specific configurations that might not be handled correctly by the code.

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

### Test Case 1: Simple Tree with a Degree 4 Node
- This tree will have a node with degree 4 to test the degree checking logic.
- The tree will be linear initially and then form a star-like structure.

### Test Case 2: Tree with a Long Path and a Branch
- This tree will have a long path and a branch to test the LCA and diameter calculations.

### Test Case 3: Tree with Multiple Branches
- This tree will have multiple branches to test the segment tree and LCA logic.

### Test Input
```plaintext
3
6
1 2 2 3 4
10
1 1 2 2 4 4 5 5 7 6
10
1 1 3 3 1 5 4 4 6 8
```

### Explanation
1. **Test Case 1**:
   - The tree is constructed as follows:
     - 1 -> 2
     - 2 -> 3
     - 2 -> 4
     - 3 -> 5
     - 4 -> 6
   - Node 2 has a degree of 4, which should trigger the `bad` condition.

2. **Test Case 2**:
   - The tree is constructed as follows:
     - 1 -> 2
     - 1 -> 3
     - 2 -> 4
     - 2 -> 5
     - 4 -> 6
     - 4 -> 7
     - 5 -> 8
     - 5 -> 9
     - 7 -> 10
   - This tree has a long path and a branch, which should test the LCA and diameter calculations.

3. **Test Case 3**:
   - The tree is constructed as follows:
     - 1 -> 2
     - 1 -> 3
     - 3 -> 4
     - 3 -> 5
     - 1 -> 6
     - 5 -> 7
     - 4 -> 8
     - 4 -> 9
     - 6 -> 10
   - This tree has multiple branches, which should test the segment tree and LCA logic.

This test input should help expose any bugs in the provided buggy code by covering various edge cases and complex tree structures.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Iris's Full Binary Tree
time_limit_ms: 4000
memory_limit_mb: 1024
Description: Iris likes full binary trees.

Let's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.

Iris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.

Since performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.

Iris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \leq i \leq n$), she'll give you an integer $p_i$ ($1 \leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.

Iris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \le i \le n$. Can you tell her the answer?

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 5 \cdot 10^5$) — the final size of the tree.

The second line of each test case contains $n - 1$ integers $p_2, p_3, \ldots, p_n$ ($1 \leq p_i < i$) — descriptions of all edges of the tree.

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

Output: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.

Examples:
input:
7
3
1 1
6
1 2 3 4 5
7
1 1 3 2 5 1
10
1 1 2 1 4 2 4 5 8
10
1 1 3 1 3 2 2 2 6
20
1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12
25
1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23
output:
1 2 2
1 2 2 3 3 4
1 2 2 3 3 4 4
1 2 2 3 3 3 4 4 5 5
1 2 2 3 3 4 4 4 -1 -1
1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7
1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8

Note: In the first test case, the final tree is shown below:

![](CDN_BASE_URL/5045337022386beaeb8585d242e8d14a)

  * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$).    * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$).    * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).

In the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):

![](CDN_BASE_URL/47a803b975dced95812a171507c56a8b)

The depth of the formed full binary tree is $4$.

In the fifth test case, the final tree is shown below:

![](CDN_BASE_URL/ffe91eb3e0db4e64598df8a805ba1e4a)

It can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.

Buggy Code:
#include<bits/stdc++.h>
#define Pr pair<int,int>
#define fi first
#define nd second
#define psb push_back
using namespace std;
typedef long long LL;
inline int read(){
	char ch=getchar();
	while(!isdigit(ch) && ch!='-') ch=getchar();
	int x=0,ff=1; if(ch=='-') ff=-1,ch=getchar();
	while(isdigit(ch)) x=(x<<3) + (x<<1) + (ch^48),ch=getchar();
	return x*ff;
}
const int N=5e5+5,inf=1e9;
int n,p[N],b[N],ld[N],tot,son[N],fa[N],top[N],dep[N],sz[N],ru,rv,rds,ans,ind[N],gfa[N],rT; bool vis[N]; vector<int> e[N]; set<Pr> st[N];
void dfs1(int u,int Fa){
	sz[u]=1; fa[u]=Fa; dep[u]=dep[Fa]+1;
	for(int v:e[u]) if(v!=Fa){
		dfs1(v,u);  sz[u]+=sz[v];
		if(sz[v]>sz[son[u]]) son[u]=v;
	}
}
void dfs2(int u,int tpp){
	top[u]=tpp; b[ld[u]=++tot]=u;
	if(son[u]){
		dfs2(son[u],tpp);
		for(int v:e[u]) if(v!=son[u] && v!=fa[u]) dfs2(v,v);
	}
}
int LCA(int u,int v){
	while(top[u]!=top[v]){
		if(dep[top[u]]<dep[top[v]]) swap(u,v);
		u=fa[top[u]];
	} return dep[u]<dep[v]?u:v;
}
int K_fa(int u,int kx){
	kx=dep[u]-kx; while(dep[top[u]]>kx) u=fa[top[u]];
	return b[ld[u]-dep[u]+kx];
}
int gdis(int x,int y) {return dep[x]+dep[y]-2*dep[LCA(x,y)];}
void getrt(int u,int Fa,int S){
	sz[u]=1; int mx=0;
	for(int v:e[u]) if(v!=Fa && !vis[v]) {
		getrt(v,u,S); sz[u]+=sz[v];
		mx=max(mx,sz[v]);
	} mx=max(mx,S-sz[u]); rT=mx*2<=S?u:rT;
}
void dfz(int u,int Fa,int S){
	getrt(u,0,S); u=rT; gfa[u]=Fa; vis[u]=1; getrt(u,0,S);
	for(int v:e[u]) if(!vis[v]) dfz(v,u,sz[v]);
}
void addx(int x,int o){ int u=x;
	while(u){
		Pr bf={gdis(x,u),x};
		if(o) st[u].insert(bf);
		else st[u].erase(bf);
		u=gfa[u];
	}
}
int qry(int x){
	int u=x; Pr z={inf,0};
	while(u){
		if(st[u].size()){
			Pr bf=*st[u].begin(); bf.fi+=gdis(x,u);
			z=min(z,bf);
		}
		u=gfa[u];
	} return z.nd;
}
void vmain(){
	n=read(); tot=0; for(int i=1;i<=n;i++) son[i]=ind[i]=gfa[i]=vis[i]=0,e[i].clear(),st[i].clear();
	for(int i=2;i<=n;i++) p[i]=read(),e[p[i]].push_back(i),e[i].push_back(p[i]);
	dfs1(1,0); dfs2(1,1); dfz(1,0,n); ans=0;
	for(int i=1;i<=n;i++){
		if(~ans){
			addx(i,1);
			if(i==1) ans=1,ru=rv=1,rds=0;
			else { int x=p[i]; ++ind[x]; ind[i]=1;
				if(ind[x]<=3){
					if(ind[x]==3) addx(x,0);
					if(gdis(ru,i)>rds) ++rds,rv=i;
					if(gdis(rv,i)>rds) ++rds,ru=i;
					int lc=LCA(ru,rv),px=0,z=rds/2; px=(z<=dep[ru]-dep[lc])?K_fa(ru,z):K_fa(rv,rds-z);
					px=qry(px); ans=max(gdis(ru,px),gdis(rv,px))+1;
					if(rds&1){
						++z; px=(z<=dep[ru]-dep[lc])?K_fa(ru,z):K_fa(rv,rds-z); px=qry(px);
						ans=min(ans,max(gdis(ru,px),gdis(rv,px))+1);
					}
				} else ans=-1;
			}
		}
		printf("%d ",ans);
	} puts("");
}
int main(){
	int T=read(); while(T--) vmain();
	return 0;
}
/*
g++ 1.cpp -o 1 -std=c++20 -Wall -Wl,-stack=999999999
*/

**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 2 1 4 2 4 5 8
```

**Explanation:**

1. **Test Case Structure**: The test input consists of one test case with 10 vertices.
2. **Edge Cases**:
   - **Branching Early**: The tree branches early, which can expose issues in how the algorithm handles early branching.
   - **Deep Branching**: The tree has a deep branch, which can test the depth calculation and the ability to handle deep trees.
   - **Multiple Branches**: The tree has multiple branches, which can test the algorithm's ability to handle multiple branches and their interactions.
3. **Specific Edges**:
   - `1 1 2 1 4 2 4 5 8`:
     - Vertex 1 is the root.
     - Vertex 2 and 3 are children of vertex 1.
     - Vertex 4 is a child of vertex 2.
     - Vertex 5 and 6 are children of vertex 4.
     - Vertex 7 and 8 are children of vertex 5.
     - Vertex 9 and 10 are children of vertex 8.
4. **Expected Output**:
   - The binary depth should be calculated for each step, ensuring that the algorithm correctly handles the branching and depth calculations.

This test case is designed to challenge the algorithm by creating a tree with early and deep branching, which can help expose bugs in the implementation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Iris's Full Binary Tree
time_limit_ms: 4000
memory_limit_mb: 1024
Description: Iris likes full binary trees.

Let's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.

Iris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.

Since performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.

Iris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \leq i \leq n$), she'll give you an integer $p_i$ ($1 \leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.

Iris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \le i \le n$. Can you tell her the answer?

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 5 \cdot 10^5$) — the final size of the tree.

The second line of each test case contains $n - 1$ integers $p_2, p_3, \ldots, p_n$ ($1 \leq p_i < i$) — descriptions of all edges of the tree.

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

Output: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.

Examples:
input:
7
3
1 1
6
1 2 3 4 5
7
1 1 3 2 5 1
10
1 1 2 1 4 2 4 5 8
10
1 1 3 1 3 2 2 2 6
20
1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12
25
1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23
output:
1 2 2
1 2 2 3 3 4
1 2 2 3 3 4 4
1 2 2 3 3 3 4 4 5 5
1 2 2 3 3 4 4 4 -1 -1
1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7
1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8

Note: In the first test case, the final tree is shown below:

![](CDN_BASE_URL/5045337022386beaeb8585d242e8d14a)

  * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$).    * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$).    * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).

In the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):

![](CDN_BASE_URL/47a803b975dced95812a171507c56a8b)

The depth of the formed full binary tree is $4$.

In the fifth test case, the final tree is shown below:

![](CDN_BASE_URL/ffe91eb3e0db4e64598df8a805ba1e4a)

It can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.

Buggy Code:
#include<bits/stdc++.h>
#define Pr pair<int,int>
#define fi first
#define nd second
#define psb push_back
using namespace std;
typedef long long LL;
inline int read(){
	char ch=getchar();
	while(!isdigit(ch) && ch!='-') ch=getchar();
	int x=0,ff=1; if(ch=='-') ff=-1,ch=getchar();
	while(isdigit(ch)) x=(x<<3) + (x<<1) + (ch^48),ch=getchar();
	return x*ff;
}
const int N=5e5+5,inf=1e9;
int n,p[N],b[N],ld[N],tot,son[N],fa[N],top[N],dep[N],sz[N],ru,rv,rds,ans,ind[N],gfa[N],rT; bool vis[N]; vector<int> e[N]; set<Pr> st[N];
void dfs1(int u,int Fa){
	sz[u]=1; fa[u]=Fa; dep[u]=dep[Fa]+1;
	for(int v:e[u]) if(v!=Fa){
		dfs1(v,u);  sz[u]+=sz[v];
		if(sz[v]>sz[son[u]]) son[u]=v;
	}
}
void dfs2(int u,int tpp){
	top[u]=tpp; b[ld[u]=++tot]=u;
	if(son[u]){
		dfs2(son[u],tpp);
		for(int v:e[u]) if(v!=son[u] && v!=fa[u]) dfs2(v,v);
	}
}
int LCA(int u,int v){
	while(top[u]!=top[v]){
		if(dep[top[u]]<dep[top[v]]) swap(u,v);
		u=fa[top[u]];
	} return dep[u]<dep[v]?u:v;
}
int K_fa(int u,int kx){
	kx=dep[u]-kx; while(dep[top[u]]>kx) u=fa[top[u]];
	return b[ld[u]-dep[u]+kx];
}
int gdis(int x,int y) {return dep[x]+dep[y]-2*dep[LCA(x,y)];}
void getrt(int u,int Fa,int S){
	sz[u]=1; int mx=0;
	for(int v:e[u]) if(v!=Fa && !vis[v]) {
		getrt(v,u,S); sz[u]+=sz[v];
		mx=max(mx,sz[v]);
	} mx=max(mx,S-sz[u]); rT=mx*2<=S?u:rT;
}
void dfz(int u,int Fa,int S){
	getrt(u,0,S); u=rT; gfa[u]=Fa; vis[u]=1; getrt(u,0,S);
	for(int v:e[u]) if(!vis[v]) dfz(v,u,sz[v]);
}
void addx(int x,int o){ int u=x;
	while(u){
		Pr bf={gdis(x,u),x};
		if(o) st[u].insert(bf);
		else st[u].erase(bf);
		u=gfa[u];
	}
}
int qry(int x){
	int u=x; Pr z={inf,0};
	while(u){
		if(st[u].size()){
			Pr bf=*st[u].begin(); bf.fi+=gdis(x,u);
			z=min(z,bf);
		}
		u=gfa[u];
	} return z.nd;
}
void vmain(){
	n=read(); tot=0; for(int i=1;i<=n;i++) son[i]=ind[i]=gfa[i]=vis[i]=0,e[i].clear(),st[i].clear();
	for(int i=2;i<=n;i++) p[i]=read(),e[p[i]].push_back(i),e[i].push_back(p[i]);
	dfs1(1,0); dfs2(1,1); dfz(1,0,n); ans=0;
	for(int i=1;i<=n;i++){
		if(~ans){
			addx(i,1);
			if(i==1) ans=1,ru=rv=1,rds=0;
			else { int x=p[i]; ++ind[x]; ind[i]=1;
				if(ind[x]<=3){
					if(ind[x]==3) addx(x,0);
					if(gdis(ru,i)>rds) ++rds,rv=i;
					if(gdis(rv,i)>rds) ++rds,ru=i;
					int lc=LCA(ru,rv),px=0,z=rds/2; px=(z<=dep[ru]-dep[lc])?K_fa(ru,z):K_fa(rv,rds-z);
					px=qry(px); ans=max(gdis(ru,px),gdis(rv,px))+1;
					if(rds&1){
						++z; px=(z<=dep[ru]-dep[lc])?K_fa(ru,z):K_fa(rv,rds-z); px=qry(px);
						ans=min(ans,max(gdis(ru,px),gdis(rv,px))+1);
					}
				} else ans=-1;
			}
		}
		printf("%d ",ans);
	} puts("");
}
int main(){
	int T=read(); while(T--) vmain();
	return 0;
}
/*
g++ 1.cpp -o 1 -std=c++20 -Wall -Wl,-stack=999999999
*/

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given algorithm, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are some key points to consider:

1. **Small Scale**: The test input should be small enough to be manageable but still complex enough to challenge the algorithm.
2. **Edge Cases**: Consider cases where the tree structure is unusual or where the binary depth is particularly difficult to compute.
3. **Maximizing Failure Probability**: Focus on cases where the binary depth changes frequently or where the tree structure is not straightforward.

Given these points, let's create a test input that includes a few test cases with different tree structures:

1. A simple tree with a linear structure.
2. A tree with a branching structure that is close to being a full binary tree.
3. A tree that cannot be made into a full binary tree.

Here is the test input:

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

### Explanation:

1. **Test Case 1:**
   - Tree with 3 vertices: 1 -> 2, 1 -> 3
   - This is a simple tree that is already a full binary tree of depth 2.
   - Expected output: `1 2 2`

2. **Test Case 2:**
   - Tree with 5 vertices: 1 -> 2, 2 -> 3, 3 -> 4, 4 -> 5
   - This is a linear tree where each vertex has exactly one child.
   - Expected output: `1 2 3 4 5`

3. **Test Case 3:**
   - Tree with 6 vertices: 1 -> 2, 1 -> 3, 2 -> 4, 2 -> 5, 4 -> 6
   - This tree is close to being a full binary tree but has an extra vertex that makes it impossible to form a full binary tree.
   - Expected output: `1 2 2 3 3 4`

This test input should be challenging enough to expose any bugs in the given algorithm, especially in handling cases where the tree structure is not straightforward and where the binary depth changes frequently.


**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:
// Online C++ compiler to run C++ program online
#include <iostream>
#include <cstring>

using namespace std;
long long val;
long parent[200002],dp[200002][11],sizes[200002],ind[200002][11];
long par(long k){
    if(parent[k]==k)
        return k;
    return parent[k]=par(parent[k]);
}
void unionFind(long  a,long b){
    a=par(a);
    b=par(b);
    if(a==b)
        return;
    if(sizes[b]>sizes[a]){
            swap(a,b);
    }

    parent[b]=a;
         sizes[a]+=sizes[b];

    return;
}
void solve(){
        long long n,m;
       cin>>n>>m;
      // cout<<n<<endl;
      val=n;
     for(int i=1;i<=n;i++){
        sizes[i]=1;
        parent[i]=i;
        for(int j=1;j<=10;j++)
            ind[i][j]=i;
     }

      memset(dp,0,sizeof(dp));

     for(int i=0;i<m;i++){
        long a,d,k;
        cin>>a>>d>>k;
        dp[a][d]++;
        long p=a+((k)*d);
        if(p>n+1)
            p=n+1;
        dp[p][d]--;
     }
     for(long i=1;i<=n;i++){
        for(long j=1;j<=10;j++){
            if(i-j>0){
                dp[i][j]=dp[i-j][j]+dp[i][j];

            if(dp[i-j][j]>0)
                unionFind(ind[i-j][j],i);
                ind[i][j]=ind[i-j][j];
            }

        }
     }
     val=0;
     for(int i=1;i<=n;i++){
        if(parent[i]==i)
            val++;
     }

       cout<<val<<"\n";
}
int main() {
    // Write C++ code here
   long long t;
   ios_base::sync_with_stdio(0);
cin.tie(0);
   cin>>t;
   for(int ii=0;ii<t;ii++){
    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 given code, we need to craft a test case that highlights potential issues in the algorithm. The key points to consider are:

1. **Edge Cases**: Test with the smallest and largest possible values for `n` and `m`.
2. **Overlap and Connectivity**: Ensure that the operations create complex connected components, especially where the `a_i`, `d_i`, and `k_i` values lead to overlapping arcs.
3. **Boundary Conditions**: Test with `a_i` and `k_i` values that push the limits of the array indices.

Given the constraints and the nature of the problem, a good test case would involve a small number of points and operations that create a non-trivial connectivity pattern.

### Test Case

Let's create a test case with `n = 10` and `m = 3` operations that create overlapping arcs:

1. Connect points `1, 3, 5, 7, 9` (using `a_i = 1`, `d_i = 2`, `k_i = 4`).
2. Connect points `2, 4, 6, 8, 10` (using `a_i = 2`, `d_i = 2`, `k_i = 4`).
3. Connect points `1, 4, 7, 10` (using `a_i = 1`, `d_i = 3`, `k_i = 3`).

This test case will ensure that some points are part of multiple connected components, which should help expose any issues in the union-find implementation.

### Test Input

```plaintext
1
10 3
1 2 4
2 2 4
1 3 3
```

### Explanation

- **First Operation**: Connects `1, 3, 5, 7, 9`.
- **Second Operation**: Connects `2, 4, 6, 8, 10`.
- **Third Operation**: Connects `1, 4, 7, 10`.

After these operations, the points `1, 4, 7, 10` will form a single connected component, and the points `2, 5, 6, 8, 9` will form another connected component. Therefore, the expected output is `2`.

This test case should help expose any issues in the union-find logic, especially in handling overlapping arcs and maintaining the correct connected components.


**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>
#define endl "\n"
using namespace std;
#define int long long

#define mod 1000000007

class DSU{
public:
	int n, *par, *rank;
public:
	DSU(int n){
		this->n=n;
		par=(int*)malloc(sizeof(int)*(n+1));
		rank=(int*)malloc(sizeof(int)*(n+1));
		memset(rank, 0, sizeof(rank));
		for(int i=1;i<=n;i++)par[i]=i;
	}
	int find(int x){
		if(par[x]==x)return x;
		return find(par[x]);
	}
	void unionSet(int a, int b){
		a=find(a), b=find(b);
		if(a!=b){
			if(rank[a]<rank[b])swap(a, b);
			par[b]=a;
			if(rank[a]==rank[b])rank[a]++;
		}
	}
};

/*
INIT
DSU dsu(len)

FUNCS
find(x), unionSet(x, y)
*/

int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    #ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif

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

    	int n,m; cin>>n>>m;

    	DSU dsu(n);
    	int dp[n+1][11];

    	memset(dp,0,sizeof(dp));

    	for(int i=0;i<m;i++){
    		int a,d,k; cin>>a>>d>>k;
    		for(int j=1;j<=k;j++){
    			if(dp[a+j*d][j])break;
    			dp[a+j*d][j]=1;
    			dsu.unionSet(a+j*d,a);
    		}

    	}

    	set<int>st;

    	for(int i=1;i<=n;i++){
    		st.insert(dsu.find(dsu.par[i]));
    	}

    	cout<<st.size()<<endl;

    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
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:
//libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

//namespaces
using namespace std;
using namespace __gnu_pbds;
using namespace chrono;

//utility variables
long long int _debug = 0, _nested = 0, _t = 1, _random_setup=0;
time_point <high_resolution_clock> _code_start, _code_end;
duration<double> _code_duration;

// macros-------------------------------------------------------------------------------
#define ll long long
#define int ll
#define all(x) x.begin(), x.end()
#define rep(i, x, n) for (int(i) = (x); (i) < (n); (i)++)
#define endl "\n"
#define nl        \
    cout << endl; \
    if (_debug)   \
        cerr << endl;

//Random number generator
int getRandomNumber(int start, int end) {
	static random_device rd; // Random device to generate a seed
	static mt19937 rng(rd()); // Mersenne Twister generator with random seed
	uniform_int_distribution<> dist(start, end); // Uniform distribution
	return dist(rng); // Generate and return a random number
}

// print
// helper-------------------------------------------------------------------------------
template <typename Primitive>
void print(Primitive &&P)
{
	cout << P << " ";
	if (_debug)
		cerr << P << " ";
	if (!_nested)
	{
		nl;
	}
}
template <typename Complex>
void print(vector<Complex> &Vector)
{
	_nested = 1;
	for (Complex &C : Vector)
		print(C);
	nl;
	_nested = 0;
}
template <typename Complex, typename... ComplexObjects>
void print(Complex &&C, ComplexObjects &&...list)
{
	_nested = 1, print(C);
	_nested = 0, print(list...);
}

// debug-------------------------------------------------------------------------------
#define bug(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
	cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char *names, Arg1 &&arg1, Args &&...args)
{
#ifndef ONLINE_JUDGE

	const char *comma = strchr(names + 1, ',');
	cout.write(names, comma - names) << ":" << arg1 << "|";
	__f(comma + 1, args...);

#endif
}

// policy based data
// structures---------------------------------------------------------------------------
#define indexed_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
#define indexed_map tree<int, int, less<int>, rb_tree_tag, tree_order_statistics_node_update>

// code----------------------------------------------------------------------------------

struct DSU {

	vector<int> parent;

	DSU(int n)
	{
		parent.resize(n+1,0);
		iota(parent.begin(), parent.end(), 0);
	}

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

	bool unite(int a, int b)
	{
		auto p1 = find(a);
		auto p2 = find(b);

		if(p1==p2)
		{
			return false;
		}

		if(p1<p2)
		{
			swap(p1,p2);
		}
		parent[p2]=p1;
		return true;
	}

};

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

	vector<DSU> dus;
	for(int i = 0; i<=10; i++)
	{
		DSU temp(n+1);
		dus.push_back(temp);
	}

	for(int i = 0; i<m; i++)
	{
		int a, d, k;
		cin>>a>>d>>k;

		int last = a+k*d;
		int first = dus[d].parent[a];
		int second = first+d;

		while(second<=last)
		{
			dus[d].unite(first, second);
			first = dus[d].parent[second];
			second = first+d;
		}
	}

	bool isRoot[n+1];
	memset(isRoot, true, sizeof(isRoot));

	for(int i = 1; i<=10; i++)
	{
		for(int j = 1; j+i<=n; j++)
		{
			if(isRoot[j] && isRoot[j+i])
			{
				if(dus[i].find(j)==dus[i].find(j+i))
				{
					isRoot[j]=false;
				}
			}
		}
	}

	int ans =0;
	for(int i = 1;i<=n;i++)
	{
	    ans+= isRoot[i]==true;
	}
	print(ans);

}

signed main()
{

	// _code_start =high_resolution_clock::now();
// #ifndef ONLINE_JUDGE
//     freopen("input.txt", "r", stdin);
//     freopen("output.txt", "w", stdout);
// #endif

	ios_base::sync_with_stdio(false), cin.tie(NULL);
	cin >> _t;

	for (int tc = 1; tc <= _t; tc++)
	{
		// cout << "Case " << tc << ":"
		//       << " ";
		solve();
	}

	// _code_end =high_resolution_clock::now();
	// _code_duration = _code_end - _code_start;
	//  cerr << "Code executed in " << _code_duration.count() << " seconds." <<endl;

}

// 1) cin does not move the input cursor to the next line, use cin.ignore() to
// move it to the next line;

// 2) GCD(a, b, c , d...) = GCD(a,b-a,c-a,d-a...);

// 3) A reverse iterator moves backwards while doing it++

**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 3
1 2 4
3 2 3
5 2 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:
t = int(input())
for _ in range(t):
    n, m = map(int, input().split())
    qrs = []
    for _ in range(m):
        a, d, k = map(int, input().split())
        qrs.append((a, d, k))
    tab = [[0]*11 for _ in range(n + 1)]

    for i in range(m):
        a, d, k = qrs[i]
        tab[a][d] += 1
        tab[a + k * d][d] -= 1
    for j in range(1, 11):
        for z in range(0, j):
            cur = 0
            for i in range(1 + z, n + 1, j):
                if tab[i][j] == 0:
                    tab[i][j] = 1 if cur > 0 else 0
                elif tab[i][j] > 0:
                    cur += tab[i][j]
                    tab[i][j] = 1 if cur > 0 else 0
                else:
                    c = tab[i][j]
                    tab[i][j] = 1 if cur > 0 else 0
                    cur += c

    # for j in range(1, 11):
    #     if sum(tab[i][j] for i in range(1, n + 1)) == 0:
    #         continue
    #     for i in range(1, n + 1):
    #         print(tab[i][j], end=' ')
    #     print()

    res = 0
    # find connected components with dfs:
    low = 1
    vis = [0] * (n + 1)
    vis_cn = 0
    while low < n + 1:
        stk = [low]
        while stk:
            x = stk.pop()
            vis[x] = 1
            vis_cn += 1
            for j in range(1, 11):
                if x + j < n + 1 and tab[x][j] == 1 and tab[x + j][j] == 1 and vis[x + j] == 0:
                    stk.append(x + j)
        while low <=n and vis[low] == 1:
            low += 1
        res += 1

    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
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:
t=int(input())
for _ in range(t):
    n,m = map(int,input().split(" "))
    lista = [i for i in range(1,n+1)]
    arch=0
    for __ in range(m):
        a,d,k = map(int,input().split(" "))
        allista = [a+i*d for i in range(k+1)]
        benne=True
        for v in allista:
            if v in lista:
                lista.remove(v)
            else:
                benne= False
        if benne:
            arch+=1
    print(arch+len(lista))

**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 3
1 1 9
2 2 4
3 3 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:
for _ in range(int(input())):
    n, m = map(int, input().split())
    queries = [[] for i in range(11)]
    for _ in range(m):
        a, d, k = map(int, input().split())
        queries[d].append([a, k])
    for i in range(11):
        queries[i].sort()
    nodes = [False for i in range(n+1)]

    ans = 0
    for i in range(1, 11):
        visited_till = [0 for i in range(i)]
        for j in queries[i]:
            a = j[0]
            k = j[1] + 1
            if a < visited_till[a%i]:
                k -= (visited_till[a%i] - a)//i
                a = visited_till[a%i]
            if k<=0:
                continue
            found = False
            while k>0:
                if nodes[a]:
                    found = True
                else:
                    nodes[a] = True
                k-=1
                a+=i
            if not found:
                ans += 1
    print(ans+nodes.count(False)-1)

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

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

Think step by step.

Response:
```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 javax.print.attribute.HashPrintServiceAttributeSet;
import java.io.BufferedInputStream;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.*;

public class Main {
        public static void main(String[] args) {
            FastScanner in = new FastScanner();
            int testCount = in.nextInt();
            for (int i = 1; i <= testCount; i++) {
                Solver solver = new Solver();
                try {
                    solver.solve(i, in, System.out);
                } catch (IOException e) {
                    e.printStackTrace(); // Handle exception appropriately
                }
            }
        }

        static class Solver {
            int visited[];
            public int dfs(ArrayList<Integer> graph[]){
                int comp=0;
                for(int i=1;i<graph.length;i++){
                    if(visited[i]==0){
                        hlpr(graph,i);
                        comp++;
                    }
                }
                return comp;
            }
            public void hlpr(ArrayList<Integer> graph[],int i){
                Queue<Integer> q = new LinkedList<>();
                q.add(i);
                visited[i]=1;
                while(!q.isEmpty()){
                    for(int temp:graph[q.remove()]){
                        if(visited[temp]==0){
                            visited[temp]=1;
                            q.add(temp);
                        }
                    }
                }
            }
            public void solve(int testNumber, FastScanner sc, Appendable out) throws IOException {
                // Solution logic goes here
                int n=sc.nextInt();
                int m=sc.nextInt();
                int arr[][]=new int[n+1][11];
                visited=new int[n+1];
                ArrayList<Integer> graph[]=new ArrayList[n+1];
                for(int i=0;i<graph.length;i++){
                    graph[i]=new ArrayList<Integer>();
                }

                for(int i=0;i<m;i++){
                    int a=sc.nextInt();
                    int d=sc.nextInt();
                    int k=sc.nextInt();
                    arr[a][d]=Math.max(k,arr[a][d]);
                }
                for(int i=1 ;i<=n;i++){
                     for(int j=1;j<=10;j++){
                        int iter=arr[i][j];
                        arr[i][j]=0;
                        int tempj=j;
                        while(iter>0&&tempj<n){
                            graph[i].add(i+tempj);
                            graph[i+tempj].add(i);
                            tempj+=j;
                            iter=Math.max(iter,arr[tempj][j]);
                            arr[tempj][j]=0;
                            iter--;
                        }
                    }
                }
                System.out.println(dfs(graph));
            }
        }

        static class FastScanner {
            private int BS = 1 << 16;
            private char NC = (char) 0;
            private byte[] buf = new byte[BS];
            private int bId = 0, size = 0;
            private char c = NC;
            private double cnt = 1;
            private BufferedInputStream in;

            public FastScanner() {
                in = new BufferedInputStream(System.in, BS);
            }

            public int nextInt() {
                return (int) nextLong();
            }

            public int[] nextInts(int N) {
                int[] res = new int[N];
                for (int i = 0; i < N; i++) {
                    res[i] = (int) nextLong();
                }
                return res;
            }

            public long nextLong() {
                cnt = 1;
                boolean neg = false;
                if (c == NC) c = getChar();
                for (; (c < '0' || c > '9'); c = getChar()) {
                    if (c == '-') neg = true;
                }
                long res = 0;
                for (; c >= '0' && c <= '9'; c = getChar()) {
                    res = (res << 3) + (res << 1) + c - '0';
                    cnt *= 10;
                }
                return neg ? -res : res;
            }

            public double nextDouble() {
                double cur = nextLong();
                return c != '.' ? cur : cur + nextLong() / cnt;
            }

            public String next() {
                StringBuilder res = new StringBuilder();
                while (c <= 32) c = getChar();
                while (c > 32) {
                    res.append(c);
                    c = getChar();
                }
                return res.toString();
            }

            public String nextLine() {
                StringBuilder res = new StringBuilder();
                while (c <= 32) c = getChar();
                while (c != '\n') {
                    res.append(c);
                    c = getChar();
                }
                return res.toString();
            }

            public boolean hasNext() {
                if (c > 32) return true;
                while (true) {
                    c = getChar();
                    if (c == NC) return false;
                    else if (c > 32) return true;
                }
            }

            public int[] sortedIntArray(int n) {
                int[] arr = new int[n];
                for (int i = 0; i < n; i++) {
                    arr[i] = nextInt();
                }
                Arrays.sort(arr);
                return arr;
            }

            public long[] sortedLongArray(int n) {
                long[] arr = new long[n];
                for (int i = 0; i < n; i++) {
                    arr[i] = nextLong();
                }
                Arrays.sort(arr);
                return arr;
            }

            public long gcd(long a, long b) {
                while (b != 0) {
                    long temp = b;
                    b = a % b;
                    a = temp;
                }
                return a;
            }

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

            public int[] prefixSum(int[] arr) {
                int n = arr.length;
                int[] prefixSum = new int[n];
                prefixSum[0] = arr[0];
                for (int i = 1; i < n; i++) {
                    prefixSum[i] = prefixSum[i - 1] + arr[i];
                }
                return prefixSum;
            }
            public static void sort(int[] arr)
            {
                //because Arrays.sort() uses quicksort which is dumb
                //Collections.sort() uses merge sort
                ArrayList<Integer> ls = new ArrayList<Integer>();
                for(int x: arr)
                    ls.add(x);
                Collections.sort(ls);
                for(int i=0; i < arr.length; i++)
                    arr[i] = ls.get(i);
            }
            public static long power(long x, long y, long p)
            {
                //0^0 = 1
                long res = 1L;
                x = x%p;
                while(y > 0)
                {
                    if((y&1)==1)
                        res = (res*x)%p;
                    y >>= 1;
                    x = (x*x)%p;
                }
                return res;
            }
            public static void iniDP(int dp[][]){
                for(int arr[]:dp) Arrays.fill(arr,-1);
            }
            private char getChar() {
                while (bId == size) {
                    try {
                        size = in.read(buf);
                    } catch (Exception e) {
                        return NC;
                    }
                    if (size == -1) return NC;
                    bId = 0;
                }
                return (char) buf[bId++];
            }
        }
    }

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. Here are some considerations:

1. **Edge Cases:**
   - Test with the minimum and maximum limits for `n` and `m`.
   - Test with `k` values that are zero or very large.
   - Test with overlapping and non-overlapping ranges of points.

2. **Faulty Behavior:**
   - The code might incorrectly handle overlapping ranges.
   - The code might not correctly update the graph for points that are connected multiple times.
   - The code might fail to handle large `k` values correctly.

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

### 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, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79
     - `7 2 6` connects points: 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79
     - `17 2 31` connects points: 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79
   - Expected output: 61 (one connected component of size 40, and 60 single-point components)

This test input is designed to check the algorithm's ability to handle overlapping ranges and large `k` values, which are potential sources of bugs in the provided code.


**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 {

    static class DSU {
        DSU(int n) {
            init(n);
        }

        private int[] p, sz;

        void init(int n) {
            p = new int[n];
            sz = new int[n];
            for (int i = 0; i < n; i++) {
                p[i] = i;
                sz[i] = 1;
            }
        }

        //用于连接两个连通分量
        void union(int x, int y) {
            x = find(x);
            y = find(y);
            if (x == y) return;
            sz[x] += sz[y];
            p[y] = x;
        }

        //用于查询某个元素所在连通分量的根节点
        int find(int x) {
            while (p[x] != x) {
                p[x] = p[p[x]];
                x = p[x];
            }
            return x;
        }

        //查询两个元素是否在同一连通分量中
        boolean same(int x, int y) {
            return find(x) == find(y);
        }

        //返回 x 所在连通分量的元素个数
        int size(int x) {
            return sz[find(x)];
        }
    }

    public static void main(String[] args) {
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt(), q = sc.nextInt();
            //a[i],d,k
            DSU dsu = new DSU(n);
            List<int[]>[][] query = new List[11][10];//d,idx%d
            for (int i = 0; i < 11; i++) {
                for (int j = 0; j < 10; j++) {
                    query[i][j] = new ArrayList<>();
                }
            }
            for (int i = 0; i < q; i++) {
                int[] a = new int[3];
                a[0] = sc.nextInt() - 1;
                a[1] = sc.nextInt();
                a[2] = sc.nextInt();
                query[a[1]][a[0] % a[1]].add(a);
            }
            int[] diff = new int[n + 1];
            for (int i = 1; i < 11; i++) {//枚举 d
                for (int j = 0; j < i; j++) {//枚举 idx % d
                    if (query[i][j].size() > 0) {
                        Arrays.fill(diff, 0);
                        for (int[] z : query[i][j]) {
                            int ai = z[0], d = z[1], k = z[2];
                            int idx = ai / d;
                            diff[idx]++;
                            diff[idx + k + 1]--;
                        }
                        for (int k = 1; k <= n; k++) {
                            diff[k] += diff[k - 1];
                        }
                        for (int k = 1; k < n; k++) {
                            if (diff[k] > 0 && diff[k - 1] > 0) {
                                dsu.union(k * i + j, (k - 1) * i + j);
                            }
                        }

                    }

                }
            }

            int ans = 0;
            for (int i = 0; i < n; i++) {
                if (dsu.find(i) == i) ans++;
            }
            out.println(ans);

        }
        out.close();
    }

    static Kattio sc = new Kattio();
    static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

    static class Kattio {
        static BufferedReader r;
        static StringTokenizer st;

        public Kattio() {
            r = new BufferedReader(new InputStreamReader(System.in));
        }

        public String next() {
            try {
                while (st == null || !st.hasMoreTokens()) {
                    st = new StringTokenizer(r.readLine());
                }
                return st.nextToken();
            } catch (Exception e) {
                return null;
            }
        }

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

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

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

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

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

Think step by step.

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

1. **Understand the Bug**: The buggy code uses a difference array to manage the union-find operations. The bug might be in how it handles the union operations or in the way it processes the queries.

2. **Identify Edge Cases**:
   - **Small `n` and `m`**: Test with a small number of points and operations to ensure the logic is correct.
   - **Large `n` and `m`**: Test with a large number of points and operations to check the performance and correctness.
   - **Single Operation**: Test with a single operation that connects multiple points.
   - **Multiple Operations**: Test with multiple operations that might overlap or have different step sizes.
   - **Edge Points**: Test with points at the edges of the range (1 and n).

3. **Generate a Small-Scale Test Input**:
   - Choose a small `n` and `m` to make the test input manageable.
   - Ensure the operations are designed to test the union-find logic thoroughly.

Given these considerations, let's create a test input that includes a single test case with a small `n` and `m`, but with operations that can expose the bug:

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

### Explanation:
- **Test Case 1**:
  - **n = 10**: There are 10 points.
  - **m = 3**: There are 3 operations.
  - **Operation 1**: Connect points `1, 3, 5, 7, 9` (1 + 2 * k for k = 0 to 4).
  - **Operation 2**: Connect points `2, 4, 6, 8, 10` (2 + 2 * k for k = 0 to 4).
  - **Operation 3**: Connect points `5, 8` (5 + 3 * k for k = 0 to 1).

### Expected Output:
- **Connected Components**:
  - Points `1, 3, 5, 7, 9` form one connected component.
  - Points `2, 4, 6, 8, 10` form another connected component.
  - Points `5` and `8` are already connected by the third operation, so they do not form a new component.

Thus, the expected output is:
```plaintext
2
```

This test input is small and manageable but includes operations that should expose any issues in the union-find logic and the difference array management.


**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<set>
#include<queue>
#include<set>
#include<map>
#include<algorithm>
#include<unordered_map>
#include<queue>

using namespace std;

vector<int> dsu,ranked;

int parent(int root){
    if(dsu[root]==root)return root;
    return dsu[root]= parent(dsu[root]);
}

void unite(int c1, int c2 ){
    int a = parent(c1);
    int b = parent(c2);
    if(a==b)return;
    if(ranked[b]>ranked[a])swap(a,b);
    ranked[a]+=ranked[b];
    dsu[b]=a;
}

void solve(){
    int n ,m;
    cin>>n>>m;
    dsu.resize(n+1),ranked.resize(n+1);
    for(int i = 0;i<=n;i++){
        dsu[i]=i;
        ranked[i]=1;
    }
    vector<vector<int> > q(m,vector<int>(3));
    for(int i = 0;i<m;i++){cin>>q[i][0]>>q[i][1]>>q[i][2];
    // q[i][0]--;
    }
    sort(q.begin(),q.end());
    vector<vector<int> >  lst(10,vector<int>(10,-1)),p(10,vector<int>(10,-1));
    int idx = 0;
    for(int i = 0;i<=n;i++){

        for(int j=0;j<10;j++){
            int x = (i+j)%10, y = (10-j)%10;
            if(p[x][y]!=-1)
                if(lst[x][y]>=i){
                    unite(i,p[x][y]);
                }
        }

        while(idx<m&&q[idx][0]==i){
            lst[i%10][q[idx][1]%10]=max(lst[i%10][q[idx][1]%10],q[idx][0]+q[idx][1]*q[idx][2]);
            p[i%10][q[idx][1]%10]=parent(i);
            for(int l=1;l<=q[idx][2]&&l<10&&q[idx][0]%10 != (q[idx][0] + q[idx][1]*l)%10;l++){
                int nxt = (q[idx][0] + q[idx][1]*l)%10;
                lst[nxt][q[idx][1]%10]=max(lst[nxt][q[idx][1]%10],q[idx][0]+q[idx][1]*q[idx][2]);
                if(p[nxt][q[idx][1]%10]!=-1)
                unite(p[nxt][q[idx][1]%10],i);
                p[nxt][q[idx][1]%10]=parent(i);
            }
            idx++;
        }

    }
    set<int> s;

    for(int i =1;i<=n;i++){
        s.insert(parent(i));
    }
    cout<<s.size()<<endl;

}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    #ifndef ONLINE_JUDGE
        freopen("input.txt","r",stdin);
        freopen("output.txt","w",stdout);
    #endif
    int t=1;
    cin>>t;
    while(t--){
        solve();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
10 3
1 1 1
2 1 1
3 1 1
```