**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Paths on a Tree
time_limit_ms: 4000
memory_limit_mb: 1024
Description: Note the unusual definition of $\text{MEX}$ in this problem.

Piggy gave Turtle a binary tree$^{\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.

If a set of paths $P = \\{(x_i, y_i)\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.

Turtle defines the value of a set of paths as $\sum\limits_{(x, y) \in P} f(x, y)$, where $f(x, y)$ denotes the $\text{MEX}^{\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).

Turtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!

$^{\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.

$^{\ddagger}\text{MEX}$ of a collection of integers $c_1, c_2, \ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\text{MEX}$ of $[2, 3]$ is $1$.

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 a single integer $n$ ($2 \le n \le 2.5 \cdot 10^4$) — the number of vertices in the tree.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the sequence $a$.

The third line of each test case contains $n - 1$ integers $p_2, p_3, \ldots, p_n$ ($1 \le p_i < i$) — the parent of each vertex in the tree.

Additional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.

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

Output: For each test case, output a single integer — the minimum value over all good sets of paths.

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

Note: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:

![](CDN_BASE_URL/45a1635df0fcab1b711d2165163f0dfb)

The good set of paths with the minimum value is $\\{(2, 3), (4, 5)\\}$.

Note that in this test case $\\{(4, 5)\\}$ and $\\{(3, 4), (4, 5)\\}$ are not good sets of paths because each edge should be covered exactly once.

In the second test case, the tree is as follows:

![](CDN_BASE_URL/a41b8a69adf0471a8353a08785fa6375)

The set of good paths with the minimum value is $\\{(1, 2), (1, 3), (4, 5)\\}$.

In the third test case, the tree is as follows:

![](CDN_BASE_URL/7e62b6f397447336b5c5732c0d915ebd)

The set of good paths with the minimum value is $\\{(1, 6), (3, 5)\\}$.

Buggy Code:
from collections import defaultdict, deque
import sys
input = sys.stdin.read
sys.setrecursionlimit(30000)

inf = float('inf')

def dfs(u, m, a, G, f):
    if not G[u]:
        for i in range(1, m + 1):
            f[u][i] = inf if i == a[u] else 0
        return

    if len(G[u]) == 1:
        x = G[u][0]
        dfs(x, m, a, G, f)
        mn = inf
        for i in range(1, m + 1):
            if i != a[u]:
                mn = min(mn, f[x][i] + i)
        if u == 1:
            print(mn)
            return
        for i in range(1, m + 1):
            f[u][i] = inf if i == a[u] else min(f[x][i], mn)
        return

    x, y = G[u][0], G[u][1]
    mnx, mny, k = inf, inf, inf
    dfs(x, m, a, G, f)
    dfs(y, m, a, G, f)
    for i in range(1, m + 1):
        if i != a[u]:
            mnx = min(mnx, f[x][i] + i)
            mny = min(mny, f[y][i] + i)
            k = min(k, f[x][i] + f[y][i] + i)
    k = min(k, mnx + mny)
    if u == 1:
        print(k)
        return
    for i in range(1, m + 1):
        f[u][i] = inf if i == a[u] else min(f[x][i] + mny, f[y][i] + mnx, k)

def solve():
    data = input().split()
    idx = 0
    T = int(data[idx])
    idx += 1
    results = []
    for _ in range(T):
        n = int(data[idx])
        idx += 1
        m = min(n + 1, 3863)
        a = [0] * (n + 1)
        G = defaultdict(list)
        f = [[0] * (m + 1) for _ in range(n + 1)]
        for i in range(1, n + 1):
            a[i] = int(data[idx])
            idx += 1
            G[i] = []
        for i in range(2, n + 1):
            p = int(data[idx])
            idx += 1
            G[p].append(i)
        dfs(1, m, a, G, f)
    for result in results:
        print(result)

if __name__ == "__main__":
    solve()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases or scenarios where the code might fail. The key areas to consider include:

1. **Tree Structure**: Ensure the tree is a valid binary tree with each non-leaf vertex having at most 2 children.
2. **Sequence Values**: Use values for the sequence \(a\) that can cause issues with the MEX calculation.
3. **Edge Cases**: Consider small trees and specific configurations that might not be handled correctly by the code.

Let's create a small-scale test input that includes a tree with a few vertices and a sequence that might expose the bug.

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

### Explanation
- **Test Case**: 1
- **Number of Vertices**: 6
- **Sequence \(a\)**: [1, 2, 1, 2, 1, 3]
- **Parent Array**: [1, 2, 3, 3, 4]

### Tree Structure
- Vertex 1 (root) has children 2 and 3.
- Vertex 2 has no children.
- Vertex 3 has children 4 and 5.
- Vertex 4 has no children.
- Vertex 5 has no children.
- Vertex 6 is a child of vertex 3.

### Expected Output
The expected output for this test case is 6, as the minimum value over all good sets of paths is 6.

### Why This Test Case is Challenging
- **MEX Calculation**: The sequence values are chosen to ensure that the MEX calculation is non-trivial. For example, the MEX of the path from 1 to 6 includes the values [1, 2, 1, 2, 1, 3], which should result in an MEX of 4.
- **Tree Structure**: The tree is structured in a way that requires the algorithm to correctly handle paths that cover each edge exactly once. This includes paths that involve the root and its children, as well as paths that involve the children of the root's children.

This test case should help expose any issues in the provided code, particularly in how it handles the MEX calculation and the structure of the tree.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Paths on a Tree
time_limit_ms: 4000
memory_limit_mb: 1024
Description: Note the unusual definition of $\text{MEX}$ in this problem.

Piggy gave Turtle a binary tree$^{\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.

If a set of paths $P = \\{(x_i, y_i)\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.

Turtle defines the value of a set of paths as $\sum\limits_{(x, y) \in P} f(x, y)$, where $f(x, y)$ denotes the $\text{MEX}^{\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).

Turtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!

$^{\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.

$^{\ddagger}\text{MEX}$ of a collection of integers $c_1, c_2, \ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\text{MEX}$ of $[2, 3]$ is $1$.

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 a single integer $n$ ($2 \le n \le 2.5 \cdot 10^4$) — the number of vertices in the tree.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the sequence $a$.

The third line of each test case contains $n - 1$ integers $p_2, p_3, \ldots, p_n$ ($1 \le p_i < i$) — the parent of each vertex in the tree.

Additional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.

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

Output: For each test case, output a single integer — the minimum value over all good sets of paths.

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

Note: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:

![](CDN_BASE_URL/45a1635df0fcab1b711d2165163f0dfb)

The good set of paths with the minimum value is $\\{(2, 3), (4, 5)\\}$.

Note that in this test case $\\{(4, 5)\\}$ and $\\{(3, 4), (4, 5)\\}$ are not good sets of paths because each edge should be covered exactly once.

In the second test case, the tree is as follows:

![](CDN_BASE_URL/a41b8a69adf0471a8353a08785fa6375)

The set of good paths with the minimum value is $\\{(1, 2), (1, 3), (4, 5)\\}$.

In the third test case, the tree is as follows:

![](CDN_BASE_URL/7e62b6f397447336b5c5732c0d915ebd)

The set of good paths with the minimum value is $\\{(1, 6), (3, 5)\\}$.

Buggy Code:
def solve():
    import sys
    input = sys.stdin.read
    data = input().split()

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

    results = []

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

        a = list(map(int, data[index:index + n]))
        index += n

        p = [-1] * n
        for i in range(1, n):
            p[i] = int(data[index]) - 1
            index += 1

        adj = [[] for _ in range(n)]
        for i in range(1, n):
            adj[p[i]].append(i)

        lo, hi = 1, n + 1
        while lo < hi:
            m = (lo + hi + 1) // 2
            ans = 0
            ans += (m - 1) // 3 + 1
            for k in range(2, (m + 1) // 2 + 1):
                ans += (m - k) // k
            if ans <= n:
                lo = m
            else:
                hi = m - 1

        m = lo
        inf = int(1E9)
        dp = [[0] * (m + 1) for _ in range(n)]

        def get(x):
            ans = inf
            for i in range(1, m + 1):
                ans = min(ans, dp[x][i] + i)
            return ans

        ans = inf

        def dfs(x):
            nonlocal ans
            if len(adj[x]) == 0:
                if a[x] <= m:
                    dp[x][a[x]] = inf
            elif len(adj[x]) == 1:
                y = adj[x][0]
                dfs(y)
                if a[x] <= m:
                    dp[y][a[x]] = inf
                gy = get(y)
                for i in range(1, m + 1):
                    if i == a[x]:
                        dp[x][i] = inf
                    else:
                        dp[x][i] = min(gy, dp[y][i])
            else:
                y = adj[x][0]
                z = adj[x][1]
                dfs(y)
                dfs(z)
                if a[x] <= m:
                    dp[y][a[x]] = inf
                    dp[z][a[x]] = inf
                gy = get(y)
                gz = get(z)
                gyz = inf
                for i in range(1, m + 1):
                    gyz = min(gyz, dp[y][i] + dp[z][i] + i)
                for i in range(1, m + 1):
                    if i == a[x]:
                        dp[x][i] = inf
                    else:
                        dp[x][i] = min(gy + gz, dp[y][i] + gz, gy + dp[z][i], gyz)
                        if x == 0:
                            ans = min(ans, gyz)

        dfs(0)

        ans = min(ans, get(0))
        results.append(ans)

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

if __name__ == "__main__":
    import sys
    input = sys.stdin.read
    solve()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Paths on a Tree
time_limit_ms: 4000
memory_limit_mb: 1024
Description: Note the unusual definition of $\text{MEX}$ in this problem.

Piggy gave Turtle a binary tree$^{\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.

If a set of paths $P = \\{(x_i, y_i)\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.

Turtle defines the value of a set of paths as $\sum\limits_{(x, y) \in P} f(x, y)$, where $f(x, y)$ denotes the $\text{MEX}^{\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).

Turtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!

$^{\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.

$^{\ddagger}\text{MEX}$ of a collection of integers $c_1, c_2, \ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\text{MEX}$ of $[2, 3]$ is $1$.

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 a single integer $n$ ($2 \le n \le 2.5 \cdot 10^4$) — the number of vertices in the tree.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the sequence $a$.

The third line of each test case contains $n - 1$ integers $p_2, p_3, \ldots, p_n$ ($1 \le p_i < i$) — the parent of each vertex in the tree.

Additional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.

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

Output: For each test case, output a single integer — the minimum value over all good sets of paths.

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

Note: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:

![](CDN_BASE_URL/45a1635df0fcab1b711d2165163f0dfb)

The good set of paths with the minimum value is $\\{(2, 3), (4, 5)\\}$.

Note that in this test case $\\{(4, 5)\\}$ and $\\{(3, 4), (4, 5)\\}$ are not good sets of paths because each edge should be covered exactly once.

In the second test case, the tree is as follows:

![](CDN_BASE_URL/a41b8a69adf0471a8353a08785fa6375)

The set of good paths with the minimum value is $\\{(1, 2), (1, 3), (4, 5)\\}$.

In the third test case, the tree is as follows:

![](CDN_BASE_URL/7e62b6f397447336b5c5732c0d915ebd)

The set of good paths with the minimum value is $\\{(1, 6), (3, 5)\\}$.

Buggy Code:
//RCh - F
#include<bits/stdc++.h>
#define INF LONG_LONG_MAX
#define inf 1e9
#define bctz __builtin_ctz
#define bclz __builtin_clz
#define bpt __builtin_popcount
#define pb push_back
#define x first
#define y second

using namespace std;

typedef long long INT;
typedef pair<int, int> pii;
typedef pair<int, pii> pi3;

template <class T> inline void smax(T &a, T b){if(a < b)a = b;}
template <class T> inline void smin(T &a, T b){if(a > b)a = b;}

const int NN = 3e4 + 10;
const int mod = 998244353;

int A[NN], B[NN], C[NN], ans[NN];
int pa[NN], ls[NN], rs[NN];
int n;

vector <int> dfs(int u){
	ans[u] = inf;
	vector <int> dp(n + 2, 0);
	if(ls[u] == 0){
		dp[A[u]] = inf;
		return dp;
	}
	for(int i = 0; i <= n + 1; i++) dp[i] = inf;
	if(rs[u] == 0){
		dp = dfs(ls[u]);
		dp[A[u]] = inf;
		for(int i = 1; i <= n + 1; i++) smin(ans[u], i + dp[i]);
		for(int i = 1; i <= n + 1; i++){
			if(i == A[u]) continue;
			smin(dp[i], ans[u]);
		}
		return dp;
	}
	vector <int> dpL = dfs(ls[u]);
	vector <int> dpR = dfs(rs[u]);
	int mnL = inf, mnR = inf, mn = inf;
	for(int i = 1; i <= n + 1; i++){
		if(i == A[u]) continue;
		smin(mnL, dpL[i] + i);
		smin(mnR, dpR[i] + i);
		smin(mn, dpL[i] + dpR[i] + i);
	}
	for(int i = 1; i <= n + 1; i++){
		if(i == A[u]) continue;
		smin(ans[u], mn);
		smin(ans[u], mnL + mnR);
		smin(dp[i], dpL[i] + mnR);
		smin(dp[i], dpR[i] + mnL);
		smin(dp[i], mnL + mnR);
		smin(dp[i], mn);
	}
	return dp;
}

int main(){
#ifndef ONLINE_JUDGE
	freopen("in.in", "r", stdin);
	freopen("out.out", "w", stdout);
#endif
	int t;
	scanf("%d", &t);
	while(t--){
		scanf("%d", &n);
		for(int i = 1; i <= n; i++) scanf("%d", A + i);
		for(int i = 2; i <= n; i++) scanf("%d", pa + i);
		for(int i = 1; i <= n; i++) ls[i] = rs[i] = 0;
		for(int i = 2; i <= n; i++){
			if(ls[pa[i]]) rs[pa[i]] = i;
			else ls[pa[i]] = i;
		}
		dfs(1);
		printf("%d\n", ans[1]);
	}
	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
1 2 3 1
1 1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Paths on a Tree
time_limit_ms: 4000
memory_limit_mb: 1024
Description: Note the unusual definition of $\text{MEX}$ in this problem.

Piggy gave Turtle a binary tree$^{\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.

If a set of paths $P = \\{(x_i, y_i)\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.

Turtle defines the value of a set of paths as $\sum\limits_{(x, y) \in P} f(x, y)$, where $f(x, y)$ denotes the $\text{MEX}^{\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).

Turtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!

$^{\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.

$^{\ddagger}\text{MEX}$ of a collection of integers $c_1, c_2, \ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\text{MEX}$ of $[2, 3]$ is $1$.

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 a single integer $n$ ($2 \le n \le 2.5 \cdot 10^4$) — the number of vertices in the tree.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the sequence $a$.

The third line of each test case contains $n - 1$ integers $p_2, p_3, \ldots, p_n$ ($1 \le p_i < i$) — the parent of each vertex in the tree.

Additional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.

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

Output: For each test case, output a single integer — the minimum value over all good sets of paths.

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

Note: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:

![](CDN_BASE_URL/45a1635df0fcab1b711d2165163f0dfb)

The good set of paths with the minimum value is $\\{(2, 3), (4, 5)\\}$.

Note that in this test case $\\{(4, 5)\\}$ and $\\{(3, 4), (4, 5)\\}$ are not good sets of paths because each edge should be covered exactly once.

In the second test case, the tree is as follows:

![](CDN_BASE_URL/a41b8a69adf0471a8353a08785fa6375)

The set of good paths with the minimum value is $\\{(1, 2), (1, 3), (4, 5)\\}$.

In the third test case, the tree is as follows:

![](CDN_BASE_URL/7e62b6f397447336b5c5732c0d915ebd)

The set of good paths with the minimum value is $\\{(1, 6), (3, 5)\\}$.

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

namespace SOL {

using i64 = long long;
void debug(const char *msg, ...) {
#ifdef CLESIP
    va_list arg;
    static char pbString[512];
    va_start(arg,msg);
    vsprintf(pbString,msg,arg);
    cerr << "[DEBUG] " << pbString << "\n";
    va_end(arg);
#endif
}
template<typename T, typename L>
bool chkmax(T &x, L y) { if (x < y) return x = y, true; return false; }
template<typename T, typename L>
bool chkmin(T &x, L y) { if (y < x) return x = y, true; return false; }

const int N = 2.5e4 + 10;
const int K = 3900;
const int inf = 1e9;
int n, f[N][K], a[N], lim;
vector<int> G[N];

void dfs(int x) {
	for (int i = 1; i <= lim; i ++)
		f[x][i] = 0;
	if (a[x] <= lim)
		f[x][a[x]] = inf;
	for (int y : G[x])
		dfs(y);
	if ((int) G[x].size() == 1) {
		int minx = inf;
		for (int i = 1; i <= lim; i ++)
			if (i != a[x])
				chkmin(minx, f[G[x][0]][i] + i);
		for (int i = 1; i <= lim; i ++)
			if (i != a[x])
				f[x][i] = min(minx, f[G[x][0]][i]);
		if (x == 1)
			cout << minx << "\n";
	} else if ((int) G[x].size() == 2) {
		int minx = inf, miny = inf, minm = inf;
		for (int i = 1; i <= lim; i ++) {
			if (i == a[x])
				continue;
			chkmin(minx, f[G[x][0]][i] + i);
			chkmin(miny, f[G[x][1]][i] + i);
			chkmin(minm, f[G[x][0]][i] + f[G[x][1]][i] + i);
		}
		for (int i = 1; i <= lim; i ++)
			if (a[x] != i)
				f[x][i] = min({ f[G[x][0]][i] + miny, f[G[x][1]][i] + minx, minm, minx + miny });
		if (x == 1)
			cout << min(minm, minx + miny) << "\n";
	} else if (x == 1)
		cout << "0\n";
}

void solve() {
	cin >> n;
	for (int i = 1; i <= n; i ++)
		G[i].clear();
	for (int i = 1; i <= n; i ++)
		cin >> a[i];
	for (int i = 2, p; i <= n; i ++) {
		cin >> p;
		G[p].push_back(i);
	}
	int l = 1, r = n; lim = n;
	auto calc = [&](int t) {
		int ret = 0;
		for (int i = 1; i <= t; i ++) {
			if (i <= 3)
				ret += (t + i + 3) / (i + 4) - 1;
			else
				ret += (t + i + 2) / (i + 3) - 1;
		}
		return ret;
	};
	while (l <= r) {
		int mid = (l + r) >> 1;
		if (calc(mid) <= n) {
			lim = mid;
			l = mid + 1;
		} else {
			r = mid - 1;
		}
	}
	lim = n + 1;
	dfs(1);
}

}
int main() {
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	int tt; cin >> tt;
	while (tt --)
		SOL::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
5
5
3 2 2 1 1
1 1 2 2
5
3 2 1 1 1
1 1 2 2
6
1 2 1 2 1 3
1 2 3 3 4
7
2 1 2 3 1 2 1
1 1 2 2 3 3
10
1 2 2 1 4 2 3 1 2 1
1 1 2 2 3 3 4 5 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rectangle Arrangement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.

You will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.

What is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$).

The $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \le w_i, h_i \le 100$).

Output: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.

Examples:
input:
5
5
1 5
2 4
3 3
4 2
5 1
3
2 2
1 1
1 2
1
3 2
3
100 100
100 100
100 100
4
1 4
2 3
1 5
3 2
output:
20
8
10
400
16

Note: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.

![](CDN_BASE_URL/877686333ced56f689851506a90eefb8)

After all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.

In the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.

Buggy Code:
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp>
#include <functional> // for less

using namespace __gnu_pbds;
using namespace std;

//----------------------------DEBUG--------------------------------------------------------------------------
// #include "template.cpp"
//-----------------------------------------------------------------------------------------------------------
typedef tree<int, null_type, less<int>, rb_tree_tag,
            tree_order_statistics_node_update>
    ordered_set;

//-------------------------MACROS-------------------------------------

#define _GLIBCXX_DEBUG
#define int long long
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define F0R(i, a) FOR(i, 0, a)
#define scan(arr) for(auto &i:arr)cin>>i;
#define scan_mp(arr,map) for(auto &i:arr){cin>>i;mp[i]++;}
#define print(arr,n) for(int i=0;i<n;i++)cout<<arr[i]<<" \n"[i==n-1];
#define print2d(arr,n,m) FOR(i,0,n)FOR(j,0,m)cout<<arr[i][j]<<" \n"[j==m-1];
#define VI vector<int>
#define VVI vector<VI>
#define PII pair<int, int>
#define SI set<int>
#define MSI multiset<int>
#define OS ordered_set
#define MII map<int, int>
#define all(x) (x).begin(), (x).end()
#define lcm(a,b) a/__gcd(a,b)*b
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define setBits(x) __builtin_popcountll(x)

//--------------------------CONSTANTS---------------------------------

const int N=3001;
const int M=1e9+7;
const int M2=998244353;
const int inf=1e18;

//------------------------------------------------------------------------------------------------------------------------------------------------------//

int binexp(int a,int b){
    if(b==0)return 1;
    int res=binexp(a,b/2);
    if(b&1)return (a*res*res)%M;
    else return (res*res)%M;
}

int binexp2(int a,int b){
    int ans=1;
    while(b){
        if(b&1)ans=(ans * a) % M;
        a=(a * a) % M;
        b>>=1;
    }
    return ans;
}

int binMul(int a,int b){
    int ans=0;
    while(b){
        if(b&1)ans=(ans+a)%M;
        a=(a+a)%M;
        b>>=1;
    }
    return ans;
}

int gcd(int a,int b){
    if(b==0)return a;
    return gcd(b,a%b);
}

//------------------------------------------------------------------------------------------------------------------------------------------------------//

vector<bool>prime;
VI hp;
void sieve(){
    hp.resize(N,0);
    prime.resize(N,0);
    FOR(i,0,N)prime[i]=1,hp[i]=0;
    prime[0]=prime[1]=0;
    FOR(i,2,N){
        if(!hp[i]){
            for(int j=i;j<N;j+=i){
                hp[j]=i;
                if(j>i)
                prime[j]=0;
            }
        }
    }
}
VI fact;
void pre(){
    fact.resize(N,0);
    fact[0]=fact[1]=1;
    FOR(i,2,N){
        fact[i]=fact[i-1]*i;
        fact[i]%=M;
    }
}
void prefix(int *pref,int *arr,int n,int i){
    pref[i]=arr[0];
    FOR(j,i+1,n+i){
        pref[j]=pref[j-1]+arr[j-i];
    }
}

//function<return_type(data_type,...)> name ;

//----------SOLVE------------------------------
//----------HERE-------------------------------

int lb(int *arr,int n,int x){
    int l=-1,r=n;
    while(l<r-1){
        int m=(l+r)/2;
        if(arr[m]>=x)r=m;
        else l=m;
    }
    return r;
}

int ub(int *arr,int n,int x){
    int l=-1,r=n;
    while(l<r-1){
        int m=(l+r)/2;
        if(arr[m]<=x)l=m;
        else r=m;
    }
    return l;
}
struct hash_pair {
    template <class T1, class T2>
    size_t operator()(const pair<T1, T2>& p) const
    {
        // Hash the first element
        size_t hash1 = hash<T1>{}(p.first);
        // Hash the second element
        size_t hash2 = hash<T2>{}(p.second);
        // Combine the two hash values
        return hash1
               ^ (hash2 + 0x9e3779b9 + (hash1 << 6)
                  + (hash1 >> 2));
    }
};

bool solver(int *arr,int i,int n,int sum){
    int ans=0;
    if(i==n){
        if(sum%360==0)return 1;
        else return 0;
    }
    ans|=solver(arr,i+1,n,sum+arr[i]);
    ans|=solver(arr,i+1,n,sum-=arr[i]);
    return ans;
}
void solve1() {

    int n;
    cin>>n;
    PII p[n];
    int mxh=0;
    FOR(i,0,n){
        cin>>p[i].ff>>p[i].ss;
    }
    sort(p,p+n);
    int ans=2*(p[0].ff+p[0].ss);
    int w=p[0].ff;
    int h=p[0].ss;
    FOR(i,1,n){
        if(p[i].ff>w)ans+=2*(p[i].ff-w);
        if(p[i].ss>h)ans+=2*(p[i].ss-h);
        w=p[i].ff;
        h=p[i].ss;
    }
    cout<<ans<<endl;

}
//----------------MAIN FUNCTION--------------
int32_t main(){
    #ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t=1;
    cin>>t;
    while(t--){
        solve1();
    }
    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
5
1
100 1
1
1 100
1
1 1
1
100 100
1
100 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rectangle Arrangement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.

You will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.

What is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$).

The $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \le w_i, h_i \le 100$).

Output: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.

Examples:
input:
5
5
1 5
2 4
3 3
4 2
5 1
3
2 2
1 1
1 2
1
3 2
3
100 100
100 100
100 100
4
1 4
2 3
1 5
3 2
output:
20
8
10
400
16

Note: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.

![](CDN_BASE_URL/877686333ced56f689851506a90eefb8)

After all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.

In the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.

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

using i32 = int;
using i64 = long long;
const i32 INF = 1e9;

void solve() {
  i32 n;
  cin >> n;

  vector <i32> H (102);
  for (i32 i = 0; i < n; ++i) {
    i32 w, h;
    cin >> w >> h;
    H[w] = h;
  }

  for (i32 i = 100; i >= 1; i--) {
    H[i] = max(H[i], H[i + 1]);
  }

  i32 ans = 0;
  for (i32 i = 1; i <= 101; ++i) {
    // cout << H[i] << ' ';
    ans += 2 * (H[i] != 0);
    ans += abs(H[i] - H[i - 1]);
  }
  cout << ans << '\n';
}

int main() {
  ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
  i32 tests = 1;
  cin >> tests;
  for (i32 test = 0; test < tests; ++test) {
    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
3
1 100
1 100
1 100
2
1 1
100 1
1
100 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rectangle Arrangement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.

You will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.

What is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$).

The $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \le w_i, h_i \le 100$).

Output: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.

Examples:
input:
5
5
1 5
2 4
3 3
4 2
5 1
3
2 2
1 1
1 2
1
3 2
3
100 100
100 100
100 100
4
1 4
2 3
1 5
3 2
output:
20
8
10
400
16

Note: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.

![](CDN_BASE_URL/877686333ced56f689851506a90eefb8)

After all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.

In the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int                 long long
#define fios();             ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

void solution();

signed main() {
    fios();
    int t = 1;
    cin >> t;
    while(t--){
        solution();
        cout<<endl;
    }
}

void solution() {
    int n, maxi=0, x, y;
    cin>>n;
    map<int, int>m;
    for(int i=0; i<n; i++){
        cin>>x;
        cin>>y;
        m[x]=(max(m[x],y));
        //m.insert({x, y});
        maxi=max(maxi, x);
    }
    int ans=0, i=0, tem1=0, tem2=0;
    for(auto val: m){
        ans+=abs(val.first-tem1);
        ans+=abs(val.second-tem2);
        tem1=val.first;
        tem2=val.second;
        i++;
    }
    auto it=m.end();
    it--;
    ans+=(it->first+it->second);
    cout<<ans;
}

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

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

Think step by step.

Response:
```plaintext
5
5
1 5
2 4
3 3
4 2
5 1
3
2 2
1 1
1 2
1
3 2
3
100 100
100 100
100 100
4
1 4
2 3
1 5
3 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rectangle Arrangement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.

You will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.

What is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$).

The $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \le w_i, h_i \le 100$).

Output: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.

Examples:
input:
5
5
1 5
2 4
3 3
4 2
5 1
3
2 2
1 1
1 2
1
3 2
3
100 100
100 100
100 100
4
1 4
2 3
1 5
3 2
output:
20
8
10
400
16

Note: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.

![](CDN_BASE_URL/877686333ced56f689851506a90eefb8)

After all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.

In the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.

Buggy Code:
t=int(input())
for _ in range(t):
    n=int(input())
    s=[0]*100
    for x in range(n):
        h,w=map(int,input().split())
        for i in range(w):
            s[i]=max(h,s[i])
    c=s[0]+2
    for i in range(1,100):
        if s[i]==0:
            break
        c+=abs(s[i-1]-s[i])+2
#         print(c)
    print(c+s[i-1])

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases where the code might fail. The buggy code calculates the perimeter of the black regions by maintaining an array `s` that keeps track of the maximum height at each column. However, it might not handle cases where the stamps are not optimally placed to minimize the perimeter.

Let's consider a test case where the stamps are of different sizes and orientations, and the arrangement of the stamps can significantly affect the perimeter.

Here is a small-scale test input that can expose the bug:

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

### Explanation:

1. **Test Case 1:**
   - **Input:** 4 stamps, each of size 1x1.
   - **Expected Output:** 8
   - **Explanation:** All stamps can be placed in a 2x2 grid, forming a single connected region with a perimeter of 8.

2. **Test Case 2:**
   - **Input:** 3 stamps, with sizes 1x2, 2x1, and 1x2.
   - **Expected Output:** 10
   - **Explanation:** The stamps can be placed in a 3x2 grid, forming a single connected region with a perimeter of 10.

3. **Test Case 3:**
   - **Input:** 5 stamps, with sizes 2x3, 3x2, 2x2, 1x4, and 4x1.
   - **Expected Output:** 24
   - **Explanation:** The stamps can be placed in a 5x4 grid, forming a single connected region with a perimeter of 24.

These test cases are designed to check if the code correctly handles:
- Small and identical stamps.
- Stamps of different sizes and orientations.
- Stamps that can be optimally placed to minimize the perimeter.

By using these test cases, we can verify if the buggy code correctly calculates the minimum perimeter of the connected regions of black squares.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rectangle Arrangement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.

You will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.

What is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$).

The $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \le w_i, h_i \le 100$).

Output: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.

Examples:
input:
5
5
1 5
2 4
3 3
4 2
5 1
3
2 2
1 1
1 2
1
3 2
3
100 100
100 100
100 100
4
1 4
2 3
1 5
3 2
output:
20
8
10
400
16

Note: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.

![](CDN_BASE_URL/877686333ced56f689851506a90eefb8)

After all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.

In the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.

Buggy Code:
def min_total_perimeter(rectangles):
    # Step 3: Calculate max possible overlap for each pair of rectangles
    overlaps = []
    for i in range(len(rectangles)):
        for j in range(i+1, len(rectangles)):
            wi, hi = rectangles[i]
            wj, hj = rectangles[j]
            overlap_x = min(wi, wj)
            overlap_y = min(hi, hj)
            overlaps.append(overlap_x * 2 + overlap_y * 2)

    # Step 4 & 5: Sort overlaps and initialize perimeter counter
    overlaps.sort(reverse=True)
    initial_perimeter = sum([2*(w + h) for w, h in rectangles])

    # Step 6: Subtract effective overlaps from the perimeter count
    for k in range(min(len(overlaps), len(rectangles)-1)):
        initial_perimeter -= overlaps[k]

    return initial_perimeter

# Input reading function
def main():
    t = int(input())
    for _ in range(t):
        n = int(input())
        rectangles = [tuple(map(int, input().split())) for __ in range(n)]
        print(min_total_perimeter(rectangles))

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
3
1
100 100
2
100 1
1 100
3
100 100
1 1
100 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rectangle Arrangement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.

You will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.

What is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$).

The $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \le w_i, h_i \le 100$).

Output: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.

Examples:
input:
5
5
1 5
2 4
3 3
4 2
5 1
3
2 2
1 1
1 2
1
3 2
3
100 100
100 100
100 100
4
1 4
2 3
1 5
3 2
output:
20
8
10
400
16

Note: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.

![](CDN_BASE_URL/877686333ced56f689851506a90eefb8)

After all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.

In the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.

Buggy Code:
def min_perimeter(rectangles):
    # Sort rectangles by their perimeter in descending order
    rectangles.sort(key=lambda x: 2 * (x[0] + x[1]), reverse=True)

    # Initialize total perimeter with the perimeter of the largest rectangle
    total_perimeter = 2 * (rectangles[0][0] + rectangles[0][1])

    # For subsequent rectangles, add only the unique edges that don't overlap
    for i in range(1, len(rectangles)):
        total_perimeter += 2 * (rectangles[i][0] + rectangles[i][1])
        # Subtract the overlapped part (assumption: minimal overlap)
        total_perimeter -= 2 * min(rectangles[i-1][0], rectangles[i][0])
        total_perimeter -= 2 * min(rectangles[i-1][1], rectangles[i][1])

    return total_perimeter

# Read input and process each test case
def main():
    import sys
    input = sys.stdin.read
    data = input().split()

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

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

        for _ in range(n):
            w = int(data[index])
            h = int(data[index + 1])
            rectangles.append((w, h))
            index += 2

        result = min_perimeter(rectangles)
        results.append(result)

    for res in results:
        print(res)

if __name__ == "__main__":
    main()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rectangle Arrangement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.

You will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.

What is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$).

The $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \le w_i, h_i \le 100$).

Output: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.

Examples:
input:
5
5
1 5
2 4
3 3
4 2
5 1
3
2 2
1 1
1 2
1
3 2
3
100 100
100 100
100 100
4
1 4
2 3
1 5
3 2
output:
20
8
10
400
16

Note: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.

![](CDN_BASE_URL/877686333ced56f689851506a90eefb8)

After all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.

In the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.

Buggy Code:
import java.util.*;

public class A_Rectangle_Arrangement {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for (int z = 0; z < t; z++) {
            int n;
            n = sc.nextInt();
            int arr[][] = new int[n][2];
            for (int i = 0; i < n; i++) {
                arr[i][0] = sc.nextInt();
                arr[i][1] = sc.nextInt();
            }

            Arrays.sort(arr,(a,b)->{
                if(a[0]!=b[0]) return a[0]-b[0];
                return b[1]-a[1];
            });

            int ans=0;
            ans+=arr[0][0]+arr[n-1][0]+arr[0][1]+arr[n-1][1];

            int lw=arr[0][0],lh=arr[0][1];
            for(int i=1;i<n;i++){
                if(arr[i][0]>lw){
                    ans+=arr[i][0]-lw;
                    lw=arr[i][0];
                    ans+=lh-arr[i][1];
                    lh=arr[i][1];
                }
            }
            System.out.println(ans);
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5
1 1
1 1
1 1
1 1
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rectangle Arrangement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.

You will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.

What is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$).

The $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \le w_i, h_i \le 100$).

Output: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.

Examples:
input:
5
5
1 5
2 4
3 3
4 2
5 1
3
2 2
1 1
1 2
1
3 2
3
100 100
100 100
100 100
4
1 4
2 3
1 5
3 2
output:
20
8
10
400
16

Note: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.

![](CDN_BASE_URL/877686333ced56f689851506a90eefb8)

After all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.

In the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.

Buggy Code:
import com.sun.source.tree.Tree;

import java.awt.*;
import java.awt.desktop.SystemEventListener;
import java.util.*;
import java.util.List;

public class Main {
    public static int mod=(int)1e9+7;
    public static class pair{
        int key;
        int val;
        pair(int key,int val){
            this.key=key;
            this.val=val;
        }
        void dis(){
            System.out.println(this.key+" "+this.val);
        }
    }
    static void dis(long arr[],int st,int end){
        if(st>end){
            for(int i=st-1;i>=end;i--){
                System.out.print(arr[i]+" ");
            }
            System.out.println();
        }
        else{
            for(int i=st;i<end;i++){
                System.out.print(arr[i]+" ");
            }
            System.out.println();
        }
    }
    public static long solve(int n) {
        if (n == 1) return 0;
        if (n == 2) return 1;
        long[] dp = new long[n + 1];
        dp[1] = 0;
        dp[2] = 1;
        for (int i = 3; i <= n; i++) {
            dp[i] = ((i - 1) * (dp[i - 1] + dp[i -   2]) % mod) % mod;
        }
        return dp[n];
    }
    static int gcd(int a,int b){
        if(b==0) return a;
        else return gcd(b,a%b);
    }
    static int justSmaller(List<Integer> l,int x){
        int st=0;
        int n=l.size();
        int end=n-1;
        int ans=0;
        while(st<=end){
            int mid=st+(end-st)/2;
            if(l.get(mid)<=x){
                ans=l.get(mid);
                st=mid+1;
            }
            else end=mid-1;
        }
        return ans;
    }
    static int justGreater(long arr[],int x){
        int st=0;
        int end=arr.length-1;
        int ans=arr.length-1;
        while(st<=end){
            int mid=st+(end-st)/2;
            if(arr[mid]>x){
                end=mid-1;
                ans=mid;
            }
            else st=mid+1;
        }
        return ans;
    }
    static int bp(int a, int b) {
        if (b == 0) return 1;
        int res = bp(a, b / 2);
        res = (int) ((long) res * res % mod);
        if (b % 2 == 1) return (int) ((long) res * a % mod);
        else return res;
    }
    static Comparator<pair> comp=new Comparator<pair>() {
        @Override
        public int compare(pair o1, pair o2) {
            if(o1.val>o2.val) return 1;
            else if(o1.val==o2.val) return 0;
            else return -1;
        }
    };
    static long binexp(long a,long b){
        if(b==0) return (long)1;
        if(b==1) return a%mod;
        long res=binexp(a,b/2);
        if(b%2==0) return (res*res)%mod;
        else return (((res * res) % mod) * a) % mod;
    }
    //Arrays.sort(array,(a,b)->Integer.compare(a[0],b[0]));

    //Arrays.sort(arr_name, Comparator.comparingInt(i -> arr[i]));
    static boolean check(char arr[][],int r,int c){
        int n=8;
        for(int i=0;i<n;i++){
            if(arr[r][i]=='#') return false;
        }
        for(int i=0;i<n;i++){
            if(arr[i][c]=='#') return false;
        }
        return true;
    }
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0){
            int n=sc.nextInt();
            int sum=0;
            int maxa=0;
            int maxb=0;
            int preva=0;
            int prevb=0;
            while(n-->0){
                int a=sc.nextInt();
                int b=sc.nextInt();
                if(a>=maxa && b>=maxb){
                    sum=2*(a+b);
                    maxa=a;
                    maxb=b;
                    preva=a;
                    prevb=b;
                }
                else if(a>maxa){
                    sum+=2*(a-maxa);
                    preva=maxa;
                    maxa=a;
                }
                else if(b>maxb){
                    sum+=2*(b-maxb);
                    prevb=maxb;
                    maxb=b;
                }
                else if(a<maxa && b<maxb){
                    if(maxb!=maxa){
                        sum+=(a-preva+b-prevb);
                    }
                }

            }
            System.out.println(sum);
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
3
2 2
1 1
1 1
2
100 1
1 100
1
50 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rectangle Arrangement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.

You will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.

What is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$).

The $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \le w_i, h_i \le 100$).

Output: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.

Examples:
input:
5
5
1 5
2 4
3 3
4 2
5 1
3
2 2
1 1
1 2
1
3 2
3
100 100
100 100
100 100
4
1 4
2 3
1 5
3 2
output:
20
8
10
400
16

Note: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.

![](CDN_BASE_URL/877686333ced56f689851506a90eefb8)

After all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.

In the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std ;
int main ()
{
    int t ;
    cin>> t ;
    while(t-->0)
    {
        int n ; cin>> n ;
        int w , h ; cin >>w>>h ;
        int a = w , b = h ;
        int p = 2*w + 2*h ;
            for(int i = 1 ; i < n ; i++ ) {
                cin >>w>>h ;
        if(w>a){
            p+=2;
        }
        if(h>b){
            p+=2;
        }
            }
        cout<<p <<endl;
    }

}

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

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

Think step by step.

Response:
```plaintext
3
2
1 1
100 100
2
100 100
1 1
4
1 1
1 1
100 100
100 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rectangle Arrangement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.

You will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.

What is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$).

The $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \le w_i, h_i \le 100$).

Output: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.

Examples:
input:
5
5
1 5
2 4
3 3
4 2
5 1
3
2 2
1 1
1 2
1
3 2
3
100 100
100 100
100 100
4
1 4
2 3
1 5
3 2
output:
20
8
10
400
16

Note: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.

![](CDN_BASE_URL/877686333ced56f689851506a90eefb8)

After all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.

In the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.

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

using namespace std;

bool myComparator1(pair<int, int> a, pair<int, int> b) {
    return (a.first > b.first);
}

bool myComparator2(pair<int, int> a, pair<int, int> b) {
    return (a.second > b.second);
}

int main() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
        int a;
        cin >> a;
        if (a != 1) {
            vector<pair<int, int>> v1;
            vector<pair<int, int>> v2;
            int count1 = 0;
            for (int j = 0; j < a; j++) {
                int b, c;
                cin >> b >> c;
                if (b == c) {
                    count1++;
                }
                v1.push_back({b, c});
                v2.push_back({b, c});
            }
            if (count1 == v1.size()) {
                cout << v1[0].first * 4 << endl;
            } else {
                sort(v1.begin(), v1.end(), myComparator1);
                sort(v2.begin(), v2.end(), myComparator2);
                int peri = 0;
                if (v2[0].second > v1[0].second) {
                    peri = v1[0].first + v2[0].second + v2[0].first + v1[0].second;
                } else {
                    peri = v1[0].first + v2[0].second + v1[0].second;
                }

                for (int j = 0; j < (a - 1); j++) {
                    peri += (v1[j].first - v1[j + 1].first) + (v2[a - j - 2].second - v2[a - j - 1].second);
                }
                cout << peri << endl;
            }
        } else {
            int b, c;
            cin >> b >> c;
            cout << 2 * (b + c) << endl;
        }
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rectangle Arrangement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.

You will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.

What is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$).

The $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \le w_i, h_i \le 100$).

Output: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.

Examples:
input:
5
5
1 5
2 4
3 3
4 2
5 1
3
2 2
1 1
1 2
1
3 2
3
100 100
100 100
100 100
4
1 4
2 3
1 5
3 2
output:
20
8
10
400
16

Note: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.

![](CDN_BASE_URL/877686333ced56f689851506a90eefb8)

After all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.

In the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.

Buggy Code:
def findCmn(curr, maxW):
    wL = [0 for _ in range(maxW)]
    for w,h in curr:
        wL[w-1] = max(h,wL[w-1])
    per = maxW
    pH = 0
    for i in range(maxW-1, -1, -1):
        if wL[i] == 0:
            per += 1
        else:
            per += abs(wL[i] - pH) + 1
            pH = wL[i]
    return per + pH

for _ in range(int(input())):
    n = int(input())
    curr = []
    maxW = 0
    for i in range(n):
        w, h = [int(s) for s in input().split()]
        maxW = max(w,maxW)
        curr.append([w,h])
    res = findCmn(curr, maxW)
    print(res)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rectangle Arrangement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.

You will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.

What is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$).

The $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \le w_i, h_i \le 100$).

Output: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.

Examples:
input:
5
5
1 5
2 4
3 3
4 2
5 1
3
2 2
1 1
1 2
1
3 2
3
100 100
100 100
100 100
4
1 4
2 3
1 5
3 2
output:
20
8
10
400
16

Note: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.

![](CDN_BASE_URL/877686333ced56f689851506a90eefb8)

After all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.

In the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.

Buggy Code:
import itertools

def c(f, x, y):
    e = 0

    if f[x][y]:
        e += 1 if x == 0 or not f[x - 1][y] else 0
        e += 1 if y == 0 or not f[x][y - 1] else 0
        e += 1 if x == 99 or not f[x + 1][y] else 0
        e += 1 if y == 99 or not f[x][y + 1] else 0

    return e

p = list(itertools.product(range(100), repeat=2))

for _ in range(int(input())):
    n = int(input())

    f = [[False for _ in range(100)] for _ in range(100)]

    for _ in range(n):
        a, b = list(map(int, input().split()))

        for i in range(a):
            for j in range(b):
                f[i][j] = True

    print(sum(map(lambda a, b : c(f, a, b), [p[i][0] for i in range(len(p))], [p[i][1] for i in range(len(p))])))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rectangle Arrangement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.

You will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.

What is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$).

The $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \le w_i, h_i \le 100$).

Output: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.

Examples:
input:
5
5
1 5
2 4
3 3
4 2
5 1
3
2 2
1 1
1 2
1
3 2
3
100 100
100 100
100 100
4
1 4
2 3
1 5
3 2
output:
20
8
10
400
16

Note: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.

![](CDN_BASE_URL/877686333ced56f689851506a90eefb8)

After all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.

In the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.

Buggy Code:
import java.io.Writer;
import java.io.BufferedWriter;
import java.io.OutputStreamWriter;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.io.InputStream;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.ArrayList;
import java.util.InputMismatchException;
import java.util.*;
import java.io.*;
import java.math.*;
import java.util.concurrent.TimeUnit;
public class Main{
    public static MyScanner sc;
    public static int solve(int n){
        int a[]=new int[n];
        int b[]=new int[n];
        for(int i=0;i<n;i++){
            int w=sc.nextInt();
            int h=sc.nextInt();
            a[i]=w;
            b[i]=h;
        }
        int ans=b[0]+b[n-1];
        for(int i=1;i<n;i++){
            if(a[i-1]<a[i]){
                ans+=Math.abs(b[i-1]-b[i]);
            }
        }
        Arrays.sort(a);
        ans+=a[n-1]*2;
        return ans;
        }
   public static void main(String[] args) {
      sc = new MyScanner();
      out = new PrintWriter(new BufferedOutputStream(System.out));
        int t=sc.nextInt();
        while(t-->0){
            int n=sc.nextInt();
            out.println(solve(n));
        }
      // Start writing your solution here. -------------------------------------

      /*
      int n      = sc.nextInt();        // read input as integer
      long k     = sc.nextLong();       // read input as long
      double d   = sc.nextDouble();     // read input as double
      String str = sc.next();           // read input as String
      String s   = sc.nextLine();       // read whole line as String

      int result = 3*n;
      out.println(result);                    // print via PrintWriter
      */

      // Stop writing your solution here. -------------------------------------
      out.close();
   }

   //-----------PrintWriter for faster output---------------------------------
   public static PrintWriter out;

   //-----------MyScanner class for faster input----------
   public static class MyScanner {
      BufferedReader br;
      StringTokenizer st;

      public MyScanner() {
         br = new BufferedReader(new InputStreamReader(System.in));
      }

      String next() {
          while (st == null || !st.hasMoreElements()) {
              try {
                  st = new StringTokenizer(br.readLine());
              } catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }

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

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

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

      String nextLine(){
          String str = "";
      try {
         str = br.readLine();
      } catch (IOException e) {
         e.printStackTrace();
      }
      return str;
      }

   }
   //--------------------------------------------------------
}

**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 the following:

1. **Edge Cases**: The code might not handle edge cases correctly, such as when all stamps have the same dimensions or when the stamps are very small or very large.
2. **Non-Optimal Perimeter Calculation**: The code seems to be incorrectly calculating the perimeter of the connected regions. It only considers the difference in heights and does not account for the widths of the stamps.

Let's create a test case that highlights these issues:

- **Test Case 1**: All stamps have the same dimensions.
- **Test Case 2**: Stamps with very different dimensions to see if the code handles the perimeter calculation correctly.
- **Test Case 3**: Stamps that can be placed inside each other to minimize the perimeter.

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

```plaintext
3
3
10 10
10 10
10 10
3
1 100
100 1
100 100
4
5 5
5 10
10 5
10 10
```

Explanation:
1. **Test Case 1**: All stamps are 10x10. The expected output should be 40 (since they can be placed in a single region).
2. **Test Case 2**: Stamps with very different dimensions. The expected output should be 402 (since the smallest stamp can be placed inside the largest one, and the middle one can be placed next to the largest one).
3. **Test Case 3**: Stamps that can be placed inside each other to minimize the perimeter. The expected output should be 40 (since they can be placed in a single region).

This test input should help expose the bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rectangle Arrangement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.

You will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.

What is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$).

The $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \le w_i, h_i \le 100$).

Output: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.

Examples:
input:
5
5
1 5
2 4
3 3
4 2
5 1
3
2 2
1 1
1 2
1
3 2
3
100 100
100 100
100 100
4
1 4
2 3
1 5
3 2
output:
20
8
10
400
16

Note: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.

![](CDN_BASE_URL/877686333ced56f689851506a90eefb8)

After all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.

In the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.

Buggy Code:
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        int t = scanner.nextInt();

        for (int testCase = 0; testCase < t; testCase++) {
            int n = scanner.nextInt();
            int[][] stamps = new int[n][2];

            // Read width and height for each stamp
            for (int i = 0; i < n; i++) {
                stamps[i][0] = scanner.nextInt();
                stamps[i][1] = scanner.nextInt();
            }
            Arrays.sort(stamps, (a, b) -> Integer.compare(a[0], b[0]));
            if(n == 1){
                System.out.println(2 * (stamps[0][0] + stamps[0][1]));
            }else if(n == 2){
                if(stamps[0][1] < stamps[1][1]){
                    System.out.println(2 * (stamps[1][0] + stamps[1][1]));
                } else {
                    System.out.println(2*stamps[1][0] + stamps[0][1] + stamps[1][1]);
                }

            } else {
                // Sort by max dimension (width or height)

            int result = 0, i;
            // System.out.println("Stamps after sorting:");
            // for (i = 0; i < n; i++) {
            //     System.out.println("Width: " + stamps[i][0] + ", Height: " + stamps[i][1]);
            // }

            // Print all stamps after sorting
            int start = 0;
            while (start != n-1 && stamps[start][0] == stamps[start + 1][0]) start ++;
            if (start == n - 1){
                result = 2*(stamps[start][0] + stamps[start][1]);
                System.out.println(result);
            } else {
                result = 2*stamps[start][0] + stamps[start][1] + stamps[start][1] - stamps[start + 1][1]  ;

            // System.out.println("result 1: "+result);
            int thisMaxHeight = stamps[n-1][1];
            for (i = n-2; i >= 0; i--) {
               thisMaxHeight = Math.max(stamps[i][1], thisMaxHeight);
               stamps[i][1]  = thisMaxHeight;
            }

            // System.out.println("Stamps after normalizing:");
            // for (i = 0; i < n; i++) {
            //     System.out.println("Width: " + stamps[i][0] + ", Height: " + stamps[i][1]);
            // }
            for (i = start + 1; i < n-1; i++) {
               result += 2*(stamps[i][0] - stamps[i-1][0]);
               result += stamps[i][1] - stamps[i+1][1];
            //   System.out.println("result: "+result);
            }
            result += 2*(stamps[n-1][0] - stamps[n-2][0]);
            result += stamps[n-1][1];

            System.out.println(result);
            // System.out.println("=======");
            }
            }

        }

        scanner.close();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider edge cases and scenarios that are likely to cause issues. The code has several potential problems, including:

1. **Incorrect Handling of Edge Cases**: The code might not handle cases where all stamps have the same width or height correctly.
2. **Incorrect Perimeter Calculation**: The code might not correctly calculate the perimeter when stamps are stacked or placed in a way that minimizes the perimeter.
3. **Sorting and Normalization**: The code might not correctly sort and normalize the stamps, leading to incorrect perimeter calculations.

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

1. **All Stamps with the Same Width**: This will test the code's ability to handle a uniform width.
2. **Stamps with Varying Heights but the Same Width**: This will test the code's ability to handle a sequence of stamps with the same width but different heights.
3. **Stamps with Varying Widths and Heights**: This will test the code's ability to handle a more general case.

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

```plaintext
3
3
5 5
5 5
5 5
4
5 5
5 10
5 15
5 20
5
1 10
2 20
3 30
4 40
5 50
```

### Explanation:
1. **Test Case 1**: All stamps have the same width (5) and height (5). The expected output is 20, as all stamps will form a single connected region.
2. **Test Case 2**: All stamps have the same width (5) but different heights (5, 10, 15, 20). The expected output is 60, as the stamps will form a single connected region with a height of 50.
3. **Test Case 3**: Stamps have varying widths and heights. The expected output is 120, as the stamps will form a single connected region with a width of 15 and a height of 150.

This test input should expose any issues in the code's handling of uniform and varying dimensions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rectangle Arrangement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.

You will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.

What is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$).

The $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \le w_i, h_i \le 100$).

Output: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.

Examples:
input:
5
5
1 5
2 4
3 3
4 2
5 1
3
2 2
1 1
1 2
1
3 2
3
100 100
100 100
100 100
4
1 4
2 3
1 5
3 2
output:
20
8
10
400
16

Note: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.

![](CDN_BASE_URL/877686333ced56f689851506a90eefb8)

After all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.

In the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.

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

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(NULL);
    int t;
    cin >> t;

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

        vector<int> w(n), h(n);
        for(int i=0; i<n; i++){
            cin>>w[i]>>h[i];
        }

        int ans = 0;

        if(n==1){
            ans = 2*(w[0]+h[0]);
        }
        else {
            int max_h = *max_element(h.begin(), h.end());
            ans+=2*max_h;

            int max_w = *max_element(w.begin(), w.end());
            int min_w = *min_element(w.begin(), w.end());

            for(int i=0; i<n; i++){
                if(w[i] != min_w){
                    ans+=2*min_w;
                }
            }
            ans+=2*min_w;
        }

        cout<<ans<<endl;

    }
}
// (◔◡◔) (◔◡◔) (◔◡◔)

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

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

Think step by step.

Response:
```plaintext
5
2
1 1
100 100
3
1 1
1 1
100 100
1
100 100
3
1 1
1 1
1 1
4
1 1
1 1
1 1
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Rectangle Arrangement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.

You will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.

What is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$).

The $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \le w_i, h_i \le 100$).

Output: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.

Examples:
input:
5
5
1 5
2 4
3 3
4 2
5 1
3
2 2
1 1
1 2
1
3 2
3
100 100
100 100
100 100
4
1 4
2 3
1 5
3 2
output:
20
8
10
400
16

Note: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.

![](CDN_BASE_URL/877686333ced56f689851506a90eefb8)

After all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.

In the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.

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

#define fastio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define MOD 1000000007
#define MOD1 998244353
#define INF 1e18
#define nline "\n"
#define pb push_back
#define ppb pop_back
#define mp make_pair
#define ff first
#define ss second
#define PI 3.141592653589793238462
#define set_bits __builtin_popcountll
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()

typedef long long ll;
typedef unsigned long long ull;
typedef long double lld;
// typedef tree<pair<int, int>, null_type, less<pair<int, int>>, rb_tree_tag, tree_order_statistics_node_update > pbds; // find_by_order, order_of_key
typedef __gnu_pbds::tree<int, __gnu_pbds::null_type, less<int>, __gnu_pbds::rb_tree_tag, __gnu_pbds::tree_order_statistics_node_update> ordered_set;

#ifndef ONLINE_JUDGE
#define debug(x) cerr << #x <<" "; _print(x); cerr << endl;
#else
#define debug(x)
#endif

void _print(ll t) {cerr << t;}
void _print(int t) {cerr << t;}
void _print(string t) {cerr << t;}
void _print(char t) {cerr << t;}
void _print(lld t) {cerr << t;}
void _print(double t) {cerr << t;}
void _print(ull t) {cerr << t;}

template <class T, class V> void _print(pair <T, V> p);
template <class T> void _print(vector <T> v);
template <class T> void _print(set <T> v);
template <class T, class V> void _print(map <T, V> v);
template <class T> void _print(multiset <T> v);
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.ff); cerr << ","; _print(p.ss); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}

bool cmp(pair<int, int> a, pair<int, int> b) {
    if(a.first == b.first) return a.second >= b.second;
    return a.first >= b.first;

}

void solve() {
    int n; cin>>n;
    vector<pair<int, int>> arr(n);
    for(int i = 0; i<n;i++) {
        cin>>arr[i].first>>arr[i].second;
    }
    if(n == 1) {
        int val = 2*(arr[0].first + arr[0].second);
        cout<<val<<endl; return;
    }
    sort(all(arr), cmp);

    ll ans = 0;
    ans += arr[0].first; ans += arr[0].second;
    int wid = arr[0].first; int hei = arr[0].second;
    for(int i = 1; i<n;i++) {
        if(arr[i].first == arr[i-1].first) continue;
        if(arr[i].second <= hei) continue;
        ans += (wid - arr[i].first); ans += (arr[i].second - hei);
        wid = arr[i].first; hei = arr[i].second;
    }
    ans += hei; ans += wid;
    cout<<ans<<endl;
 }

int main() {
#ifndef ONLINE_JUDGE
    freopen("Error.txt", "w", stderr);
#endif

    int t; cin>>t;
    // int t = 1;
    while(t--) solve();

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider the following:

1. **Edge Cases**: Test cases where the dimensions of the stamps are such that they should ideally be able to fit within each other or align in a specific way to minimize the perimeter.
2. **Sorting and Comparison**: The code sorts the stamps based on their dimensions and then processes them. We need to create a scenario where this sorting and processing logic fails to produce the correct minimum perimeter.
3. **Overlapping and Non-overlapping Stamps**: Ensure that the test case includes stamps that can overlap in a way that the code might not handle correctly.

Given these considerations, let's create a test case:

```plaintext
5
3
1 2
2 1
2 2
3
2 2
1 1
1 2
2
2 3
3 2
4
1 1
2 2
1 1
1 1
1
100 100
```

### Explanation of the Test Case:

1. **Test Case 1**:
   - Stamps: (1, 2), (2, 1), (2, 2)
   - The correct minimum perimeter should be 8, but the code might fail to handle the overlap correctly.

2. **Test Case 2**:
   - Stamps: (2, 2), (1, 1), (1, 2)
   - The correct minimum perimeter should be 8, but the code might not handle the smaller stamp fitting inside the larger one correctly.

3. **Test Case 3**:
   - Stamps: (2, 3), (3, 2)
   - The correct minimum perimeter should be 10, but the code might not handle the different dimensions correctly.

4. **Test Case 4**:
   - Stamps: (1, 1), (2, 2), (1, 1), (1, 1)
   - The correct minimum perimeter should be 10, but the code might not handle multiple small stamps correctly.

5. **Test Case 5**:
   - Stamp: (100, 100)
   - The correct minimum perimeter should be 400, which is a simple case but ensures the code handles large dimensions correctly.

This test case is designed to challenge the sorting and processing logic of the code, particularly in scenarios where stamps can overlap or fit within each other.