**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Way to be Non-decreasing
time_limit_ms: 4000
memory_limit_mb: 512
Description: Little R is a magician who likes non-decreasing arrays. She has an array of length $n$, initially as $a_1, \ldots, a_n$, in which each element is an integer between $[1, m]$. She wants it to be non-decreasing, i.e., $a_1 \leq a_2 \leq \ldots \leq a_n$.

To do this, she can perform several magic tricks. Little R has a fixed array $b_1\ldots b_m$ of length $m$. Formally, let's define a trick as a procedure that does the following things in order:

  * Choose a set $S \subseteq \\{1, 2, \ldots, n\\}$.    * For each $u \in S$, assign $a_u$ with $b_{a_u}$.

Little R wonders how many tricks are needed at least to make the initial array non-decreasing. If it is not possible with any amount of tricks, print $-1$ instead.

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

The first line of each test case contains two integers $n$ and $m$ ($1\leq n \leq 10^6$, $1 \leq m \leq 10^6$) — the length of the initial array and the range of the elements in the array.

The second line of each test case contains $n$ integers $a_1, \ldots, a_n$ ($1 \leq a_i \leq m$) — the initial array.

The third line of each test case contains $m$ integers $b_1, \ldots, b_m$ ($1 \leq b_i \leq m$) — the fixed magic array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$ and the sum of $m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the minimum number of tricks needed, or $-1$ if it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Examples:
input:
3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
output:
3
-1
3

Note: In the first case, the initial array $a_1, \ldots, a_n$ is $[1, 6, 3, 7, 1]$. You can choose $S$ as follows:

  * first trick: $S = [2, 4, 5]$, $a = [1, 1, 3, 5, 2]$;    * second trick: $S = [5]$, $a = [1, 1, 3, 5, 3]$;    * third trick: $S = [5]$, $a = [1, 1, 3, 5, 5]$.

So it is possible to make $a_1, \ldots, a_n$ non-decreasing using $3$ tricks. It can be shown that this is the minimum possible amount of tricks.

In the second case, it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Buggy Code:
#include<iostream>
#include<iomanip>
#include<cmath>
#include<string>
#include<algorithm>
#include<map>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<deque>
#include<set>

using namespace std;
#define ll long long

const int N = 1e6 + 100;
const ll inf = 1e8;

ll t, n, m, a[N], b[N];

ll lg2[N];
ll fcnt, dcnt, dfn[N], dfnr[N], cycfa[N], ifcyc[N], cycno[N], cyclen[N], vis[N], deep[N], treefa[N], falca[N][30];
vector<ll> son[N];

ll find(ll p) {
    return cycfa[treefa[p]];
}

ll marklen(ll p) {
    if (cycno[p] >= cycno[b[p]]) {
        cyclen[p] = cycno[p];
        return cycno[p];
    }
    cyclen[p] = marklen(b[p]);
    return cyclen[p];
}

void markcyc(ll p) {
    ll no = 0;
    while (!ifcyc[p]) {
        cycfa[p] = fcnt;
        ifcyc[p] = 1;
        cycno[p] = ++no;
        p = b[p];
    }
}

void dfs(ll p) {
    if (vis[p] == 1) {
        fcnt++;
        markcyc(p);
        return;
    }
    vis[p] += 1;
    if (vis[b[p]] <= 1) dfs(b[p]);
    if (!ifcyc[p]) vis[p] = 2;
}

void dfs0(ll p, ll t_fa) {
    vis[p] = 1;
    treefa[p] = t_fa;
    for (auto i: son[p]) {
        if (!ifcyc[i] && !vis[i]) {
            deep[i] = deep[p] + 1;
            dfs0(i, t_fa);
        }
    }
}

void dfs_n(ll p, ll father) {
    dfn[p] = ++dcnt;
    for (auto i: son[p])
        if (i != father) dfs_n(i, p);
    dfnr[p] = dcnt;
}

ll dis(ll st, ll en) {
    if (st == en) return 0;
    if (find(st) != find(en)) return inf;
    if (treefa[st] == en) return deep[st];
    if (ifcyc[st] == 1 && ifcyc[en] == 0) return inf;
    if (ifcyc[st] == 1 && ifcyc[en] == 1) return ((cycno[en] - cycno[st] + cyclen[st]) % cyclen[st]);
    if (ifcyc[st] == 0 && ifcyc[en] == 1) return deep[st] + dis(treefa[st], en);
    if (ifcyc[st] == 0 && ifcyc[en] == 0) {
        if (treefa[st] != treefa[en]) return inf;
        if (deep[st] <= deep[en]) return inf;
        if (dfn[st] > dfnr[en]) return inf;
        return (deep[st] - deep[en]);
    }
    return inf;
}

bool check(ll mid) {
    ll now = 1;
    for (int i = 1; i <= n; i++) {
        while (dis(a[i], now) > mid) {
            now++;
            if (now >= m + 1) return 0;
        }
    }
    return 1;
}

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);

    cin >> t;
    for (int i = 1; i <= 1e6; ++i)
        lg2[i] = lg2[i - 1] + (1 << lg2[i - 1] == i);
    while (t--) {
        cin >> n >> m;
        for (int i = 1; i <= n; i++) cin >> a[i];
        for (int i = 1; i <= m; i++) cin >> b[i];
        fcnt = 0, dcnt = 0;
        for (int i = 1; i <= m; i++) {
            son[i].clear();
            cycfa[i] = ifcyc[i] = cycno[i] = cyclen[i] = vis[i] = deep[i] = treefa[i] = dfn[i] = dfnr[i] = 0;
        }

        for (int i = 1; i <= m; i++) {
            son[b[i]].push_back(i);
        }
        for (int i = 1; i <= m; i++) {
            if (!vis[i]) dfs(i);
        }
        for (int i = 1; i <= m; i++) {
            if (vis[i] >= 2) ifcyc[i] = 0;
        }
        for (int i = 1; i <= m; i++) {
            if (cycno[i] == 1) cyclen[i] = marklen(i);
        }

        for (int i = 1; i <= m; i++) vis[i] = 0;
        for (int i = 1; i <= m; i++) {
            if (ifcyc[i])
                dfs0(i, i);
        }
        for (int i = 1; i <= m; i++)
            if (ifcyc[i])
                for (auto j: son[i])
                    if (!ifcyc[j]) dfs_n(j, i);

        for (int i = 1; i <= m; i++) if (ifcyc[i]) treefa[i] = i;

//        for (int i = 1; i <= m; i++)
//            printf("%d : ifcyc:%lld c.no:%lld c.len:%lld  deep:%lld tfa:%lld\n", i, ifcyc[i], cycno[i],
//                   cyclen[i], deep[i], treefa[i]);
//        for (int i = 1; i <= m; i++) {
//            for (int j = 1; j <= m; j++) {
//                if (dis(i, j) >= inf) printf("-1 ");
//                else printf(" %lld ", dis(i, j));
//            }
//            printf("\n");
//        }

        ll l = 0, r = m;
        if (n < 90000000) {
            while (l < r) {
                ll mid = (l + r) >> 1;
                if (check(mid)) r = mid;
                else l = mid + 1;
            }
        }
        if (!check(l)) l = -1;
        cout << l << "\n";
    }

    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to consider edge cases and scenarios that challenge the algorithm's assumptions. The code appears to handle the transformation of the array using a complex graph-based approach, which might have issues with specific patterns in the input. Here are some steps to generate a challenging test input:

1. **Check for Cycles and Loops**: The code uses a graph to detect cycles and loops in the transformation array `b`. We can create a test case where the transformation array has a complex cycle that might not be handled correctly.

2. **Non-decreasing Sequence**: Ensure the test case is designed such that the transformation might require multiple steps or might be impossible to achieve a non-decreasing sequence.

3. **Edge Cases**: Consider edge cases where the transformation array `b` is either the identity transformation or has a single element.

4. **Large Input Size**: Although we need a small-scale test input, we can still create a test case that is large enough to challenge the algorithm but not too large to exceed the output limit.

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

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

### Explanation:
1. **First Test Case**:
   - Initial array: `[5, 4, 3, 2, 1]`
   - Transformation array: `[2, 3, 4, 5, 1]`
   - This case is challenging because the initial array is in reverse order, and the transformation array has a cycle. The algorithm should correctly determine the minimum number of tricks needed to make the array non-decreasing.

2. **Second Test Case**:
   - Initial array: `[1, 2, 3, 4, 5]`
   - Transformation array: `[1, 2, 3, 4, 5]`
   - This case is simple and should return `0` because the initial array is already non-decreasing and the transformation array is the identity transformation.

3. **Third Test Case**:
   - Initial array: `[1, 2, 3, 4, 5]`
   - Transformation array: `[2, 3, 4, 5, 1]`
   - This case is challenging because the transformation array has a cycle, and the initial array is already non-decreasing. The algorithm should correctly determine the minimum number of tricks needed, which should be `1`.

This test input should help expose any bugs in the code by challenging its ability to handle cycles and transformations correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Way to be Non-decreasing
time_limit_ms: 4000
memory_limit_mb: 512
Description: Little R is a magician who likes non-decreasing arrays. She has an array of length $n$, initially as $a_1, \ldots, a_n$, in which each element is an integer between $[1, m]$. She wants it to be non-decreasing, i.e., $a_1 \leq a_2 \leq \ldots \leq a_n$.

To do this, she can perform several magic tricks. Little R has a fixed array $b_1\ldots b_m$ of length $m$. Formally, let's define a trick as a procedure that does the following things in order:

  * Choose a set $S \subseteq \\{1, 2, \ldots, n\\}$.    * For each $u \in S$, assign $a_u$ with $b_{a_u}$.

Little R wonders how many tricks are needed at least to make the initial array non-decreasing. If it is not possible with any amount of tricks, print $-1$ instead.

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

The first line of each test case contains two integers $n$ and $m$ ($1\leq n \leq 10^6$, $1 \leq m \leq 10^6$) — the length of the initial array and the range of the elements in the array.

The second line of each test case contains $n$ integers $a_1, \ldots, a_n$ ($1 \leq a_i \leq m$) — the initial array.

The third line of each test case contains $m$ integers $b_1, \ldots, b_m$ ($1 \leq b_i \leq m$) — the fixed magic array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$ and the sum of $m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the minimum number of tricks needed, or $-1$ if it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Examples:
input:
3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
output:
3
-1
3

Note: In the first case, the initial array $a_1, \ldots, a_n$ is $[1, 6, 3, 7, 1]$. You can choose $S$ as follows:

  * first trick: $S = [2, 4, 5]$, $a = [1, 1, 3, 5, 2]$;    * second trick: $S = [5]$, $a = [1, 1, 3, 5, 3]$;    * third trick: $S = [5]$, $a = [1, 1, 3, 5, 5]$.

So it is possible to make $a_1, \ldots, a_n$ non-decreasing using $3$ tricks. It can be shown that this is the minimum possible amount of tricks.

In the second case, it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Buggy Code:
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <cmath>

using namespace std;

typedef long long lld;

const int M = 2e6 + 10;
const int N = 2e6 + 10;
const int bitL = 22;
int head[N];
int to[M * 2], nextEdge[M * 2];
int len;
int dfn[N], sz[N];
int h[N];

void initPara(int n)
{
	len = 0;
	for (int i = 0; i < n; i++)
	{
		head[i] = -1;
	}
}

void add(int a, int b)
{
	to[len] = b;
	nextEdge[len] = head[a];
	head[a] = len++;
	//cout << a << "->" << b << endl;
}

int dfnnum = 0;

int dfs(int x, int fa)
{
	dfn[x] = dfnnum++;
	sz[x] = 1;
	if (fa == -1)h[x] = 0;
	else h[x] = h[fa] + 1;
	int i;
	for (i = head[x]; i != -1; i = nextEdge[i])
	{
		int j = to[i];
		if (fa == j)continue;
		sz[x] += dfs(j, x);
	}

	return sz[x];
}

int arra[N];
int arrb[N];
int ind[M];
int indlen;

int getInd(int i, int j) {
	if (ind[i * 2 + j] == -1) {
		ind[i * 2 + j] = indlen++;
	}
	return ind[i * 2 + j];
}
// dfs ��ʱ����
int temp[N];
int tlen;

bool vis[M];

int dfs(int i) {
	if (vis[i])return i;
	temp[tlen++] = i;
	vis[i] = true;

	auto k = dfs(arrb[i]);
	return k;
}

bool reachable(int u, int v, int num) {
	if (u == v)return true;

	int uu = getInd(u, 0);
	int vv = getInd(v, 0);
	if (dfn[vv] < dfn[uu] && dfn[uu] < dfn[vv] + sz[vv] && h[uu] - h[vv] + 1 <= num)return true;

	//  �����ڻ��ϣ��п���Ҫ�绷
	vv = getInd(v, 1);
	if (vv > -1) {
		if (dfn[vv] < dfn[uu] && dfn[uu] < dfn[vv] + sz[vv] && h[uu] - h[vv] + 1 <= num)return true;
	}
	return false;
}

bool check(int num, int n, int m) {
	int minv = 0;
	for (int i = 0; i < n; ++i) {
		while (minv < m && !reachable(arra[i], minv, num))minv++;
		if (minv >= m)return false;
	}

	return true;
}

int roots[M];
int rlen;

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

		// ��ʼ������
		dfnnum = 0;
		indlen = 0;
		rlen = 0;
		for (int i = 0; i < m; ++i) {
			ind[i * 2] = -1;
			ind[i * 2 + 1] = -1;
			vis[i] = 0;
		}
		initPara(m * 2);
		for (int i = 0; i < n; ++i) {
			cin >> arra[i];
			--arra[i];
		}
		for (int i = 0; i < m; ++i) {
			cin >> arrb[i];
			--arrb[i];
		}

		for (int i = 0; i < m; ++i) {
			if (vis[i])continue;
			tlen = 0;
			auto k = dfs(i);
			int j = 0;
			int beforid = -1;

			// �������
			for (; j < tlen && temp[j] != k; ++j) {
				auto id = getInd(temp[j], 0);
				if (beforid >= 0)add(id, beforid);
				beforid = id;
			}
			for (int s = j; s < tlen; s++) {
				auto id = getInd(temp[s], 0);
				if (beforid >= 0)add(id, beforid);
				beforid = id;
			}
			// ����ѭ��
			for (int s = j; s < tlen; s++) {
				auto id = getInd(temp[s], 1);
				if (beforid >= 0)add(id, beforid);
				beforid = id;
			}

			// û��ѭ�����һ��Ҫ���Ѿ���������ϲ�
			if (j == tlen) {
				auto id = getInd(k, 0);
				if (beforid >= 0)add(id, beforid);
				beforid = id;
			}
			else {
				roots[rlen++] =getInd(temp[tlen - 1], 1);
			}
		}

		/*for (int i = 0; i < m; ++i) {
			cout << i << "'s ids: ";
			auto id = ind[2 * i];
			cout << id << " ";
			id = ind[2 * i+1];
			if(id>-1)cout << id << " ";
			cout << endl;
		}*/

		// ��lca
		for (int i = 0; i < rlen; ++i) {
			//cout << "root: " << roots[i] << endl;
			dfs(roots[i], -1);
		}

		/*if (n == 1000000 && m == 1000000) {
			cout << N << endl;
			continue;
		}*/
		// cout << father[0][getInd(3, 1)] << endl;
		//cout << lca(getInd(0, 0), getInd(1, 0)) << endl;
		/*cout << reachable(7, 3, 3) << endl;

		continue;*/
		//cout << check(0 + 1, n, m) << endl;
		int l = 0, r = m + 2;
		int best = -1;
		while (l <= r) {
			int mid = (l + r) / 2;

			if (!check(mid + 1, n, m)) {
				l = mid + 1;
			}
			else {
				best = mid;
				r = mid-1;
			}
		}
		cout << best << endl;
	}

}

int main() {
	solve();
	return 0;
}
/*

3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5

1
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6

1
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5

1
10 10
5 3 9 5 9 10 10 10 9 10
6 1 3 1 4 6 8 10 4 7

1
3 3
1 2 3
1 2 3

1
5 5
1 2 3 4 5
5 1 2 3 4
1
5 5
5 4 3 2 1
5 1 2 3 4

1
5 5
5 4 3 2 1
5 4 1 2 3
 */

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Way to be Non-decreasing
time_limit_ms: 4000
memory_limit_mb: 512
Description: Little R is a magician who likes non-decreasing arrays. She has an array of length $n$, initially as $a_1, \ldots, a_n$, in which each element is an integer between $[1, m]$. She wants it to be non-decreasing, i.e., $a_1 \leq a_2 \leq \ldots \leq a_n$.

To do this, she can perform several magic tricks. Little R has a fixed array $b_1\ldots b_m$ of length $m$. Formally, let's define a trick as a procedure that does the following things in order:

  * Choose a set $S \subseteq \\{1, 2, \ldots, n\\}$.    * For each $u \in S$, assign $a_u$ with $b_{a_u}$.

Little R wonders how many tricks are needed at least to make the initial array non-decreasing. If it is not possible with any amount of tricks, print $-1$ instead.

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

The first line of each test case contains two integers $n$ and $m$ ($1\leq n \leq 10^6$, $1 \leq m \leq 10^6$) — the length of the initial array and the range of the elements in the array.

The second line of each test case contains $n$ integers $a_1, \ldots, a_n$ ($1 \leq a_i \leq m$) — the initial array.

The third line of each test case contains $m$ integers $b_1, \ldots, b_m$ ($1 \leq b_i \leq m$) — the fixed magic array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$ and the sum of $m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the minimum number of tricks needed, or $-1$ if it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Examples:
input:
3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
output:
3
-1
3

Note: In the first case, the initial array $a_1, \ldots, a_n$ is $[1, 6, 3, 7, 1]$. You can choose $S$ as follows:

  * first trick: $S = [2, 4, 5]$, $a = [1, 1, 3, 5, 2]$;    * second trick: $S = [5]$, $a = [1, 1, 3, 5, 3]$;    * third trick: $S = [5]$, $a = [1, 1, 3, 5, 5]$.

So it is possible to make $a_1, \ldots, a_n$ non-decreasing using $3$ tricks. It can be shown that this is the minimum possible amount of tricks.

In the second case, it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

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

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

	vector<int> a(n);
	vector<int> b(m);

	for(int i = 0;i < n;i++){
		cin >> a[i], --a[i];
	}

	for(int i = 0;i < m;i++){
		cin >> b[i], --b[i];
	}

	// i -> b[i]
	vector<bool> in_cycle(m);
	vector<int> id_cycle(m);
	vector<int> siz_cycle(m);
	vector<int> id(m);
	vector<int> vis(m);
	int cur_id = 0;

	for(int i = 0;i < m;i++){
		if(vis[i] != 0) continue;
		int u = i;
		while(!vis[u]){
			vis[u] = i + 1;
			u = b[u];
		}
		if(vis[u] != i + 1) continue; // already processed.
		vector<int> c;
		int x = u;
		cur_id += 1;
		do{
			c.push_back(x);
			in_cycle[x] = true;
			id_cycle[x] = cur_id;
			x = b[x];
		} while(x != u);
		int siz = (int)c.size();
		for(int j = 1;j <= (int)c.size();j++){
			int v = c[j - 1];
			id[v] = j;
			siz_cycle[v] = siz;
		}
	}

	vector<int> my_root(m);
	vector<vector<int>> g(m);
	for(int i = 0;i < m;i++){
		if(in_cycle[i]) continue;
		int par = b[i];
		g[par].push_back(i);
		if(in_cycle[par]) my_root[par] = par;
	}

	vector<int> tin(m), tout(m), dep(m);
	int T = 0;
	auto dfs = [&](auto &&dfs, int u, int par)->void{
		tin[u] = T++;
		for(int v : g[u]){ // no - parent check
			my_root[v] = my_root[u];
			dep[v] = dep[u] + 1;
			dfs(dfs, v, u);
		}
		tout[u] = T++;
	};

	for(int i = 0;i < m;i++){
		if(in_cycle[i]){
			dfs(dfs, i, i);
		}
	}

	const int inf = (int)1e9;

	auto dCycle = [&](int from, int to)->int{
		assert(id_cycle[from] == id_cycle[to]);
		if(id[from] <= id[to]) return id[to] - id[from];
		return siz_cycle[from] - id[to] + id[from];
	};

	auto isAnc = [&](int a, int b)->bool{
		return tin[a] <= tin[b] and tout[b] <= tout[a];
	};

	auto Moves = [&](int from, int to)->int{ // boild down to distance in functional - graph.
		if(from == to) return 0;
		if(in_cycle[from] and id_cycle[from] != id_cycle[to]) return inf;
		if(in_cycle[from] and in_cycle[to]){
			return dCycle(from, to);
		}
		if(in_cycle[to]){
			int root = my_root[from];
			assert(in_cycle[root]);
			int dist = dep[from];
			if(id_cycle[root] != id_cycle[to]){
				return inf;
			}
			return dist + dCycle(root, to);
		}
		if(dep[to] >= dep[from]) return inf;
		if(isAnc(to, from)){
			return dep[from] - dep[to];
		}
		return inf;
	};

	auto ok = [&](int cap)->bool{
		int num = 0, ind = 0;
		while(ind < n){
				bool found = false;
			while(num < m){
				if(Moves(a[ind], num) <= cap){
					ind += 1;
					found = true;
					break;
				} else {
					num += 1;
				}
			}
			if(!found) break;
		}
		return ind == n;
	};

	int lo = 0, hi = m - 1, mid, ans = -1;
	while(lo <= hi){
		mid = lo + (hi - lo) / 2;
		if(ok(mid)){
			ans = mid;
			hi = mid - 1;
		} else {
			lo = mid + 1;
		}
	}

	cout << ans << '\n';
}

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

	int T = 1;
	cin >> T;

	while(T--){
		solve();
	}

	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Way to be Non-decreasing
time_limit_ms: 4000
memory_limit_mb: 512
Description: Little R is a magician who likes non-decreasing arrays. She has an array of length $n$, initially as $a_1, \ldots, a_n$, in which each element is an integer between $[1, m]$. She wants it to be non-decreasing, i.e., $a_1 \leq a_2 \leq \ldots \leq a_n$.

To do this, she can perform several magic tricks. Little R has a fixed array $b_1\ldots b_m$ of length $m$. Formally, let's define a trick as a procedure that does the following things in order:

  * Choose a set $S \subseteq \\{1, 2, \ldots, n\\}$.    * For each $u \in S$, assign $a_u$ with $b_{a_u}$.

Little R wonders how many tricks are needed at least to make the initial array non-decreasing. If it is not possible with any amount of tricks, print $-1$ instead.

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

The first line of each test case contains two integers $n$ and $m$ ($1\leq n \leq 10^6$, $1 \leq m \leq 10^6$) — the length of the initial array and the range of the elements in the array.

The second line of each test case contains $n$ integers $a_1, \ldots, a_n$ ($1 \leq a_i \leq m$) — the initial array.

The third line of each test case contains $m$ integers $b_1, \ldots, b_m$ ($1 \leq b_i \leq m$) — the fixed magic array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$ and the sum of $m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the minimum number of tricks needed, or $-1$ if it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Examples:
input:
3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
output:
3
-1
3

Note: In the first case, the initial array $a_1, \ldots, a_n$ is $[1, 6, 3, 7, 1]$. You can choose $S$ as follows:

  * first trick: $S = [2, 4, 5]$, $a = [1, 1, 3, 5, 2]$;    * second trick: $S = [5]$, $a = [1, 1, 3, 5, 3]$;    * third trick: $S = [5]$, $a = [1, 1, 3, 5, 5]$.

So it is possible to make $a_1, \ldots, a_n$ non-decreasing using $3$ tricks. It can be shown that this is the minimum possible amount of tricks.

In the second case, it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
inline long long read(){
	long long x=0; char ch; bool f=0;
	while(((ch=getchar())<'0'||ch>'9')&&ch!='-') ;
	if(ch=='-') f=1;
	else x=ch^48;
	while((ch=getchar())>='0'&&ch<='9') x=(x<<1)+(x<<3)+(ch^48);
	return f?-x:x;
}
bool st2;
const int N=1e6+5;
int n,m,head[N],tot;
struct edge{
	int nex,y;
}e[N];
void add(int x,int y){
	e[++tot]={head[x],y}; head[x]=tot;
}
int a[N],b[N],deg[N];
queue<int> q;
vector<int> s[N],h[N];
int sy[N],d[N],cnt,fa[22][N],c2,ps[N];
vector<int> v;
bool vis[N];
void dfs(int x){
	vis[x]=1,v.push_back(x);
	for(auto y:s[x]) if(!vis[y]) dfs(y);
}
int root[N];
struct tree{
	int ls,rs,sum;
}t[N*22];
void ins(int &p,int q,int l,int r,int x){
	p=++cnt,t[p]=t[q];
	t[p].sum++;
	if(l==r) return;
	int mid=(l+r)/2;
	if(x<=mid) ins(t[p].ls,t[q].ls,l,mid,x);
	else ins(t[p].rs,t[q].rs,mid+1,r,x);
}
int ask(int p,int q,int w,int l,int r){
	if(r<w||!(t[q].sum-t[p].sum)) return 1e9;
	if(l==r) return l;
	int mid=(l+r)/2;
	return min(ask(t[p].ls,t[q].ls,w,l,mid),ask(t[p].rs,t[q].rs,w,mid+1,r));
}
void dfs2(int x,int f){
	ps[x]=h[c2].size()-1,sy[x]=c2;
	if(!deg[x]){
		d[x]=d[f]+1;
		fa[0][x]=f;
		for(int i=1;i<=20;i++) fa[i][x]=fa[i-1][fa[i-1][x]];
	}
	for(int i=head[x];i;i=e[i].nex){
		int y=e[i].y;
		if(!deg[y]){
			ins(root[y],root[x],1,m,y);
			dfs2(y,x);
		}
	}
}
int lg[(1<<21)+5];
void sv(int rt){
	v.clear();
	dfs(rt);
	c2++;
	h[c2].clear();
	for(auto x:v){
		if(deg[x]){
			h[c2].push_back(x),dfs2(x,0);
			if(h[c2].size()==1) ins(root[x],root[0],1,m,x);
			else ins(root[x],root[h[c2][h[c2].size()-2]],1,m,x);
		}
	}
}
int get(int x,int k,int w){
	int tx=x; k++;
	if(d[x]>=k){
		for(int i=k;i;i^=i&-i) x=fa[lg[i&-i]][x];
		if(d[tx]==k) return ask(0,root[tx],w,1,m);
		else return ask(root[x],root[tx],w,1,m);
	}k-=d[x];
	int t=1e9;
//	cout<<x<<": "<<d[x]<<"??\n";
	if(d[x]) t=ask(0,root[tx],w,1,m);
	int sy=::sy[x];
	x=ps[x];
	if(k>=h[sy].size()){
		return min(t,ask(0,root[h[sy][h[sy].size()-1]],w,1,m));
	}
	if(x+k-1<h[sy].size()) return min(t,ask(!x?0:root[h[sy][x-1]],root[h[sy][x+k-1]],w,1,m));
	else{
		return min(min(t,ask(0,root[h[sy][x+k-1-h[sy].size()]],w,1,m)),ask(!x?0:root[h[sy][x-1]],root[h[sy][h[sy].size()-1]],w,1,m));
	}
}
bool pd(int md){
//	cout<<md<<": ------";
	int lst=get(a[1],md,0);
	for(int i=2;i<=n;i++){
		int x=get(a[i],md,lst);
		if(x>m) return 0;
//		cout<<i<<": "<<x<<"??\n";
		lst=x;
	}
	return 1;
}
void solve(){
	n=read(),m=read();
	tot=cnt=c2=0;
	for(int i=1;i<=m;i++) head[i]=deg[i]=d[i]=0,s[i].clear(),vis[i]=ps[i]=sy[i]=root[i]=0;
	for(int i=1;i<=m;i++) s[i].clear();
	for(int i=1;i<=n;i++) a[i]=read();
	for(int i=1;i<=m;i++) b[i]=read(),add(b[i],i),deg[b[i]]++,s[i].push_back(b[i]),s[b[i]].push_back(i);
	for(int i=1;i<=m;i++) if(!deg[i]) q.push(i);
	while(q.size()){
		int x=q.front(); q.pop();
		if(!(--deg[b[x]])) q.push(b[x]);
	}
//	for(int i=1;i<=m;i++) cerr<<deg[i]<<" "; puts("");
//	cerr<<2<<"??\n";
	for(int i=1;i<=m;i++){
		if(!vis[i]){
			sv(i);
		}
	}
//	while(1){
//		int x=read(),k=read(),lst=read();
//		cout<<get(x,k,lst)<<"??\n";
//	}
//	cerr<<1<<"??\n";
	int l=0,r=m+5;
	while(l<=r){
		int mid=(l+r)/2;
		if(pd(mid)) r=mid-1;
		else l=mid+1;
	}
//	cout<<pd(2)<<" ";
	if(l>m) puts("-1");
	else printf("%d\n",l);
}
bool st1;
int main(){
//	cout<<(&st1-&st2)/1024/1024<<"\n";
	for(int i=2;i<=(1<<21);i++) lg[i]=lg[i>>1]+1;
//	freopen("t1.out","w",stdout);
	int T=read();
	while(T--) solve();
	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Way to be Non-decreasing
time_limit_ms: 4000
memory_limit_mb: 512
Description: Little R is a magician who likes non-decreasing arrays. She has an array of length $n$, initially as $a_1, \ldots, a_n$, in which each element is an integer between $[1, m]$. She wants it to be non-decreasing, i.e., $a_1 \leq a_2 \leq \ldots \leq a_n$.

To do this, she can perform several magic tricks. Little R has a fixed array $b_1\ldots b_m$ of length $m$. Formally, let's define a trick as a procedure that does the following things in order:

  * Choose a set $S \subseteq \\{1, 2, \ldots, n\\}$.    * For each $u \in S$, assign $a_u$ with $b_{a_u}$.

Little R wonders how many tricks are needed at least to make the initial array non-decreasing. If it is not possible with any amount of tricks, print $-1$ instead.

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

The first line of each test case contains two integers $n$ and $m$ ($1\leq n \leq 10^6$, $1 \leq m \leq 10^6$) — the length of the initial array and the range of the elements in the array.

The second line of each test case contains $n$ integers $a_1, \ldots, a_n$ ($1 \leq a_i \leq m$) — the initial array.

The third line of each test case contains $m$ integers $b_1, \ldots, b_m$ ($1 \leq b_i \leq m$) — the fixed magic array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$ and the sum of $m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the minimum number of tricks needed, or $-1$ if it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Examples:
input:
3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
output:
3
-1
3

Note: In the first case, the initial array $a_1, \ldots, a_n$ is $[1, 6, 3, 7, 1]$. You can choose $S$ as follows:

  * first trick: $S = [2, 4, 5]$, $a = [1, 1, 3, 5, 2]$;    * second trick: $S = [5]$, $a = [1, 1, 3, 5, 3]$;    * third trick: $S = [5]$, $a = [1, 1, 3, 5, 5]$.

So it is possible to make $a_1, \ldots, a_n$ non-decreasing using $3$ tricks. It can be shown that this is the minimum possible amount of tricks.

In the second case, it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

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

template<class T> bool cmin(T &i,T j) { return i > j ? i = j,true: false; }
template<class T> bool cmax(T &i,T j) { return i < j ? i = j,true: false; }

constexpr int nax = 1e6+69;
int a[nax],b[nax],n,m;

bool can(int mid) {
    int mn = 0;
    for (int i=1;i<=n;i++) {
        int &x = a[i];
        int c = mid, nmn = (x >= mn ? x : nax), p = x;
        while(c) {
            p = b[p];
            if (p >= mn) {
                cmin(nmn,p);
            }
            c--;
        }
        if (nmn == nax) {
            return false;
        }
        mn = nmn;
    }
    return true;
}

// for the successor graph: say the node was in the cycle. Then I want to process the cycle alone.

bool in_cycle[nax];
int component_sz[nax];
bool vst[nax];
int time_visit_cycle[nax];

int t[nax];
int in[nax];
int where[nax];
int nodes[nax];
int go_cycle[nax];

vector<int> st[4*nax];

void build(int u=1,int tl=1,int tr=nax) {
    st[u].clear();
    if (tl == tr) {
        st[u] = vector<int>(1,nodes[tl]);
        return;
    }
    int mid=(tl+tr)/2;
    build(2*u,tl,mid);
    build(2*u+1,mid+1,tr);
    merge(st[2*u].begin(),st[2*u].end(),st[2*u+1].begin(),st[2*u+1].end(),back_inserter(st[u]));
}

void clear() {
    for (int i=1;i<=m;i++) {
        in_cycle[i] = component_sz[i] = vst[i] = time_visit_cycle[i] = t[i] = in[i] = where[i] = nodes[i] = go_cycle[i] = 0;
    }
}

int qry2(int l,int len,int x) {
    int res = nax;
    int r = l+len-1;
    for (;l<=r;l++) {
        if (nodes[l] >= x) {
            cmin(res,nodes[l]);
        }
    }
    return res;
}

int qry(int l,int r,int x,int u,int tl,int tr) {
    if (l > r) { return nax; }
    if (l == tl && r == tr) {
        auto it = lower_bound(st[u].begin(),st[u].end(),x);
        if (it == st[u].end()) { return nax; }
        return *it;
    }
    int mid=(tl+tr)/2;
    int R=qry(l,min(mid,r),x,2*u,tl,mid);
    if (R != nax) { return R; }
    return qry(max(mid+1,l),r,x,2*u+1,mid+1,tr);
}

int qry(int l,int len,int x) {
    return qry(l,l+len-1,x,1,1,nax);
}

int get_line(int x,int len,int mn) {
    cmin(len,component_sz[x]-t[x]);
    return qry(where[x],len,mn);
}

int get_cycle(int x,int len,int mn) {
    if (t[x] + len > component_sz[x]) {
        int rem = min(component_sz[x],t[x] + len - component_sz[x]);
        int r = min(qry(where[x],component_sz[x]-t[x],mn),qry(where[x]-t[x],rem,mn));
        /*
        cout << "x = " << x << endl;
        cout << "len = " << len << endl;
        cout << "mn = " << mn << endl;
        cout << "res = " << r << endl;
        */
        return r;
    } else {
        return qry(where[x],len,mn);
    }
}

bool can2(int len) {
    len++;
    int mn = 0;
    for (int i=1;i<=n;i++) {
        int &x = a[i];
        int nmn = nax;
        if (in_cycle[x]) {
            cmin(nmn,get_cycle(x,len,mn));
        } else {
            if (t[x] + len > component_sz[x]) {
                int rem = t[x] + len - component_sz[x];
                cmin(nmn,get_line(x,component_sz[x]-t[x],mn));
                cmin(nmn,get_cycle(go_cycle[x],rem,mn));
            } else {
                cmin(nmn,get_line(x,len,mn));
            }
        }
        if (nmn == nax) {
            return false;
        }
        assert(nmn >= mn);
        mn = nmn;
    }
    return true;
}

void solve() {
    cin >> n >> m;
    for (int i=1;i<=n;i++) {
        cin >> a[i];
    }
    for (int i=1;i<=m;i++) {
        cin >> b[i];
    }

    // first step: get cycles

    int p = 1;

    for (int i=1;i<=m;i++) {
        if (vst[i]) { continue; }
        int x = i;
        while(!vst[x]) {
            vst[x] = true;
            time_visit_cycle[x]=i;
            x = b[x];
        }
        assert(time_visit_cycle[x]);
        if (time_visit_cycle[x] != i) {
            // I have already processed this part
            continue;
        }

        // cycle found!
        int y = x;
        int c = 0;
        do {
            t[y] = c++;
            in_cycle[y] = true;
            nodes[p] = y;
            where[y] = p++;
            y = b[y];
        } while(y != x);
        y = x;
        do {
            component_sz[y] = c;
            y = b[y];
        } while(y != x);
    }

    for (int i=1;i<=m;i++) {
        in[b[i]]++;
    }

    for (int i=1;i<=m;i++) {
        if (in[i]!=0) { continue; }
        int x = i, c = 0;
        while(!in_cycle[x]) {
            nodes[p] = x;
            where[x] = p++;
            t[x] = c++;
            x = b[x];
        }
        int y = x;
        x = i;
        while(!in_cycle[x]) {
            component_sz[x] = c;
            go_cycle[x] = y;
            x = b[x];
        }
    }
    /*
    for (int i=1;i<=m;i++) {
        cout << "node: " << i << endl;
        cout << "in_cycle: " << (in_cycle[i] ? "yes" : "no") << endl;
        cout << "t: " << t[i] << endl;
        cout << "component_sz: " << component_sz[i] << endl;
        cout << "where: " << where[i] << endl;
    }
    */
    // main task: do range-smallest element >= x on successor graph

    build();

    int lo = 0, hi = m;
    while(lo < hi) {
        int mid = lo+(hi-lo)/2;
        if (!can2(mid)) {
            lo = mid+1;
        } else {
            hi = mid;
        }
    }
    cout << (lo == m ? -1: lo) << "\n";
    clear();
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int _;
    cin >> _;
    while(_--) {
        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 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Way to be Non-decreasing
time_limit_ms: 4000
memory_limit_mb: 512
Description: Little R is a magician who likes non-decreasing arrays. She has an array of length $n$, initially as $a_1, \ldots, a_n$, in which each element is an integer between $[1, m]$. She wants it to be non-decreasing, i.e., $a_1 \leq a_2 \leq \ldots \leq a_n$.

To do this, she can perform several magic tricks. Little R has a fixed array $b_1\ldots b_m$ of length $m$. Formally, let's define a trick as a procedure that does the following things in order:

  * Choose a set $S \subseteq \\{1, 2, \ldots, n\\}$.    * For each $u \in S$, assign $a_u$ with $b_{a_u}$.

Little R wonders how many tricks are needed at least to make the initial array non-decreasing. If it is not possible with any amount of tricks, print $-1$ instead.

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

The first line of each test case contains two integers $n$ and $m$ ($1\leq n \leq 10^6$, $1 \leq m \leq 10^6$) — the length of the initial array and the range of the elements in the array.

The second line of each test case contains $n$ integers $a_1, \ldots, a_n$ ($1 \leq a_i \leq m$) — the initial array.

The third line of each test case contains $m$ integers $b_1, \ldots, b_m$ ($1 \leq b_i \leq m$) — the fixed magic array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$ and the sum of $m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the minimum number of tricks needed, or $-1$ if it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Examples:
input:
3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
output:
3
-1
3

Note: In the first case, the initial array $a_1, \ldots, a_n$ is $[1, 6, 3, 7, 1]$. You can choose $S$ as follows:

  * first trick: $S = [2, 4, 5]$, $a = [1, 1, 3, 5, 2]$;    * second trick: $S = [5]$, $a = [1, 1, 3, 5, 3]$;    * third trick: $S = [5]$, $a = [1, 1, 3, 5, 5]$.

So it is possible to make $a_1, \ldots, a_n$ non-decreasing using $3$ tricks. It can be shown that this is the minimum possible amount of tricks.

In the second case, it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

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

#define _FAST       ios_base::sync_with_stdio(false); cin.tie(NULL);
#define PList(V)    {for(auto x:(V)) cout<<x<<" "; cout<<endl;}
#define read(V)     {for(auto &x:(V)) cin>>x;}
#define all(V)      (V).begin(),(V).end()
#define rall(V)     (V).rbegin(),(V).rend()
#define sz(V)       int(V.size())
#define fi          first
#define se          second
#define p_b         push_back
#define int         int64_t
#define double      long double

#define MOD         1000000007
#define INF         500000000000000000
#define M_PI        3.14159265358979323846
#define ZERO        (int)0
#define ONE         (int)1

typedef vector<int>   vi;
typedef pair<int,int> ii;
typedef vector<ii>   vii;
typedef vector<vi>   vvi;
typedef vector<vii>   vvii;

inline int add(int a, int b){ a%=MOD; b%=MOD; return (a+b)%MOD; }
inline int sub(int a, int b){ a%=MOD; return (a-b+MOD)%MOD; }
inline int mul(int a, int b){ a%=MOD; b%=MOD; return (a*b)%MOD; }
int gcd(int a, int b){ return b ? gcd(b, a%b) : a; }
int log2(int N){ int ans = 0; for(int p=1; p<=N; p<<=1) ans = p; return ans; }
vi unique(vi &A){ set<int> st(all(A)); vi ans(all(st)); return ans; }

int bpow(int A, int p){
    if(p==0) return 1;
    int tmp = bpow(A, p/2);
    return mul( mul(tmp, tmp), (p%2 ? A : 1) );
}

int inv(int N){ return bpow(N, MOD-2); }
int nCr(int N, int R){ int ans=1; for(int r=1, n=N; r<=R; r++, n--) ans = mul(ans, mul(n , inv(r))); return ans; }

void solve() {

    int N, M; cin>>N>>M;
    vi A(N), B(M); read(A); read(B);
    for(auto &x : A) x--;
    for(auto &x : B) x--;

    std::vector<int> vis(M, -1);
    std::vector<int> C_pos(M, -1), branch(M), C(M), C_len(M);
    for (int i = 0; i < M; i++) {
        int s = i;
        while (vis[s] == -1) {
            vis[s] = i;
            s = B[s];
        }
        if (vis[s] == i) {
            int cur = 0;
            int u = s;
            do {
                C[u] = s;
                C_pos[u] = cur++;
                branch[u] = u;

                u = B[u];
            } while (u != s);
            C_len[s] = cur;
        }
    }

    std::vector<std::vector<int>> adj(M);
    for (int u = 0; u < M; u++) {
        adj[B[u]].push_back(u);
    }

    int cur = 0;
    std::vector<int> in(M), out(M), depth(M);
    auto dfs = [&](auto self, int u) -> void {
        in[u] = cur++;
        for (auto v : adj[u]) {
            if (C_pos[v] != -1)  continue;

            C[v] = C[u];
            branch[v] = branch[u];
            depth[v] = depth[u] + 1;
            self(self, v);
        }
        out[u] = cur;
    };

    for (int u = 0; u < M; u++) if (C_pos[u] != -1) dfs(dfs, u);

    auto reachable = [&](int u, int v, int moves) {
        if (C[u] != C[v]) {
            return false;
        }
        if (in[v] <= in[u] && in[u] < out[v] && depth[u] - depth[v] <= moves) {
            return true;
        }
        if (C_pos[v] == -1) {
            return false;
        }
        int r = branch[u];
        int dis = depth[u];
        int l = C_len[C[v]];
        dis += (C_pos[v] - C_pos[r] + l) % l;
        return dis <= moves;
    };

    auto check = [&](int k) {
        for (int i = 0, j = 0; i < N; i++) {
            while (j < M && !reachable(A[i], j, k)) {
                j++;
            }
            if (j == M) {
                return false;
            }
        }
        return true;
    };

    int lo = 0, hi = M;
    while (lo < hi) {
        int x = (lo + hi) / 2;
        if (check(x)) {
            hi = x;
        } else {
            lo = x + 1;
        }
    }
    int ans = lo;
    if (ans == M) {
        ans = -1;
    }
    std::cout << ans << "\n";
}

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

    int t;
    std::cin >> t;

    while (t--) {
        solve();
    }

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Way to be Non-decreasing
time_limit_ms: 4000
memory_limit_mb: 512
Description: Little R is a magician who likes non-decreasing arrays. She has an array of length $n$, initially as $a_1, \ldots, a_n$, in which each element is an integer between $[1, m]$. She wants it to be non-decreasing, i.e., $a_1 \leq a_2 \leq \ldots \leq a_n$.

To do this, she can perform several magic tricks. Little R has a fixed array $b_1\ldots b_m$ of length $m$. Formally, let's define a trick as a procedure that does the following things in order:

  * Choose a set $S \subseteq \\{1, 2, \ldots, n\\}$.    * For each $u \in S$, assign $a_u$ with $b_{a_u}$.

Little R wonders how many tricks are needed at least to make the initial array non-decreasing. If it is not possible with any amount of tricks, print $-1$ instead.

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

The first line of each test case contains two integers $n$ and $m$ ($1\leq n \leq 10^6$, $1 \leq m \leq 10^6$) — the length of the initial array and the range of the elements in the array.

The second line of each test case contains $n$ integers $a_1, \ldots, a_n$ ($1 \leq a_i \leq m$) — the initial array.

The third line of each test case contains $m$ integers $b_1, \ldots, b_m$ ($1 \leq b_i \leq m$) — the fixed magic array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$ and the sum of $m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the minimum number of tricks needed, or $-1$ if it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Examples:
input:
3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
output:
3
-1
3

Note: In the first case, the initial array $a_1, \ldots, a_n$ is $[1, 6, 3, 7, 1]$. You can choose $S$ as follows:

  * first trick: $S = [2, 4, 5]$, $a = [1, 1, 3, 5, 2]$;    * second trick: $S = [5]$, $a = [1, 1, 3, 5, 3]$;    * third trick: $S = [5]$, $a = [1, 1, 3, 5, 5]$.

So it is possible to make $a_1, \ldots, a_n$ non-decreasing using $3$ tricks. It can be shown that this is the minimum possible amount of tricks.

In the second case, it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Buggy Code:
#include<iostream>
#include<iomanip>
#include<cmath>
#include<string>
#include<algorithm>
#include<map>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<deque>
#include<set>

#pragma GCC optimize("Ofast", "inline", "-ffast-math")
#pragma GCC target("avx,sse2,sse3,sse4,mmx")

using namespace std;
#define ll long long

const int N = 1e6 + 100;
const ll inf = 1e8;

ll t, n, m, a[N], b[N];
ll fcnt, dcnt, dfn[N], dfnr[N], cycfa[N], ifcyc[N], cycno[N], cyclen[N], vis[N], deep[N], treefa[N];
vector<ll> son[N];

ll marklen(ll p) {
    if (cycno[p] >= cycno[b[p]]) {
        cyclen[p] = cycno[p];
        return cycno[p];
    }
    cyclen[p] = marklen(b[p]);
    return cyclen[p];
}

void markcyc(ll p) {
    ll no = 0;
    while (!ifcyc[p]) {
        cycfa[p] = fcnt;
        ifcyc[p] = 1;
        cycno[p] = ++no;
        p = b[p];
    }
}

void dfs(ll p) {
    if (vis[p] == 1) {
        fcnt++;
        markcyc(p);
        return;
    }
    vis[p] += 1;
    if (vis[b[p]] <= 1) dfs(b[p]);
    if (!ifcyc[p]) vis[p] = 2;
}

void dfs0(ll p, ll t_fa) {
    vis[p] = 1;
    treefa[p] = t_fa;
    for (auto i: son[p]) {
        if (!ifcyc[i] && !vis[i]) {
            deep[i] = deep[p] + 1;
            dfs0(i, t_fa);
        }
    }
}

void dfs_n(ll p, ll father) {
    dfn[p] = ++dcnt;
    for (auto i: son[p])
        if (i != father && !ifcyc[i]) dfs_n(i, p);
    dfnr[p] = dcnt;
}

inline ll dis(ll st, ll en) {
    if (st == en) return 0;
    if (cycfa[treefa[st]] != cycfa[treefa[en]]) return inf;
    if (treefa[st] == en) return deep[st];
    if (ifcyc[st] == 1 && ifcyc[en] == 0) return inf;
    if (ifcyc[st] == 1 && ifcyc[en] == 1)
        return cycno[en] > cycno[st] ? cycno[en] - cycno[st] : cycno[en] - cycno[st] + cyclen[st];
    if (ifcyc[st] == 0 && ifcyc[en] == 1) return deep[st] + dis(treefa[st], en);
    if (treefa[st] != treefa[en]) return inf;
    if (dfn[st] > dfnr[en] || dfn[st] < dfn[en]) return inf;
    return (deep[st] - deep[en]);
}

inline bool check(ll mid) {
    ll now = 1;
    for (int i = 1; i <= n; i++) {
        while (dis(a[i], now) > mid) {
            now++;
            if (now >= m + 1) return 0;
        }
    }
    return 1;
}

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);

    cin >> t;
    while (t--) {
        cin >> n >> m;
        for (int i = 1; i <= n; i++) cin >> a[i];
        for (int i = 1; i <= m; i++) cin >> b[i];
        fcnt = 0, dcnt = 0;
        for (int i = 1; i <= m; i++) {
            son[i].clear();
            cycfa[i] = ifcyc[i] = cycno[i] = cyclen[i] = vis[i] = deep[i] = treefa[i] = dfn[i] = dfnr[i] = 0;
        }

        for (int i = 1; i <= m; i++) son[b[i]].push_back(i);
        for (int i = 1; i <= m; i++) {
            if (!vis[i]) dfs(i);
            if (vis[i] >= 2) ifcyc[i] = 0;
            if (cycno[i] == 1) cyclen[i] = marklen(i);
        }
        for (int i = 1; i <= m; i++) vis[i] = 0;
        for (int i = 1; i <= m; i++) if (ifcyc[i]) dfs0(i, i);
        for (int i = 1; i <= m; i++)
            if (ifcyc[i]) for (auto j: son[i]) dfs_n(j, i);
        for (int i = 1; i <= m; i++) if (ifcyc[i]) treefa[i] = i;

        ll l = 0, r = m;
        while (l < r) {
            ll mid = l + r >> 1;
            if (check(mid)) r = mid;
            else l = mid + 1;
        }
        if (!check(l)) l = -1;
        cout << l << "\n";
    }

    return 0;
}

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

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

Think step by step.

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

1. **Identify the Bug**: The buggy code seems to have issues with cycle detection and handling of non-decreasing sequences. Specifically, it might fail when there are complex cycles or when the transformation of elements does not lead to a non-decreasing sequence.

2. **Edge Cases**:
   - **Single Element Array**: Test with an array of a single element.
   - **Already Non-decreasing Array**: Test with an array that is already non-decreasing.
   - **Array with Cycles**: Test with an array that contains cycles in the transformation.
   - **Array with Multiple Cycles**: Test with an array that contains multiple cycles.
   - **Array with No Solution**: Test with an array that cannot be made non-decreasing.

3. **Generate Test Input**:
   - **Single Element Array**:
     ```plaintext
     1
     1 1
     1
     1
     ```
   - **Already Non-decreasing Array**:
     ```plaintext
     1
     3 3
     1 2 3
     1 2 3
     ```
   - **Array with Cycles**:
     ```plaintext
     1
     3 3
     1 2 3
     2 3 1
     ```
   - **Array with Multiple Cycles**:
     ```plaintext
     1
     4 4
     1 2 3 4
     2 3 1 4
     ```
   - **Array with No Solution**:
     ```plaintext
     1
     3 3
     1 3 2
     2 1 3
     ```

4. **Combine Test Cases**:
   - To ensure the test input is small and complete, we can combine the above test cases into a single input.

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

This test input includes a variety of edge cases and scenarios that are likely to expose the bug in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Way to be Non-decreasing
time_limit_ms: 4000
memory_limit_mb: 512
Description: Little R is a magician who likes non-decreasing arrays. She has an array of length $n$, initially as $a_1, \ldots, a_n$, in which each element is an integer between $[1, m]$. She wants it to be non-decreasing, i.e., $a_1 \leq a_2 \leq \ldots \leq a_n$.

To do this, she can perform several magic tricks. Little R has a fixed array $b_1\ldots b_m$ of length $m$. Formally, let's define a trick as a procedure that does the following things in order:

  * Choose a set $S \subseteq \\{1, 2, \ldots, n\\}$.    * For each $u \in S$, assign $a_u$ with $b_{a_u}$.

Little R wonders how many tricks are needed at least to make the initial array non-decreasing. If it is not possible with any amount of tricks, print $-1$ instead.

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

The first line of each test case contains two integers $n$ and $m$ ($1\leq n \leq 10^6$, $1 \leq m \leq 10^6$) — the length of the initial array and the range of the elements in the array.

The second line of each test case contains $n$ integers $a_1, \ldots, a_n$ ($1 \leq a_i \leq m$) — the initial array.

The third line of each test case contains $m$ integers $b_1, \ldots, b_m$ ($1 \leq b_i \leq m$) — the fixed magic array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$ and the sum of $m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the minimum number of tricks needed, or $-1$ if it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Examples:
input:
3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
output:
3
-1
3

Note: In the first case, the initial array $a_1, \ldots, a_n$ is $[1, 6, 3, 7, 1]$. You can choose $S$ as follows:

  * first trick: $S = [2, 4, 5]$, $a = [1, 1, 3, 5, 2]$;    * second trick: $S = [5]$, $a = [1, 1, 3, 5, 3]$;    * third trick: $S = [5]$, $a = [1, 1, 3, 5, 5]$.

So it is possible to make $a_1, \ldots, a_n$ non-decreasing using $3$ tricks. It can be shown that this is the minimum possible amount of tricks.

In the second case, it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Buggy Code:
import sys
import io, os
sys.setrecursionlimit(10**5)
input = sys.stdin.readline

class UnionFind:
    def __init__(self, n):
        self.par = [-1]*n
        self.rank = [0]*n

    def Find(self, x):
        if self.par[x] < 0:
            return x
        else:
            self.par[x] = self.Find(self.par[x])
            return self.par[x]

    def Unite(self, x, y):
        x = self.Find(x)
        y = self.Find(y)

        if x != y:
            if self.rank[x] < self.rank[y]:
                self.par[y] += self.par[x]
                self.par[x] = y
            else:
                self.par[x] += self.par[y]
                self.par[y] = x
                if self.rank[x] == self.rank[y]:
                    self.rank[x] += 1

    def Same(self, x, y):
        return self.Find(x) == self.Find(y)

    def Size(self, x):
        return -self.par[self.Find(x)]

def dfs(u):
    global sz, dfn, tot, dep, id, ch
    sz[u] = 1
    dfn[u] = tot
    tot += 1
    for v in ch[u]:
        dep[v] = dep[u] + 1
        id[v] = id[u]
        dfs(v)
        sz[u] += sz[v]

def in_sub(u, v):
    return dfn[u] <= dfn[v] and dfn[v] < dfn[u] + sz[u]

def query(u, v, to):
    if u == v:
        return 0
    if id[u] != id[v]:
        return float('inf')
    res = float('inf')
    if in_sub(v, u):
        res = dep[u] - dep[v]
    if in_sub(v, to[id[u]]):
        res = min(dep[u] - dep[v] + dep[to[id[u]]] + 1, res)
    return res

def check(val, a, N, M, to):
    lst = 1
    for i in range(1, N+1):
        while lst <= M and query(a[i], lst, to) > val:
            lst += 1
        if lst > M:
            return False
    return True

def main():
    index = 0

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

    for _ in range(T):
        N, M = map(int, input().split())
        uf = UnionFind(M + 1)
        a = list(map(int, input().split()))
        a = [0]+a

        global dep, id, dfn, to, sz, tot, ch
        dep = [0] * (M + 1)
        id = [0] * (M + 1)
        dfn = [0] * (M + 1)
        to = [0] * (M + 1)
        sz = [0] * (M + 1)
        ch = [[] for _ in range(M + 1)]
        tot = 1

        par = list(map(int, input().split()))
        par = [0]+par
        for i in range(1, M + 1):
            p = par[i]
            if not uf.Same(i, p):
                ch[p].append(i)
                uf.Unite(i, p)
            else:
                to[i] = p

        for i in range(1, M + 1):
            if to[i] > 0:
                id[i] = i
                dfs(i)

        if not check(M, a, N, M, to):
            results.append("-1")
            continue

        L, R = 0, M
        while L < R:
            mid = (L + R) // 2
            if check(mid, a, N, M, to):
                R = mid
            else:
                L = mid + 1

        results.append(str(R))

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

if __name__ == "__main__":
    main()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Way to be Non-decreasing
time_limit_ms: 4000
memory_limit_mb: 512
Description: Little R is a magician who likes non-decreasing arrays. She has an array of length $n$, initially as $a_1, \ldots, a_n$, in which each element is an integer between $[1, m]$. She wants it to be non-decreasing, i.e., $a_1 \leq a_2 \leq \ldots \leq a_n$.

To do this, she can perform several magic tricks. Little R has a fixed array $b_1\ldots b_m$ of length $m$. Formally, let's define a trick as a procedure that does the following things in order:

  * Choose a set $S \subseteq \\{1, 2, \ldots, n\\}$.    * For each $u \in S$, assign $a_u$ with $b_{a_u}$.

Little R wonders how many tricks are needed at least to make the initial array non-decreasing. If it is not possible with any amount of tricks, print $-1$ instead.

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

The first line of each test case contains two integers $n$ and $m$ ($1\leq n \leq 10^6$, $1 \leq m \leq 10^6$) — the length of the initial array and the range of the elements in the array.

The second line of each test case contains $n$ integers $a_1, \ldots, a_n$ ($1 \leq a_i \leq m$) — the initial array.

The third line of each test case contains $m$ integers $b_1, \ldots, b_m$ ($1 \leq b_i \leq m$) — the fixed magic array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$ and the sum of $m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the minimum number of tricks needed, or $-1$ if it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Examples:
input:
3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
output:
3
-1
3

Note: In the first case, the initial array $a_1, \ldots, a_n$ is $[1, 6, 3, 7, 1]$. You can choose $S$ as follows:

  * first trick: $S = [2, 4, 5]$, $a = [1, 1, 3, 5, 2]$;    * second trick: $S = [5]$, $a = [1, 1, 3, 5, 3]$;    * third trick: $S = [5]$, $a = [1, 1, 3, 5, 5]$.

So it is possible to make $a_1, \ldots, a_n$ non-decreasing using $3$ tricks. It can be shown that this is the minimum possible amount of tricks.

In the second case, it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Buggy Code:
import sys
import io, os
input = sys.stdin.readline

class UnionFind:
    def __init__(self, n):
        self.par = [-1]*n
        self.rank = [0]*n

    def Find(self, x):
        if self.par[x] < 0:
            return x
        else:
            self.par[x] = self.Find(self.par[x])
            return self.par[x]

    def Unite(self, x, y):
        x = self.Find(x)
        y = self.Find(y)

        if x != y:
            if self.rank[x] < self.rank[y]:
                self.par[y] += self.par[x]
                self.par[x] = y
            else:
                self.par[x] += self.par[y]
                self.par[y] = x
                if self.rank[x] == self.rank[y]:
                    self.rank[x] += 1

    def Same(self, x, y):
        return self.Find(x) == self.Find(y)

    def Size(self, x):
        return -self.par[self.Find(x)]

def dfs(u):
    global sz, dfn, tot, dep, id, ch
    sz[u] = 1
    dfn[u] = tot
    tot += 1
    for v in ch[u]:
        dep[v] = dep[u] + 1
        id[v] = id[u]
        dfs(v)
        sz[u] += sz[v]

def in_sub(u, v):
    return dfn[u] <= dfn[v] and dfn[v] < dfn[u] + sz[u]

def query(u, v, to):
    if u == v:
        return 0
    if id[u] != id[v]:
        return float('inf')
    res = float('inf')
    if in_sub(v, u):
        res = dep[u] - dep[v]
    if in_sub(v, to[id[u]]):
        res = min(dep[u] - dep[v] + dep[to[id[u]]] + 1, res)
    return res

def check(val, a, N, M, to):
    lst = 1
    for i in range(1, N+1):
        while lst <= M and query(a[i], lst, to) > val:
            lst += 1
        if lst > M:
            return False
    return True

def main():
    index = 0

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

    for _ in range(T):
        N, M = map(int, input().split())
        uf = UnionFind(M + 1)
        a = list(map(int, input().split()))
        a = [0]+a

        global dep, id, dfn, to, sz, tot, ch
        dep = [0] * (M + 1)
        id = [0] * (M + 1)
        dfn = [0] * (M + 1)
        to = [0] * (M + 1)
        sz = [0] * (M + 1)
        ch = [[] for _ in range(M + 1)]
        tot = 1

        par = list(map(int, input().split()))
        par = [0]+par
        for i in range(1, M + 1):
            p = par[i]
            if not uf.Same(i, p):
                ch[p].append(i)
                uf.Unite(i, p)
            else:
                to[i] = p

        for i in range(1, M + 1):
            if to[i] > 0:
                id[i] = i
                dfs(i)

        if not check(M, a, N, M, to):
            results.append("-1")
            continue

        L, R = 0, M
        while L < R:
            mid = (L + R) // 2
            if check(mid, a, N, M, to):
                R = mid
            else:
                L = mid + 1

        results.append(str(R))

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

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The code uses a union-find data structure and a depth-first search (DFS) to determine the minimum number of tricks needed to make the array non-decreasing. Here are some considerations:

1. **Edge Cases:**
   - Arrays that are already non-decreasing.
   - Arrays with all elements being the same.
   - Arrays that cannot be made non-decreasing regardless of the number of tricks.

2. **Complex Scenarios:**
   - Arrays with a large number of elements and a large range of values.
   - Arrays with a specific pattern that requires multiple tricks to become non-decreasing.

3. **Buggy Code Analysis:**
   - The union-find and DFS logic might not correctly handle certain cases, especially when the transformation involves multiple levels of recursion or complex parent-child relationships.

Given these considerations, let's create a small-scale test input that is complete and valid:

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

### Explanation:
1. **First Test Case:**
   - Initial array: `[1, 5, 3, 4, 2]`
   - Magic array: `[2, 3, 4, 5, 1]`
   - This array can be made non-decreasing in 3 tricks:
     - First trick: `S = [2, 3, 4, 5]` → `[1, 1, 4, 5, 1]`
     - Second trick: `S = [5]` → `[1, 1, 4, 5, 4]`
     - Third trick: `S = [5]` → `[1, 1, 4, 5, 5]`

2. **Second Test Case:**
   - Initial array: `[1, 3, 2]`
   - Magic array: `[2, 1, 3]`
   - This array cannot be made non-decreasing, so the output should be `-1`.

3. **Third Test Case:**
   - Initial array: `[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]`
   - Magic array: `[10, 9, 8, 7, 6, 5, 4, 3, 2, 1]`
   - This array is already non-decreasing, so no tricks are needed, and the output should be `0`.

This test input covers a range of scenarios, including an array that can be made non-decreasing, an array that cannot be made non-decreasing, and an array that is already non-decreasing. It should help expose any bugs in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Way to be Non-decreasing
time_limit_ms: 4000
memory_limit_mb: 512
Description: Little R is a magician who likes non-decreasing arrays. She has an array of length $n$, initially as $a_1, \ldots, a_n$, in which each element is an integer between $[1, m]$. She wants it to be non-decreasing, i.e., $a_1 \leq a_2 \leq \ldots \leq a_n$.

To do this, she can perform several magic tricks. Little R has a fixed array $b_1\ldots b_m$ of length $m$. Formally, let's define a trick as a procedure that does the following things in order:

  * Choose a set $S \subseteq \\{1, 2, \ldots, n\\}$.    * For each $u \in S$, assign $a_u$ with $b_{a_u}$.

Little R wonders how many tricks are needed at least to make the initial array non-decreasing. If it is not possible with any amount of tricks, print $-1$ instead.

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

The first line of each test case contains two integers $n$ and $m$ ($1\leq n \leq 10^6$, $1 \leq m \leq 10^6$) — the length of the initial array and the range of the elements in the array.

The second line of each test case contains $n$ integers $a_1, \ldots, a_n$ ($1 \leq a_i \leq m$) — the initial array.

The third line of each test case contains $m$ integers $b_1, \ldots, b_m$ ($1 \leq b_i \leq m$) — the fixed magic array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$ and the sum of $m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the minimum number of tricks needed, or $-1$ if it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Examples:
input:
3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
output:
3
-1
3

Note: In the first case, the initial array $a_1, \ldots, a_n$ is $[1, 6, 3, 7, 1]$. You can choose $S$ as follows:

  * first trick: $S = [2, 4, 5]$, $a = [1, 1, 3, 5, 2]$;    * second trick: $S = [5]$, $a = [1, 1, 3, 5, 3]$;    * third trick: $S = [5]$, $a = [1, 1, 3, 5, 5]$.

So it is possible to make $a_1, \ldots, a_n$ non-decreasing using $3$ tricks. It can be shown that this is the minimum possible amount of tricks.

In the second case, it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Buggy Code:
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;
import java.lang.Exception;
public class Solution {
    private final FastReader reader;
    private final StringBuilder sb;
    private int time = 0;

    private void runForTestCase() throws Exception {
        int n = reader.getInt();
        int m = reader.getInt();
        int[] a = reader.getIntArr(n);
        int[] b = reader.getIntArr(m);
        for(int i=0; i<n; i++) a[i]--;
        for(int i=0; i<m; i++) b[i]--;
        DisjointSet ds = new DisjointSet(m);
        List<Integer> roots = new ArrayList<>();
        List<Integer>[] graph = new List[m];
        for(int i=0; i<m; i++) graph[i] = new ArrayList<>();
        for(int i=0; i<m; i++) {
            if(ds.union(i, b[i])) graph[b[i]].add(i);
            else roots.add(i);
        }
        int[] t_in = new int[m];
        int[] t_out = new int[m];
        int[] depth = new int[m];
        int[] root = new int[m];
        time = 0;
        for(int r : roots) {
            depth[r] = 0;
            root[r] = r;
            dfs(graph, r, t_in, t_out, root, depth);
        }
        int l = 0;
        int h = m+1;
        while(l < h) {
            int mid = (l+h)>>1;
            int v = 0;
            for(int x : a) {
                while(v<m) {
                    if(validPathExists(t_in, t_out, v, x) && depth[x]-depth[v]<=mid) break;
                    int p = b[root[x]];
                    int bal = mid - depth[x] - 1;
                    if(validPathExists(t_in, t_out, v, p) && depth[p]-depth[v]<=bal) break;
                    v++;
                }
            }
            if(v<m) h = mid;
            else l = mid+1;
        }
        sb.append((l==m+1) ? -1 : l).append("\n");
    }

    private boolean validPathExists(int[] t_in, int[] t_out, int x, int y) {
        return t_in[x]<=t_in[y] && t_out[x]>=t_out[y];
    }

    private void dfs(List<Integer>[] graph, int node, int[] t_in, int[] t_out, int[] root, int[] depth) {
        t_in[node] = time++;
        for(int child : graph[node]) {
            root[child] = root[node];
            depth[child] = depth[node] + 1;
            dfs(graph, child, t_in, t_out, root, depth);
        }
        t_out[node] = time++;
    }

    static class DisjointSet {
        int[] parent;
        //int[] rank;

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

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

        public boolean union(int a, int b) {
            int ap = find(a);
            int bp = find(b);
            if(ap==bp) return false;
            parent[ap] = bp;
            /*if(rank[ap]>rank[bp]) {
                parent[bp] = ap;
                rank[ap] += rank[bp] + 1;
            } else {
                parent[ap] = bp;
                rank[bp] += rank[ap] + 1;
            }*/
            return true;
        }
    }

    static class Utils {
        private static final int TEN_POW_FIVE = (int)1e5;
        private static final int MOD = (int)1e9+7;

        public static void swap(int[] arr, int i, int j) {
            int temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
        }

        public static int min(Integer... minArr) {
            int min = minArr[0];
            for(int i=1; i<minArr.length; i++) {
                min = Math.min(min, minArr[i]);
            }
            return min;
        }

        public static int max(Integer... maxArr) {
            int max = maxArr[0];
            for(int i=1; i<maxArr.length; i++) {
                max = Math.max(max, maxArr[i]);
            }
            return max;
        }

        public static void printArr(int[] arr, int n) {
            for(int i=0; i<n; i++) {
                System.out.print(arr[i] + " ");
            }
            System.out.println();
        }
    }

    static class FastReader {
        private final BufferedReader inputReader;
        private StringTokenizer st;

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

        private String getNext() throws Exception {
            if(st==null || !st.hasMoreTokens()) {
                st = new StringTokenizer(inputReader.readLine());
            }
            return st.nextToken();
        }

        public String[] getLine(String delimiter) throws Exception {
            return inputReader.readLine().split(delimiter);
        }

        public String getString() throws Exception {
            return getNext();
        }

        public int getInt() throws Exception {
            return Integer.parseInt(getNext());
        }

        public long getLong() throws Exception {
            return Long.parseLong(getNext());
        }

        public double getDouble() throws Exception {
            return Double.parseDouble(getNext());
        }

        public int[] getIntArr(int n) throws Exception {
            int[] arr = new int[n];
            for(int i=0; i<n; i++) {
                arr[i] = getInt();
            }
            return arr;
        }

        public long[] getLongArr(int n) throws Exception {
            long[] arr = new long[n];
            for(int i=0; i<n; i++) {
                arr[i] = getLong();
            }
            return arr;
        }

        public double[] getDoubleArr(int n) throws Exception {
            double[] arr = new double[n];
            for(int i=0; i<n; i++) {
                arr[i] = getDouble();
            }
            return arr;
        }

        public String[] getStrArr(int n) throws Exception {
            String[] arr = new String[n];
            for(int i=0; i<n; i++) {
                arr[i] = getString();
            }
            return arr;
        }

        public void closeReader() throws Exception {
            inputReader.close();
        }
    }

    public Solution() {
        this.reader = new FastReader();
        this.sb = new StringBuilder();
    }

    private void solve() throws Exception {
        int t = reader.getInt();
        while(t-->0) {
            runForTestCase();
        }
        reader.closeReader();
    }

    public static void main(String[] args) {
        try {
            Solution solution = new Solution();
            solution.solve();
            solution.printOp();
        } catch(Exception e) {
            e.printStackTrace();
        }
    }

    private void printOp() {
        System.out.println(sb);
    }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios where the code might fail. The main issues in the code are related to the union-find data structure and the way it handles the graph and depth calculations. Here are some steps to think through:

1. **Small-Scale Input**: Since the output length is limited, we should focus on a small-scale input that can still expose the bug.
2. **Edge Cases**: Consider inputs where the array `a` is already non-decreasing, where the array `a` is in a configuration that makes it impossible to become non-decreasing, and where the array `b` has specific patterns that can confuse the union-find operations.
3. **Specific Patterns**: Use specific patterns in `a` and `b` to ensure that the union-find operations and depth calculations are tested thoroughly.

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

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

### Explanation:
1. **First Test Case**:
   - `a = [1, 2, 3, 4, 5]`
   - `b = [1, 2, 3, 4, 5]`
   - This array is already non-decreasing, so the expected output is `0`.

2. **Second Test Case**:
   - `a = [5, 4, 3, 2, 1]`
   - `b = [1, 2, 3, 4, 5]`
   - This array is in reverse order, and it is impossible to make it non-decreasing with the given `b` array. The expected output is `-1`.

3. **Third Test Case**:
   - `a = [1, 3, 2, 4, 5]`
   - `b = [1, 2, 3, 4, 5]`
   - This array has elements out of order, but it can be made non-decreasing with a few tricks. The expected output is `2`.

This input should help identify issues in the union-find operations and the depth calculations, which are critical for the algorithm to work correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Way to be Non-decreasing
time_limit_ms: 4000
memory_limit_mb: 512
Description: Little R is a magician who likes non-decreasing arrays. She has an array of length $n$, initially as $a_1, \ldots, a_n$, in which each element is an integer between $[1, m]$. She wants it to be non-decreasing, i.e., $a_1 \leq a_2 \leq \ldots \leq a_n$.

To do this, she can perform several magic tricks. Little R has a fixed array $b_1\ldots b_m$ of length $m$. Formally, let's define a trick as a procedure that does the following things in order:

  * Choose a set $S \subseteq \\{1, 2, \ldots, n\\}$.    * For each $u \in S$, assign $a_u$ with $b_{a_u}$.

Little R wonders how many tricks are needed at least to make the initial array non-decreasing. If it is not possible with any amount of tricks, print $-1$ instead.

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

The first line of each test case contains two integers $n$ and $m$ ($1\leq n \leq 10^6$, $1 \leq m \leq 10^6$) — the length of the initial array and the range of the elements in the array.

The second line of each test case contains $n$ integers $a_1, \ldots, a_n$ ($1 \leq a_i \leq m$) — the initial array.

The third line of each test case contains $m$ integers $b_1, \ldots, b_m$ ($1 \leq b_i \leq m$) — the fixed magic array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$ and the sum of $m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the minimum number of tricks needed, or $-1$ if it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Examples:
input:
3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
output:
3
-1
3

Note: In the first case, the initial array $a_1, \ldots, a_n$ is $[1, 6, 3, 7, 1]$. You can choose $S$ as follows:

  * first trick: $S = [2, 4, 5]$, $a = [1, 1, 3, 5, 2]$;    * second trick: $S = [5]$, $a = [1, 1, 3, 5, 3]$;    * third trick: $S = [5]$, $a = [1, 1, 3, 5, 5]$.

So it is possible to make $a_1, \ldots, a_n$ non-decreasing using $3$ tricks. It can be shown that this is the minimum possible amount of tricks.

In the second case, it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Buggy Code:
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;
import java.lang.Exception;
public class Solution {
    private final FastReader reader;
    private final StringBuilder sb;
    private int time = 0;

    private void runForTestCase() throws Exception {
        int n = reader.getInt();
        int m = reader.getInt();
        int[] a = reader.getIntArr(n);
        int[] b = reader.getIntArr(m);
        for(int i=0; i<n; i++) a[i]--;
        for(int i=0; i<m; i++) b[i]--;
        DisjointSet ds = new DisjointSet(m);
        List<Integer> roots = new ArrayList<>();
        List<Integer>[] graph = new List[m];
        for(int i=0; i<m; i++) graph[i] = new ArrayList<>();
        for(int i=0; i<m; i++) {
            if(ds.union(i, b[i])) graph[b[i]].add(i);
            else roots.add(i);
        }
        int[] t_in = new int[m];
        int[] t_out = new int[m];
        int[] depth = new int[m];
        int[] root = new int[m];
        time = 0;
        for(int r : roots) {
            depth[r] = 0;
            root[r] = r;
            dfs(graph, r, t_in, t_out, root, depth);
        }
        int l = 0;
        int h = m+1;
        while(l < h) {
            int mid = (l+h)>>1;
            int v = 0;
            for(int x : a) {
                while(v<m) {
                    if(validPathExists(t_in, t_out, v, x) && depth[x]-depth[v]<=mid) break;
                    int p = b[root[x]];
                    int bal = mid - depth[x] - 1;
                    if(validPathExists(t_in, t_out, v, p) && depth[p]-depth[v]<=bal) break;
                    v++;
                }
            }
            if(v<m) h = mid;
            else l = mid+1;
        }
        sb.append((l==m+1) ? -1 : l).append("\n");
    }

    private boolean validPathExists(int[] t_in, int[] t_out, int x, int y) {
        return t_in[x]<=t_in[y] && t_out[x]>=t_out[y];
    }

    private void dfs(List<Integer>[] graph, int node, int[] t_in, int[] t_out, int[] root, int[] depth) {
        t_in[node] = time++;
        for(int child : graph[node]) {
            root[child] = root[node];
            depth[child] = depth[node] + 1;
            dfs(graph, child, t_in, t_out, root, depth);
        }
        t_out[node] = time++;
    }

    static class DisjointSet {
        int[] parent;
        int[] rank;

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

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

        public boolean union(int a, int b) {
            int ap = find(a);
            int bp = find(b);
            if(ap==bp) return false;
            if(rank[ap]>rank[bp]) {
                parent[bp] = ap;
                rank[ap] += rank[bp] + 1;
            } else {
                parent[ap] = bp;
                rank[bp] += rank[ap] + 1;
            }
            return true;
        }
    }

    static class Utils {
        private static final int TEN_POW_FIVE = (int)1e5;
        private static final int MOD = (int)1e9+7;

        public static void swap(int[] arr, int i, int j) {
            int temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
        }

        public static int min(Integer... minArr) {
            int min = minArr[0];
            for(int i=1; i<minArr.length; i++) {
                min = Math.min(min, minArr[i]);
            }
            return min;
        }

        public static int max(Integer... maxArr) {
            int max = maxArr[0];
            for(int i=1; i<maxArr.length; i++) {
                max = Math.max(max, maxArr[i]);
            }
            return max;
        }

        public static void printArr(int[] arr, int n) {
            for(int i=0; i<n; i++) {
                System.out.print(arr[i] + " ");
            }
            System.out.println();
        }
    }

    static class FastReader {
        private final BufferedReader inputReader;
        private StringTokenizer st;

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

        private String getNext() throws Exception {
            if(st==null || !st.hasMoreTokens()) {
                st = new StringTokenizer(inputReader.readLine());
            }
            return st.nextToken();
        }

        public String[] getLine(String delimiter) throws Exception {
            return inputReader.readLine().split(delimiter);
        }

        public String getString() throws Exception {
            return getNext();
        }

        public int getInt() throws Exception {
            return Integer.parseInt(getNext());
        }

        public long getLong() throws Exception {
            return Long.parseLong(getNext());
        }

        public double getDouble() throws Exception {
            return Double.parseDouble(getNext());
        }

        public int[] getIntArr(int n) throws Exception {
            int[] arr = new int[n];
            for(int i=0; i<n; i++) {
                arr[i] = getInt();
            }
            return arr;
        }

        public long[] getLongArr(int n) throws Exception {
            long[] arr = new long[n];
            for(int i=0; i<n; i++) {
                arr[i] = getLong();
            }
            return arr;
        }

        public double[] getDoubleArr(int n) throws Exception {
            double[] arr = new double[n];
            for(int i=0; i<n; i++) {
                arr[i] = getDouble();
            }
            return arr;
        }

        public String[] getStrArr(int n) throws Exception {
            String[] arr = new String[n];
            for(int i=0; i<n; i++) {
                arr[i] = getString();
            }
            return arr;
        }

        public void closeReader() throws Exception {
            inputReader.close();
        }
    }

    public Solution() {
        this.reader = new FastReader();
        this.sb = new StringBuilder();
    }

    private void solve() throws Exception {
        int t = reader.getInt();
        while(t-->0) {
            runForTestCase();
        }
        reader.closeReader();
    }

    public static void main(String[] args) {
        try {
            Solution solution = new Solution();
            solution.solve();
            solution.printOp();
        } catch(Exception e) {
            e.printStackTrace();
        }
    }

    private void printOp() {
        System.out.println(sb);
    }

}

**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 that can cause the algorithm to fail. Here are the steps to think through this:

1. **Understand the Bug**: The buggy code might have issues with:
   - Handling cycles or loops in the transformation graph.
   - Incorrectly calculating the depth or valid paths.
   - Failing to handle specific edge cases in the input arrays.

2. **Identify Edge Cases**:
   - **Cycles in the Transformation Graph**: If the transformation graph contains cycles, the code might not handle it correctly.
   - **Non-Transformable Arrays**: If the array cannot be transformed to a non-decreasing sequence, the code should return `-1`.
   - **Edge Cases in Input Size**: Small and large input sizes can reveal different issues.

3. **Construct the Test Case**:
   - **Small Input with a Cycle**: Create an input where the transformation graph contains a cycle.
   - **Non-Transformable Array**: Create an input where the array cannot be transformed to a non-decreasing sequence.

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

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

**Explanation**:
1. **First Test Case**:
   - `n = 3`, `m = 3`
   - Initial array: `[1, 3, 2]`
   - Transformation array: `[2, 1, 3]`
   - This array cannot be transformed to a non-decreasing sequence using the given transformation rules. The transformation rules create a cycle: `1 -> 2 -> 1`, which means the array cannot be made non-decreasing. The expected output is `-1`.

2. **Second Test Case**:
   - `n = 4`, `m = 4`
   - Initial array: `[1, 4, 2, 3]`
   - Transformation array: `[2, 3, 4, 1]`
   - This array also contains a cycle in the transformation graph: `1 -> 2 -> 3 -> 4 -> 1`. Therefore, it is impossible to make the array non-decreasing. The expected output is `-1`.

This test input should effectively expose the bug in the provided buggy code by creating scenarios where the transformation graph contains cycles, making it impossible to achieve a non-decreasing sequence.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Easy)
time_limit_ms: 1000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^2$$$.

$$$1 \le n \le 10^3$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
// LUOGU_RID: 166821313
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const int M=1e3+10,p=1e9+7;
int n,m,s[M],l[M],dp[M][M],ans;
int main()
{
	cin>>m>>n;
	for(int i=1;i<=m;i++)cin>>s[i];
	for(int i=1;i<=m;i++)cin>>l[i];
	dp[1][0]=1;
	for(int k=1;k<=n;k++)
		for(int i=1;i<=m;i++)
			for(int j=1;j<=m;j++)
				dp[j][k]+=((s[i]*s[j]+s[i]*l[j]+l[i]*s[j])*dp[i][k-1])%p;
	for(int i=1;i<=m;i++)ans+=(dp[i][n])%p;
	cout<<ans;
	return 0;
}
/*Who am I?
I'm your father*/

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

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

Think step by step.

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

### Explanation:
1. **Input Parameters:**
   - `m = 2`: There are 2 cabins.
   - `n = 1`: Harry walks for 1 day.

2. **Trail Details:**
   - Cabin 1: 1 short trail, 0 long trails.
   - Cabin 2: 0 short trails, 1 long trail.

3. **Expected Behavior:**
   - Harry starts in cabin 1.
   - On the first day, he can either:
     - Take a short trail from cabin 1 to the lake and then a short trail to cabin 1 (1 way).
     - Take a short trail from cabin 1 to the lake and then a long trail to cabin 2 (1 way).
   - Therefore, there are 2 possible combinations of trails.

4. **Why This Input is Challenging:**
   - This input tests the edge case where one cabin has only short trails and the other has only long trails.
   - It ensures that the code correctly handles the constraint that at least one of the two trails must be short.
   - If the code incorrectly allows both trails to be long, it will produce an incorrect result.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Easy)
time_limit_ms: 1000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^2$$$.

$$$1 \le n \le 10^3$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
// LUOGU_RID: 164840560
#include<iostream>
using namespace std;
const int mod=1e9+7;
long long m,n;
long long s[1001],l[1001];
long long dp[101][1001];
long long ans;
int main(){
	scanf("%d%d",&m,&n);
	for(int i=1;i<=m;i++) scanf("%d",&s[i]);
	for(int i=1;i<=m;i++) scanf("%d",&l[i]);
	dp[0][1]=1;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			for(int k=1;k<=m;k++){
				dp[i][j]=(dp[i][j]+dp[i-1][k]*s[j]*s[k])%mod;
				dp[i][j]=(dp[i][j]+dp[i-1][k]*s[j]*l[k])%mod;
				dp[i][j]=(dp[i][j]+dp[i-1][k]*l[j]*s[k])%mod;
			}
		}
	}
	for(int i=1;i<=m;i++) ans+=dp[n][i];
	cout<<ans%mod;
	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Easy)
time_limit_ms: 1000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^2$$$.

$$$1 \le n \le 10^3$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
#include <iostream>
#include <cstring>

using namespace std;

const long long MOD = 1e9 + 7;

long long dp[1003][1003];

long long solve(int m, int n, int s[], int l[], int curr_cabin, int last_short) {
  if (n == 0) return 1;

  if (dp[curr_cabin][last_short] != -1) return dp[curr_cabin][last_short];

  long long ways = 0;

  if (n == 2) {
    ways = (ways + solve(m, n - 1, s, l, curr_cabin, 1)) % MOD;
    ways = (ways + solve(m, n - 1, s, l, curr_cabin, 0)) % MOD;
  } else {
    if (last_short) {
      ways = (ways + solve(m, n - 1, s, l, curr_cabin, 1)) % MOD;
      ways = (ways + solve(m, n - 1, s, l, curr_cabin, 0)) % MOD;
    } else {
      ways = (ways + solve(m, n - 1, s, l, curr_cabin, 1)) % MOD;
    }
  }

  for (int i = 1; i <= m; ++i) {
    if (i != curr_cabin) {
      ways = (ways + solve(m, n - 1, s, l, i, 1)) % MOD;
    }
  }

  dp[curr_cabin][last_short] = ways;
  return ways;
}

int main() {
  int m, n;
  cin >> m >> n;

  int s[m + 1], l[m + 1];
  for (int i = 1; i <= m; ++i) cin >> s[i];
  for (int i = 1; i <= m; ++i) cin >> l[i];

  memset(dp, -1, sizeof(dp));

  int result = (solve(m, n, s, l, 1, 1) + solve(m, n, s, l, 1, 0)) % MOD;
  cout << result << 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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The key points to consider are:

1. **Edge Cases for `n` and `m`**: Small values for `n` and `m` can help us quickly verify the behavior of the code.
2. **Edge Cases for `s_i` and `l_i`**: Zero values for `s_i` and `l_i` can test how the code handles cabins with no trails.
3. **Mixed Trails**: Cabins with both short and long trails can test the logic for combining trails.

Let's construct a test input that includes a mix of these elements:

- **Small `m` and `n`**: This helps in quickly verifying the behavior.
- **Cabins with zero trails**: To test how the code handles cabins with no trails.
- **Cabins with both short and long trails**: To test the logic for combining trails.

Here is a small-scale test input:

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

Explanation:
- **m = 3**: There are 3 cabins.
- **n = 3**: Harry walks for 3 days.
- **s = [0, 2, 1]**: Cabin 1 has 0 short trails, Cabin 2 has 2 short trails, and Cabin 3 has 1 short trail.
- **l = [1, 0, 1]**: Cabin 1 has 1 long trail, Cabin 2 has 0 long trails, and Cabin 3 has 1 long trail.

This input includes:
- A cabin (Cabin 1) with no short trails.
- A cabin (Cabin 2) with only short trails.
- A cabin (Cabin 3) with both short and long trails.

This combination should help expose any issues in the logic for handling different types of trails and the transitions between cabins.