**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Omnipotent Monster Killer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You, the monster killer, want to kill a group of monsters. The monsters are on a tree with $n$ vertices. On vertex with number $i$ ($1\le i\le n$), there is a monster with $a_i$ attack points. You want to battle with monsters for $10^{100}$ rounds.

In each round, the following happens in order:

  1. All living monsters attack you. Your health decreases by the sum of attack points of all living monsters.    2. You select some (possibly all or none) monsters and kill them. After being killed, the monster will not be able to do any attacks in the future.

There is a restriction: in one round, you cannot kill two monsters that are directly connected by an edge.

If you choose what monsters to attack optimally, what is the smallest health decrement you can have after all rounds?

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

The first line of each test case contains an integer $n$ ($1\le n\le 3\cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,\ldots,a_n$ ($1\le a_i\le 10^{12}$).

The following $n-1$ lines each contain two integers $x,y$ ($1\le x,y\le n$), denoting an edge on the tree connecting vertex $x$ and $y$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, print one integer: the minimum possible health decrement.

Examples:
input:
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
output:
1000000000000
193
57

Note: In the first test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1$, so your health decreases by $10^{12}$. Then kill the monster on vertex $1$.    * In the second round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $10^{12}$.

In the second test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5$, so your health decreases by $47+15+32+29+23=146$. Then kill the monsters on vertex $1,4,5$.    * In the second round: first, receive the attack from the monster on vertex $2,3$, so your health decreases by $15+32=47$. Then kill the monsters on vertex $2,3$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $193$.

In the third test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5,6,7$, so your health decreases by $8+10+2+3+5+7+4=39$. Then kill the monsters on vertex $1,3,6,7$.    * In the second round: first, receive the attack from the monster on vertex $2,4,5$, so your health decreases by $10+3+5=18$. Then kill the monsters on vertex $2,4,5$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $57$.

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

public class Main {

	static ArrayList<Integer>[] nodes;
	static long[] vals;
	static long[][] dp;

	public static void main(String[] args) throws IOException {
		FastIO io = new FastIO();
		int T = io.nextInt();

		for (int t = 0; t < T; t++) {
			int N = io.nextInt();
			vals = new long[N + 1];
			nodes = new ArrayList[N + 1];
			dp = new long[N + 1][50];

			for (int i = 1; i <= N; i++) {
				nodes[i] = new ArrayList<>();
				vals[i] = io.nextLong();
			}

			for (int i = 0; i < N - 1; i++) {
				int a = io.nextInt();
				int b = io.nextInt();
				nodes[a].add(b);
				nodes[b].add(a);
			}

			DFS(1, 0);

			long ans = Long.MAX_VALUE;
			for (int i = 1; i < 50; i++) {
				ans = Math.min(ans, dp[1][i]);
			}

			io.println(ans);
		}

		io.close();
	}

	public static void DFS(int node, int par) {
		for (int child : nodes[node]) {
			if (child != par) DFS(child, node);
		}

		for (int i = 1; i < 50; i++) {
			dp[node][i] = vals[node] * i;
		}

		for (int child : nodes[node]) {
			if (child == par) continue;
			for (int i = 1; i < 50; i++) {
				long minChild = Integer.MAX_VALUE;
				for (int j = 1; j < 50; j++) {
					if (i == j) continue;
					minChild = Math.min(minChild, dp[child][j]);
				}
				dp[node][i] += minChild;
			}
		}
	}
}

class FastIO extends PrintWriter {

	private InputStream stream;

	private byte[] buf = new byte[1 << 16];

	private int curChar;

	private int numChars;

	// standard input

	public FastIO() { this(System.in, System.out); }

	public FastIO(InputStream i, OutputStream o) {

		super(o);

		stream = i;

	}

	// file input

	public FastIO(String i, String o) throws IOException {

		super(new FileWriter(o));

		stream = new FileInputStream(i);

	}

	// throws InputMismatchException() if previously detected end of file

	private int nextByte() {

		if (numChars == -1) { throw new InputMismatchException(); }

		if (curChar >= numChars) {

			curChar = 0;

			try {

				numChars = stream.read(buf);

			} catch (IOException e) { throw new InputMismatchException(); }

			if (numChars == -1) {

				return -1;  // end of file

			}

		}

		return buf[curChar++];

	}

	// to read in entire lines, replace c <= ' '

	// with a function that checks whether c is a line break

	public String next() {

		int c;

		do { c = nextByte(); } while (c <= ' ');

		StringBuilder res = new StringBuilder();

		do {

			res.appendCodePoint(c);

			c = nextByte();

		} while (c > ' ');

		return res.toString();

	}

	public int nextInt() {  // nextLong() would be implemented similarly

		int c;

		do { c = nextByte(); } while (c <= ' ');

		int sgn = 1;

		if (c == '-') {

			sgn = -1;

			c = nextByte();

		}

		int res = 0;

		do {

			if (c < '0' || c > '9') { throw new InputMismatchException(); }

			res = 10 * res + c - '0';

			c = nextByte();

		} while (c > ' ');

		return res * sgn;

	}

	public long nextLong() {  // nextLong() would be implemented similarly

		int c;

		do { c = nextByte(); } while (c <= ' ');

		int sgn = 1;

		if (c == '-') {

			sgn = -1;

			c = nextByte();

		}

		long res = 0;

		do {

			if (c < '0' || c > '9') { throw new InputMismatchException(); }

			res = 10 * res + c - '0';

			c = nextByte();

		} while (c > ' ');

		return res * sgn;

	}

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

}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, let's consider the following aspects:

1. **Tree Structure**: The code should handle a tree structure correctly. A common bug in tree problems is not handling the root node properly or not correctly managing the parent-child relationships.

2. **Edge Cases**: The code should handle edge cases, such as a tree with a single node, a tree with a linear structure (a path), and a tree with a balanced structure (a complete binary tree).

3. **Attack Points**: The attack points should be chosen to maximize the health decrement, especially when the code might not be handling the dynamic programming (DP) state transitions correctly.

4. **Constraints**: The input should be within the constraints provided in the problem description.

Given these considerations, let's create a test input that includes a small tree with a structure that might expose the bug in the DP state transitions.

### Test Input
```plaintext
3
1
1000000000000
4
1 2 3 4
1 2
1 3
1 4
5
1 1 1 1 1
1 2
1 3
3 4
3 5
```

### Explanation
1. **Single Node Tree**:
   - This case is straightforward and should be handled correctly. The health decrement is simply the attack points of the single node.
   - Input: `1 1000000000000`
   - Expected Output: `1000000000000`

2. **Star Tree**:
   - A tree with one central node and three children. This structure tests the ability to handle multiple children correctly.
   - Input: `4 1 2 3 4 1 2 1 3 1 4`
   - Expected Output: `10` (The optimal strategy is to kill the central node first, then the children in subsequent rounds.)

3. **Balanced Tree**:
   - A balanced tree with a root and two levels of children. This structure tests the ability to handle deeper and more balanced trees.
   - Input: `5 1 1 1 1 1 1 2 1 3 3 4 3 5`
   - Expected Output: `5` (The optimal strategy is to kill the root node first, then the children in subsequent rounds.)

This test input is designed to be small and complete, while also covering a range of tree structures that might expose the bug 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: The Omnipotent Monster Killer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You, the monster killer, want to kill a group of monsters. The monsters are on a tree with $n$ vertices. On vertex with number $i$ ($1\le i\le n$), there is a monster with $a_i$ attack points. You want to battle with monsters for $10^{100}$ rounds.

In each round, the following happens in order:

  1. All living monsters attack you. Your health decreases by the sum of attack points of all living monsters.    2. You select some (possibly all or none) monsters and kill them. After being killed, the monster will not be able to do any attacks in the future.

There is a restriction: in one round, you cannot kill two monsters that are directly connected by an edge.

If you choose what monsters to attack optimally, what is the smallest health decrement you can have after all rounds?

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

The first line of each test case contains an integer $n$ ($1\le n\le 3\cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,\ldots,a_n$ ($1\le a_i\le 10^{12}$).

The following $n-1$ lines each contain two integers $x,y$ ($1\le x,y\le n$), denoting an edge on the tree connecting vertex $x$ and $y$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, print one integer: the minimum possible health decrement.

Examples:
input:
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
output:
1000000000000
193
57

Note: In the first test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1$, so your health decreases by $10^{12}$. Then kill the monster on vertex $1$.    * In the second round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $10^{12}$.

In the second test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5$, so your health decreases by $47+15+32+29+23=146$. Then kill the monsters on vertex $1,4,5$.    * In the second round: first, receive the attack from the monster on vertex $2,3$, so your health decreases by $15+32=47$. Then kill the monsters on vertex $2,3$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $193$.

In the third test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5,6,7$, so your health decreases by $8+10+2+3+5+7+4=39$. Then kill the monsters on vertex $1,3,6,7$.    * In the second round: first, receive the attack from the monster on vertex $2,4,5$, so your health decreases by $10+3+5=18$. Then kill the monsters on vertex $2,4,5$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $57$.

Buggy Code:
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.HashMap;

public final class Solution {

    private static class FastScanner {

        private final int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        private FastScanner(boolean usingFile) throws IOException {
            if (usingFile)
                din = new DataInputStream(new FileInputStream("path"));
            else
                din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        private short nextShort() throws IOException {
            short ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do
                ret = (short) (ret * 10 + c - '0');
            while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return (short) -ret;
            return ret;
        }

        private int nextInt() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do
                ret = ret * 10 + c - '0';
            while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }

        public long nextLong() throws IOException {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do
                ret = ret * 10 + c - '0';
            while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }

        private char nextChar() throws IOException {
            byte c = read();
            while (c <= ' ')
                c = read();
            return (char) c;
        }

        private String nextString() throws IOException {
            StringBuilder ret = new StringBuilder();
            byte c = read();
            while (c <= ' ')
                c = read();
            do {
                ret.append((char) c);
            } while ((c = read()) > ' ');
            return ret.toString();
        }

        private void fillBuffer() throws IOException {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }

        private byte read() throws IOException {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }
    }

    public static long binpow(long a, long k, long p) {
        int res = 1;
        while (k > 0) {
            if ((k & 1) == 1)
                res = (int) ((long) res * a % p);
            k >>= 1;
            a = (int) ((long) a * a % p);
        }
        return res;
    }

    static int gcd(int a, int b) {
        int div = b;
        int rem = a % b;
        while (rem != 0) {
            int temp = rem;
            rem = div % rem;
            div = temp;
        }
        return div;
    }

    static void put(HashMap<Integer, Integer> cnt, int key, int val) {
        if (cnt.containsKey(key)) {
            cnt.replace(key, cnt.get(key) + val);
        } else {
            cnt.put(key, val);
        }
    }

    static class Pair<F, S> {

        F a;
        S b;

        Pair(F a, S b) {
            this.a = a;
            this.b = b;
        }
    }

    static class sortCond implements Comparator<Pair<Integer, Integer>> {
        @Override
        public int compare(Pair<Integer, Integer> p1, Pair<Integer, Integer> p2) {
            if (p1.a < p2.a) {
                return -1;
            } else if (p1.a > p2.a) {
                return 1;
            } else {
                if (p1.b > p2.b) {
                    return 1;
                } else if (p1.b < p2.b) {
                    return -1;
                } else {
                    return 0;
                }
            }
        }
    }

    public static void retAns(ArrayList<ArrayList<Integer>> tree, long[][] dp, long[] a, int root, int par) {
        int i, j, k;
        for (i = 0; i < tree.get(root).size(); i++) {
            int child = tree.get(root).get(i);
            if (child != par) {
                retAns(tree, dp, a, child, root);
            }
        }
        for (i = 0; i < 3; i++) {
            long tot = 0L;
            for (k = 0; k < tree.get(root).size(); k++) {
                int child = tree.get(root).get(k);
                if (child != par) {
                    long MINI = 3000000000001L;
                    for (j = 0; j < 3; j++) {
                        if (j != i && dp[child][j] != -1) {
                            MINI = Math.min(MINI, dp[child][j]);
                        }
                    }
                    tot += MINI;
                }
            }
            dp[root][i] = a[root] * (i + 1) + tot;
        }
    }

    public static void main(String[] args) throws IOException {
        FastScanner ip = new FastScanner(false);
        PrintWriter out = new PrintWriter(System.out);
        int cases = ip.nextInt(), n, i, j, k, l;
        while (cases-- != 0) {
            n = ip.nextInt();
            long[] a = new long[n];
            ArrayList<ArrayList<Integer>> tree = new ArrayList<>();
            for (i = 0; i < n; i++) {
                a[i] = ip.nextLong();
                tree.add(new ArrayList<>());
            }
            for (i = 0; i < n - 1; i++) {
                int u = ip.nextInt() - 1, v = ip.nextInt() - 1;
                tree.get(u).add(v);
                tree.get(v).add(u);
            }
            long[][] dp = new long[n][3];
            for (i = 0; i < n; i++) {
                for (j = 0; j < 3; j++) {
                    dp[i][j] = -1;
                }
            }
            retAns(tree, dp, a, 0, -1);
            long MINI = 1000000000000000000L;
            for (i = 0; i < 3; i++) {
                if(dp[0][i] != -1) {
                    MINI = Math.min(MINI, dp[0][i]);
                }
            }
            out.println(MINI);
        }
        out.flush();
        out.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 maximize the failure probability. The code uses a dynamic programming approach to solve the problem, but it has a potential issue in how it handles the DP transitions and the constraints of the problem.

### Key Points to Consider:
1. **Tree Structure**: The tree could be structured in a way that makes it difficult to handle the constraints of killing monsters connected by an edge.
2. **Attack Points**: The attack points of the monsters could be very high, leading to large values that need to be handled carefully.
3. **Edge Cases**: Small trees or trees with specific structures (e.g., linear, star, etc.) can expose bugs in the algorithm.

### Test Input:
Let's create a test input with a small tree that has a specific structure to challenge the algorithm. We will use a linear tree (a path) where each node has a high attack point and is connected to its adjacent nodes. This will ensure that the algorithm must handle the constraint of not killing two directly connected nodes in the same round.

```plaintext
1
5
1000000000000 1000000000000 1000000000000 1000000000000 1000000000000
1 2
2 3
3 4
4 5
```

### Explanation:
- **Number of Test Cases**: 1
- **Number of Nodes**: 5
- **Attack Points**: Each node has an attack point of \(10^{12}\).
- **Edges**: The nodes are connected in a linear fashion, forming a path from node 1 to node 5.

### Why This Input is Challenging:
- **High Attack Points**: Each monster has a very high attack point, which will make the total health decrement significant.
- **Linear Structure**: The linear structure ensures that the algorithm must carefully choose which monsters to kill in each round to minimize the health decrement. Killing a monster in one round will prevent its adjacent monsters from being killed in the same round.

This input will test the algorithm's ability to handle high values and the constraints of not killing directly connected monsters in the same round. If the algorithm fails to handle this input correctly, it will expose the bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Omnipotent Monster Killer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You, the monster killer, want to kill a group of monsters. The monsters are on a tree with $n$ vertices. On vertex with number $i$ ($1\le i\le n$), there is a monster with $a_i$ attack points. You want to battle with monsters for $10^{100}$ rounds.

In each round, the following happens in order:

  1. All living monsters attack you. Your health decreases by the sum of attack points of all living monsters.    2. You select some (possibly all or none) monsters and kill them. After being killed, the monster will not be able to do any attacks in the future.

There is a restriction: in one round, you cannot kill two monsters that are directly connected by an edge.

If you choose what monsters to attack optimally, what is the smallest health decrement you can have after all rounds?

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

The first line of each test case contains an integer $n$ ($1\le n\le 3\cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,\ldots,a_n$ ($1\le a_i\le 10^{12}$).

The following $n-1$ lines each contain two integers $x,y$ ($1\le x,y\le n$), denoting an edge on the tree connecting vertex $x$ and $y$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, print one integer: the minimum possible health decrement.

Examples:
input:
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
output:
1000000000000
193
57

Note: In the first test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1$, so your health decreases by $10^{12}$. Then kill the monster on vertex $1$.    * In the second round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $10^{12}$.

In the second test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5$, so your health decreases by $47+15+32+29+23=146$. Then kill the monsters on vertex $1,4,5$.    * In the second round: first, receive the attack from the monster on vertex $2,3$, so your health decreases by $15+32=47$. Then kill the monsters on vertex $2,3$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $193$.

In the third test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5,6,7$, so your health decreases by $8+10+2+3+5+7+4=39$. Then kill the monsters on vertex $1,3,6,7$.    * In the second round: first, receive the attack from the monster on vertex $2,4,5$, so your health decreases by $10+3+5=18$. Then kill the monsters on vertex $2,4,5$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $57$.

Buggy Code:
#pragma GCC optimize(3)  //O2优化开启
#include<bits/stdc++.h>
using namespace std;
#define int long long
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
const int mod=998244353;
const int MX=0x3f3f3f3f3f3f3f3f;
//inline int read()                     //快读
//{
//    int xr=0,F=1; char cr;
//    while(cr=getchar(),cr<'0'||cr>'9') if(cr=='-') F=-1;
//    while(cr>='0'&&cr<='9')
//        xr=(xr<<3)+(xr<<1)+(cr^48),cr=getchar();
//    return xr*F;
//}
//void write(int x)                     //快写
//{
//    if(x<0) putchar('-'),x=-x;
//    if(x>9) write(x/10); putchar(x%10+'0');
//}
// 比 unordered_map 更快的哈希表
// #include <ext/pb_ds/assoc_container.hpp>
// using namespace __gnu_pbds;
// const int RANDOM = chrono::high_resolution_clock::now().time_since_epoch().count();
// struct chash {
//     int operator()(int x) const { return x ^ RANDOM; }
// };
// typedef gp_hash_table<int, int, chash> hash_t;

int n,k;
int a[500005];
vector<int>ve[500005];
void icealsoheat(){
    cin>>n;

    for(int i=1;i<=n;i++)ve[i].clear(),cin>>a[i];

    for(int i=1;i<n;i++){
        int l,r;
        cin>>l>>r;
        ve[l].push_back(r);
        ve[r].push_back(l);
    }
    int sum1=0;
    int sum2=0;
    auto dfs=[&](auto self,int x,int fa,int p)->void{
        if(p==0)sum2+=a[x];
        else sum1+=a[x];
        for(auto i:ve[x]){
            if(i==fa)continue;
            self(self,i,x,p^1);
        }

    };
    dfs(dfs,1,0,0);

    int ans=sum1+sum2+min(sum1,sum2);
    cout<<ans<<"\n";
}
signed main(){
    ios::sync_with_stdio(false);          //int128不能用快读！！！！！！
    cin.tie();
    cout.tie();
    int _yq;
    _yq=1;
    cin>>_yq;
    while(_yq--){
        icealsoheat();
    }
}
//
//⠀⠀⠀             ⠀⢸⣿⣿⣿⠀⣼⣿⣿⣦⡀
//⠀⠀⠀⠀⠀⠀⠀⠀⠀⣀⠀⠀⠀ ⠀⢸⣿⣿⡟⢰⣿⣿⣿⠟⠁
//⠀⠀⠀⠀⠀⠀⠀⢰⣿⠿⢿⣦⣀⠀⠘⠛⠛⠃⠸⠿⠟⣫⣴⣶⣾⡆
//⠀⠀⠀⠀⠀⠀⠀⠸⣿⡀⠀⠉⢿⣦⡀⠀⠀⠀⠀⠀⠀ ⠛⠿⠿⣿⠃
//⠀⠀⠀⠀⠀⠀⠀⠀⠙⢿⣦⠀⠀⠹⣿⣶⡾⠛⠛⢷⣦⣄
//⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣿⣧⠀⠀⠈⠉⣀⡀⠀ ⠀⠙⢿⡇
//⠀⠀⠀⠀⠀⠀⢀⣠⣴⡿⠟⠋⠀⠀⢠⣾⠟⠃⠀⠀⠀⢸⣿⡆
//⠀⠀⠀⢀⣠⣶⡿⠛⠉⠀⠀⠀⠀⠀⣾⡇⠀⠀⠀⠀⠀⢸⣿⠇
//⢀⣠⣾⠿⠛⠁⠀⠀⠀⠀⠀⠀⠀⢀⣼⣧⣀⠀⠀⠀⢀⣼⠇
//⠈⠋⠁⠀⠀⠀⠀⠀⠀⠀⠀⢀⣴⡿⠋⠙⠛⠛⠛⠛⠛⠁
//⠀⠀⠀⠀⠀⠀⠀⠀⠀⣀⣾⡿⠋
//⠀⠀⠀⠀⠀⠀⠀⠀⢾⠿⠋
//

**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
3
10 10 10
1 2
2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Omnipotent Monster Killer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You, the monster killer, want to kill a group of monsters. The monsters are on a tree with $n$ vertices. On vertex with number $i$ ($1\le i\le n$), there is a monster with $a_i$ attack points. You want to battle with monsters for $10^{100}$ rounds.

In each round, the following happens in order:

  1. All living monsters attack you. Your health decreases by the sum of attack points of all living monsters.    2. You select some (possibly all or none) monsters and kill them. After being killed, the monster will not be able to do any attacks in the future.

There is a restriction: in one round, you cannot kill two monsters that are directly connected by an edge.

If you choose what monsters to attack optimally, what is the smallest health decrement you can have after all rounds?

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

The first line of each test case contains an integer $n$ ($1\le n\le 3\cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,\ldots,a_n$ ($1\le a_i\le 10^{12}$).

The following $n-1$ lines each contain two integers $x,y$ ($1\le x,y\le n$), denoting an edge on the tree connecting vertex $x$ and $y$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, print one integer: the minimum possible health decrement.

Examples:
input:
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
output:
1000000000000
193
57

Note: In the first test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1$, so your health decreases by $10^{12}$. Then kill the monster on vertex $1$.    * In the second round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $10^{12}$.

In the second test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5$, so your health decreases by $47+15+32+29+23=146$. Then kill the monsters on vertex $1,4,5$.    * In the second round: first, receive the attack from the monster on vertex $2,3$, so your health decreases by $15+32=47$. Then kill the monsters on vertex $2,3$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $193$.

In the third test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5,6,7$, so your health decreases by $8+10+2+3+5+7+4=39$. Then kill the monsters on vertex $1,3,6,7$.    * In the second round: first, receive the attack from the monster on vertex $2,4,5$, so your health decreases by $10+3+5=18$. Then kill the monsters on vertex $2,4,5$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $57$.

Buggy Code:
## Template Start ##
from functools import reduce
from math import sqrt, gcd, ceil
import heapq
from bisect import bisect_left
from copy import deepcopy
from itertools import combinations

# heap_list = [1975, 124]
# heapq.heapify(heap_list)
# print(heapq.heappop(heap_list))

def find_factors(n):
    step = 2 if n % 2 else 1
    return set(
        reduce(
            list.__add__,
            ([i, n // i] for i in range(1, int(sqrt(n)) + 1, step) if n % i == 0),
        )
    )

def SieveOfEratosthenes(num):
    prime = [True for i in range(num + 1)]
    # boolean array
    p = 2
    while p * p <= num:

        # If prime[p] is not
        # changed, then it is a prime
        if prime[p] == True:

            # Updating all multiples of p
            for i in range(p * p, num + 1, p):
                prime[i] = False
        p += 1

    # Print all prime numbers
    primes_list = []
    for p in range(2, num + 1):
        if prime[p]:
            primes_list.append(p)
    return primes_list

## END TEMPLATE ##

t = int(input())
# t = 1

def print_num_arr(arr):
    print(" ".join([str(i) for i in arr]))

def run():
    n = int(input())
    attack = [int(i) for i in input().split(" ")]
    h_map = {}
    for i in range(n):
        h_map[i] = set([])
    for i in range(n - 1):
        x, y = [int(i) for i in input().split(" ")]
        h_map[x - 1].add(y - 1)
        h_map[y - 1].add(x - 1)

    # sorted_arr = sorted(
    #     [i for i in range(n)],
    #     key=lambda x: len(h_map[x]),
    #     reverse=True,
    # )

    sum_a = 0
    sum_b = 0
    queue = [0]

    visited = set([])

    flag = True
    while len(queue) > 0:
        new_q = []
        for node in queue:
            if node not in visited:
                visited.add(node)
                if flag:
                    sum_b += attack[node]
                else:
                    sum_a += attack[node]
                new_q.extend(h_map[node])
        queue = new_q
        flag = not flag
    print(sum_a + sum_b + min(sum_b, sum_a))

for i in range(t):
    run()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Omnipotent Monster Killer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You, the monster killer, want to kill a group of monsters. The monsters are on a tree with $n$ vertices. On vertex with number $i$ ($1\le i\le n$), there is a monster with $a_i$ attack points. You want to battle with monsters for $10^{100}$ rounds.

In each round, the following happens in order:

  1. All living monsters attack you. Your health decreases by the sum of attack points of all living monsters.    2. You select some (possibly all or none) monsters and kill them. After being killed, the monster will not be able to do any attacks in the future.

There is a restriction: in one round, you cannot kill two monsters that are directly connected by an edge.

If you choose what monsters to attack optimally, what is the smallest health decrement you can have after all rounds?

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

The first line of each test case contains an integer $n$ ($1\le n\le 3\cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,\ldots,a_n$ ($1\le a_i\le 10^{12}$).

The following $n-1$ lines each contain two integers $x,y$ ($1\le x,y\le n$), denoting an edge on the tree connecting vertex $x$ and $y$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, print one integer: the minimum possible health decrement.

Examples:
input:
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
output:
1000000000000
193
57

Note: In the first test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1$, so your health decreases by $10^{12}$. Then kill the monster on vertex $1$.    * In the second round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $10^{12}$.

In the second test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5$, so your health decreases by $47+15+32+29+23=146$. Then kill the monsters on vertex $1,4,5$.    * In the second round: first, receive the attack from the monster on vertex $2,3$, so your health decreases by $15+32=47$. Then kill the monsters on vertex $2,3$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $193$.

In the third test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5,6,7$, so your health decreases by $8+10+2+3+5+7+4=39$. Then kill the monsters on vertex $1,3,6,7$.    * In the second round: first, receive the attack from the monster on vertex $2,4,5$, so your health decreases by $10+3+5=18$. Then kill the monsters on vertex $2,4,5$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $57$.

Buggy Code:
for _ in range(int(input())):
    a = int(input())
    m = list(map(int, input().split()))
    l = [[] for i in range(a)]
    for i in range(a - 1):
        x, y = map(int, input().split())
        x, y = x - 1, y - 1
        l[x].append(y)
        l[y].append(x)
    c = 20
    dp = [[m[i] * j for j in range(c)] for i in range(a)]
    def dfs(v, p):
        for h in l[v]:
            if h == p:
                continue
            dfs(h, v)
            mn1 = min(dp[h])
            mn2 = mn1 if dp[h].count(mn1) > 1 else min(i for i in dp[h] if i != mn1)
            for i in range(c):
                dp[v][i] += mn1 if dp[h][i] != mn1 else mn2

    dfs(0, -1)
    print(sum(m) + min(dp[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
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Omnipotent Monster Killer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You, the monster killer, want to kill a group of monsters. The monsters are on a tree with $n$ vertices. On vertex with number $i$ ($1\le i\le n$), there is a monster with $a_i$ attack points. You want to battle with monsters for $10^{100}$ rounds.

In each round, the following happens in order:

  1. All living monsters attack you. Your health decreases by the sum of attack points of all living monsters.    2. You select some (possibly all or none) monsters and kill them. After being killed, the monster will not be able to do any attacks in the future.

There is a restriction: in one round, you cannot kill two monsters that are directly connected by an edge.

If you choose what monsters to attack optimally, what is the smallest health decrement you can have after all rounds?

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

The first line of each test case contains an integer $n$ ($1\le n\le 3\cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,\ldots,a_n$ ($1\le a_i\le 10^{12}$).

The following $n-1$ lines each contain two integers $x,y$ ($1\le x,y\le n$), denoting an edge on the tree connecting vertex $x$ and $y$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, print one integer: the minimum possible health decrement.

Examples:
input:
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
output:
1000000000000
193
57

Note: In the first test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1$, so your health decreases by $10^{12}$. Then kill the monster on vertex $1$.    * In the second round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $10^{12}$.

In the second test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5$, so your health decreases by $47+15+32+29+23=146$. Then kill the monsters on vertex $1,4,5$.    * In the second round: first, receive the attack from the monster on vertex $2,3$, so your health decreases by $15+32=47$. Then kill the monsters on vertex $2,3$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $193$.

In the third test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5,6,7$, so your health decreases by $8+10+2+3+5+7+4=39$. Then kill the monsters on vertex $1,3,6,7$.    * In the second round: first, receive the attack from the monster on vertex $2,4,5$, so your health decreases by $10+3+5=18$. Then kill the monsters on vertex $2,4,5$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $57$.

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

public class The_Omnipotent_Monster_Killer {

    public static void main(String[] args) {
        FastReader in = new FastReader();
        int t = in.nextInt();
        while (t-- > 0) {
            int n = in.nextInt();
            long a[] = in.readArray(n);
            long s=0;
            for(long x:a)s+=x;
            ArrayList<Integer>arr[] = new ArrayList[n+1];
            for(int i=0;i<=n;i++)arr[i]=new ArrayList<>();
            for(int i=1;i<=n-1;i++){
                int x=in.nextInt();
                int y=in.nextInt();
                arr[x].add(y);
                arr[y].add(x);
            }
            long ans=s;
            while (true) {
                long dp1[][] = new long[n+1][2];
                for(int i=0;i<=n;i++){
                    dp1[i][0]=-1;
                    dp1[i][1]=-1;
                }
                long tans1 = f(1,0,arr,0,a,dp1);
                // long dp2[][] = new long[n+1][2];
                // for(int i=0;i<=n;i++){
                //     dp2[i][0]=-1;
                //     dp2[i][1]=-1;
                // }
                // long tans2 = f(1,0,arr,1,a,dp2);
                // if(tans1==0 && tans2==0)break;
                if(tans1==0)break;
                // if(tans1>tans2){
                    s-=tans1;
                    ans+=s;
                    dfs(1,0,arr,dp1,a,0);
                // }
                // else {
                //     s-=tans2;
                //     ans+=s;
                //     dfs(1,0,arr,dp2,a,1);
                // }
            }
            // for(int i=1;i<=n;i++)System.out.print(a[i-1] + " ");
            System.out.println(ans);
        }
    }

    private static void dfs(int at, int par, ArrayList<Integer>[] arr, long dp[][], long a[], int partaken){
        long ans = dp[at][partaken];
        if(ans==0)return;
        long a1=0,a2=0;
        for(int to:arr[at]){
            if(to==par)continue;
            a1+=dp[to][0];
            if(partaken==0)a2+=dp[to][1];
        }
        if(partaken==0)a2+=a[at-1];
        // System.out.println(at + " " + ans + " " + a1 + " " + a2);
        if(ans==a2){
            a[at-1]=0;
            for(int to:arr[at]){
                if(to==par)continue;
                dfs(to,at,arr,dp,a,1);
            }
        }
        else {
            for(int to:arr[at]){
                if(to==par)continue;
                dfs(to,at,arr,dp,a,0);
            }
        }
    }

    private static long f(int at, int par, ArrayList<Integer>[] arr, int partaken, long a[], long dp[][]) {
        long a1=0,a2=0;
        if(dp[at][partaken]!=-1)return dp[at][partaken];
        for(int to:arr[at]){
            if(to==par)continue;
            a1+=f(to,at,arr,0,a,dp);
            if(partaken==0)a2+=f(to,at,arr,1,a,dp);
        }
        if(partaken==0)a2+=a[at-1];
        return dp[at][partaken]=Math.max(a1,a2);
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        public FastReader() {
            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 {
                if (st.hasMoreTokens()) { str = st.nextToken("\n"); }
                else { str = br.readLine(); }
            } catch (IOException e) { e.printStackTrace(); }
            return str;
        }
        long[] readArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to create a test input that highlights the issue in the algorithm. The code seems to be using a dynamic programming approach to solve the problem, but there might be issues with how it handles the tree structure and the decision-making process for killing monsters.

### Step-by-Step Analysis:

1. **Tree Structure**:
   - The code should correctly handle the tree structure and ensure that no two directly connected monsters are killed in the same round.
   - We need to test if the code correctly identifies and processes the tree structure.

2. **Dynamic Programming**:
   - The dynamic programming (DP) arrays `dp1` and `dp2` are used to store the maximum health decrement for each node based on whether the node is killed or not.
   - We need to test if the DP arrays are correctly updated and used.

3. **Edge Cases**:
   - Test with a tree that has a linear structure (a chain of nodes).
   - Test with a tree that has a star structure (one central node connected to multiple leaf nodes).
   - Test with a tree that has a balanced structure (a binary tree or a more complex balanced tree).

### Test Input:
We will create a test input with a linear structure to see if the code correctly handles the scenario where the tree is a simple chain. This will help us identify if the code is correctly updating the DP arrays and making the right decisions.

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

### Explanation:
- **Test Case 1**:
  - **Number of Nodes**: 4
  - **Monster Attack Points**: 1, 2, 3, 4
  - **Edges**: (1, 2), (2, 3), (3, 4)

This input represents a linear tree where each node is connected to the next one in a sequence. The optimal strategy for this case would be to kill the monsters in such a way that no two directly connected monsters are killed in the same round. The expected output should be the minimum health decrement, which can be calculated as follows:

- **Round 1**: Kill nodes 1 and 3 (health decrement = 1 + 2 + 3 + 4 = 10)
- **Round 2**: Kill nodes 2 and 4 (health decrement = 2 + 4 = 6)
- **Total Health Decrement**: 10

This test input should help expose any issues in the code's handling of the tree structure and the dynamic programming approach.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Omnipotent Monster Killer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You, the monster killer, want to kill a group of monsters. The monsters are on a tree with $n$ vertices. On vertex with number $i$ ($1\le i\le n$), there is a monster with $a_i$ attack points. You want to battle with monsters for $10^{100}$ rounds.

In each round, the following happens in order:

  1. All living monsters attack you. Your health decreases by the sum of attack points of all living monsters.    2. You select some (possibly all or none) monsters and kill them. After being killed, the monster will not be able to do any attacks in the future.

There is a restriction: in one round, you cannot kill two monsters that are directly connected by an edge.

If you choose what monsters to attack optimally, what is the smallest health decrement you can have after all rounds?

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

The first line of each test case contains an integer $n$ ($1\le n\le 3\cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,\ldots,a_n$ ($1\le a_i\le 10^{12}$).

The following $n-1$ lines each contain two integers $x,y$ ($1\le x,y\le n$), denoting an edge on the tree connecting vertex $x$ and $y$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, print one integer: the minimum possible health decrement.

Examples:
input:
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
output:
1000000000000
193
57

Note: In the first test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1$, so your health decreases by $10^{12}$. Then kill the monster on vertex $1$.    * In the second round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $10^{12}$.

In the second test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5$, so your health decreases by $47+15+32+29+23=146$. Then kill the monsters on vertex $1,4,5$.    * In the second round: first, receive the attack from the monster on vertex $2,3$, so your health decreases by $15+32=47$. Then kill the monsters on vertex $2,3$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $193$.

In the third test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5,6,7$, so your health decreases by $8+10+2+3+5+7+4=39$. Then kill the monsters on vertex $1,3,6,7$.    * In the second round: first, receive the attack from the monster on vertex $2,4,5$, so your health decreases by $10+3+5=18$. Then kill the monsters on vertex $2,4,5$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $57$.

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

public class Main {
    public static void main(String[] args) {
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        Task solver = new Task();
        solver.solve(in.nextInt(), in, out);
//        solver.duiPai();
        out.close();
    }

    static class Task {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            for (int t = 0; t < testNumber; t++) {
                int n = in.nextInt();
                long[] a = new long[n];
                for (int i = 0; i < a.length; i++) {
                    a[i] = in.nextLong();
                }
                int[][] edges = new int[n - 1][2];
                for (int i = 0; i < edges.length; i++) {
                    edges[i][0] = in.nextInt() - 1;
                    edges[i][1] = in.nextInt() - 1;
                }
                out.println(new Solution().solve(a, edges));
            }
        }

        public void duiPai() {
            Random random = new Random();
            for (int t = 0; t < 100; t++) {
                int n = random.nextInt(10) + 5;
                int[] a = genArr(n, 1, 9);
                System.out.println(Arrays.toString(a));
            }
        }

        public int[] genArr(int n, int l, int r) {
            int[] arr = new int[n];
            Random random = new Random();
            for (int i = 0; i < n; i++) {
                arr[i] = random.nextInt(r - l + 1) + l;
            }
            return arr;
        }
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

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

class Solution {
    int n;
    Tree[] trees;

    public long solve(long[] a, int[][] edges) {
        n = a.length;
        trees = new Tree[n];
        for (int i = 0; i < trees.length; i++) {
            trees[i] = new Tree(a[i]);
        }
        for (int[] edge : edges) {
            trees[edge[0]].child.add(edge[1]);
            trees[edge[1]].child.add(edge[0]);
        }
        build(0, -1);
        dfs(0);
        return trees[0].min[0];
    }

    private void dfs(int v) {
        int round = trees[v].min.length;
        for (int i = 0; i < round; i++) {
            trees[v].min[i] = trees[v].val * (i + 1);
        }
        for (Integer w : trees[v].child) {
            dfs(w);
            long[][] m = new long[2][round];
            m[0][0] = trees[w].min[0];
            m[1][round - 1] = trees[w].min[round - 1];
            for (int i = 1; i < round; i++) {
                m[0][i] = Math.min(m[0][i - 1], trees[w].min[i]);
            }
            for (int i = round - 2; i >= 0; i--) {
                m[1][i] = Math.min(m[1][i + 1], trees[w].min[i]);
            }
            trees[v].min[0] += m[1][1];
            trees[v].min[round - 1] += m[0][round - 2];
            for (int i = 1; i < round - 1; i++) {
                trees[v].min[i] += Math.min(m[0][i - 1], m[1][i + 1]);
            }
        }
    }

    private void build(int v, int p) {
        trees[v].child.remove(p);
        for (Integer w : trees[v].child) {
            build(w, v);
        }
    }
}

class Tree {
    long val;
    Set<Integer> child = new HashSet<>();
    long[] min = new long[10];

    public Tree(long val) {
        this.val = val;
    }
}

**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 can maximize the failure probability. Here are the steps to think through this:

1. **Single Vertex Case**: A single vertex with a large attack value. This is straightforward and should not cause issues, but it's a good baseline.
2. **Linear Tree**: A tree where each vertex is connected to the next one in a linear fashion. This can expose issues with the way the code handles sequential attacks and kills.
3. **Star Tree**: A tree where one central vertex is connected to multiple other vertices. This can expose issues with the way the code handles a central node with multiple children.
4. **Balanced Tree**: A balanced tree where each vertex has a similar number of children. This can expose issues with the way the code handles more complex tree structures.
5. **High Attack Values**: High attack values can cause overflow issues if not handled correctly.
6. **Multiple Test Cases**: Multiple test cases to ensure the code handles different scenarios correctly.

Given the constraints and the need for a small-scale test input, let's focus on a linear tree and a star tree with moderate attack values.

### Linear Tree Example
A linear tree with 5 vertices where each vertex has an attack value of 10.

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

### Star Tree Example
A star tree with 5 vertices where the central vertex has an attack value of 10 and the other vertices have an attack value of 1.

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

### Combined Test Input
Combining both examples into a single test input with multiple test cases:

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

This test input includes both a linear tree and a star tree, which should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Omnipotent Monster Killer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You, the monster killer, want to kill a group of monsters. The monsters are on a tree with $n$ vertices. On vertex with number $i$ ($1\le i\le n$), there is a monster with $a_i$ attack points. You want to battle with monsters for $10^{100}$ rounds.

In each round, the following happens in order:

  1. All living monsters attack you. Your health decreases by the sum of attack points of all living monsters.    2. You select some (possibly all or none) monsters and kill them. After being killed, the monster will not be able to do any attacks in the future.

There is a restriction: in one round, you cannot kill two monsters that are directly connected by an edge.

If you choose what monsters to attack optimally, what is the smallest health decrement you can have after all rounds?

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

The first line of each test case contains an integer $n$ ($1\le n\le 3\cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,\ldots,a_n$ ($1\le a_i\le 10^{12}$).

The following $n-1$ lines each contain two integers $x,y$ ($1\le x,y\le n$), denoting an edge on the tree connecting vertex $x$ and $y$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, print one integer: the minimum possible health decrement.

Examples:
input:
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
output:
1000000000000
193
57

Note: In the first test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1$, so your health decreases by $10^{12}$. Then kill the monster on vertex $1$.    * In the second round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $10^{12}$.

In the second test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5$, so your health decreases by $47+15+32+29+23=146$. Then kill the monsters on vertex $1,4,5$.    * In the second round: first, receive the attack from the monster on vertex $2,3$, so your health decreases by $15+32=47$. Then kill the monsters on vertex $2,3$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $193$.

In the third test case, an optimal sequence of operations would be:

  * In the first round: first, receive the attack from the monster on vertex $1,2,3,4,5,6,7$, so your health decreases by $8+10+2+3+5+7+4=39$. Then kill the monsters on vertex $1,3,6,7$.    * In the second round: first, receive the attack from the monster on vertex $2,4,5$, so your health decreases by $10+3+5=18$. Then kill the monsters on vertex $2,4,5$.    * In the third round to the $10^{100}$-th round: all monsters have been killed, so nothing happens.

The total health decrement is $57$.

Buggy Code:
#include <iostream>
//#include <vector>
//#include <map>
#include <bits/stdc++.h>
// #include <x86intrin.h>
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>

#define S second
#define F first
#define int long long
//#define int unsigned int
//#define int unsigned long long
#define double long double
#define endl '\n'
#define pi acos(-1.0)
#define inf 3000000000000000000
#define INF 2000000000
#define random rand()^(rand()<<15)
#define fixset(x) cout<<fixed<<setprecision((int)x);

using namespace std;
// using namespace __gnu_pbds;
// template<typename T>
// using ord_set=tree<T,null_type,less_equal<T>,rb_tree_tag,tree_order_statistics_node_update>;
constexpr int mod=1e9+7;
constexpr int MOD=998244353;
//using vec = complex<double>;
// mt19937 rnd(time(nullptr));
// clock_t startTime;
// double getCurrentTime()
// {
//     return (clock() - startTime)*1.0 / CLOCKS_PER_SEC;
// }
int mul(int a, int b, int m)
{
    if(b<0) {
        return 0;
    }
    if(b==0) {
        return 1;
    }
	if(b%2==0){
		return (mul(a, b/2, m)<<1) % m;
	}
	return (mul(a, b-1, m) + a) % m;
}
int binpow(int a,int b,int mod)
{
    if(b<=0)
    {
        return 1;
    }
    if(b%2)
    {
        return a*binpow(a,b-1,mod)%mod;
    }
    return binpow(a*a%mod,b/2,mod);
}
int binpow(int a,int b)
{
    if(b<=0)
    {
        return 1;
    }
    if(b%2)
    {
        return a*binpow(a,b-1);
    }
    return binpow(a*a,b/2);
}
constexpr int SZ=5e5;
int fac[SZ+10],inv[SZ+10];
void init(int mod)
{
    fac[0]=1;
    for(int i=1; i<=SZ; i++)
    {
        fac[i]=fac[i-1]*i%mod;
    }
    inv[SZ]=binpow(fac[SZ],mod-2,mod);
    for(int i=SZ-1; i>=0; i--)
    {
        inv[i]=inv[i+1]*(i+1)%mod;
    }
}
int C(int n,int k)
{
    if(k>n||k<0)
    {
        return 0;
    }
    if(k==n||k==0)
    {
        return 1;
    }
    return ((fac[n]*inv[k])%mod*inv[n-k])%mod;
}
vector<vector<int>>g;
vector<int>w;
vector<bool>used,us1,us2;
vector<unsigned int>dp[2];
void dfs(int v)
{
    us1[v]=1;
    dp[0][v]=0;
    dp[1][v]=w[v];
    for(int&u:g[v])
    {
        if(used[u]||us1[u])continue;
        dfs(u);
        dp[0][v]+=max(dp[0][u],dp[1][u]);
        dp[1][v]+=dp[0][u];
    }
    // cout<<v<<" "<<dp[0][v]<<" "<<dp[1][v]<<" "<<used[v]<<" "<<us1[v]<<endl;
}
void dfs_rest(int v,int best)
{
    us2[v]=1;
    if(best==1)
    {
        used[v]=1;
    }
    // cout<<v<<" "<<dp[0][v]<<" "<<dp[1][v]<<" "<<best<<endl;
    for(int&u:g[v])
    {
        // cout<<u<<" "<<used[u]<<" "<<us2[u]<<endl;
        if(used[u]||us2[u])continue;
        int bbest=best;
        if(best==1)
        {
            best=0;
        }
        else
        {
            if(dp[0][u]<dp[1][u])
            {
                best=1;
            }
        }
        dfs_rest(u,best);
        best=bbest;
    }
    // cout<<v<<" "<<dp[0][v]<<" "<<dp[1][v]<<" "<<best<<endl;
}
void solve()
{
    int n;
    cin>>n;

    g.clear();
    w.clear();
    dp[0].clear();
    dp[1].clear();
    used.clear();
    us1.clear();
    us2.clear();

    g.resize(n);
    w.resize(n);
    dp[0].resize(n,0);
    dp[1].resize(n,0);
    used.resize(n,false);
    us1.resize(n,false);
    us2.resize(n,false);
    for(int i=0;i<n;i++)
    {
        cin>>w[i];
    }
    for(int i=1;i<n;i++)
    {
        int a,b;
        cin>>a>>b;
        g[a-1].push_back(b-1);
        g[b-1].push_back(a-1);
    }
    unsigned int res=0;
    while(1)
    {
        unsigned cnt=0,sum=0;
        for(int i=0;i<n;i++)
        {
            if(!used[i]){sum+=w[i];res+=w[i];cnt++;}
            us1[i]=0;dp[0][i]=dp[1][i]=0;us2[i]=0;
        }
        // cout<<sum<<" "<<cnt<<endl;
        if(cnt==0)
        {
            break;
        }
        for(int i=0;i<n;i++)
        {
            if(!us1[i]&&!used[i])
            {
                // cout<<i<<endl;
                dfs(i);
                int best=0;
                if(dp[1][i]>dp[0][i])
                {
                    best=1;
                }
                dfs_rest(i,best);
            }
        }
        // for(int i=0;i<n;i++)
        // {
        //     cout<<used[i]<<" ";
        // }
        // cout<<endl;
        // for(int i=0;i<n;i++)
        // {
        //     cout<<dp[0][i]<<" ";
        // }
        // cout<<endl;
        // for(int i=0;i<n;i++)
        // {
        //     cout<<dp[1][i]<<" ";
        // }
        // cout<<endl<<endl;
    }
    cout<<res<<endl;
}
int32_t main()
{
//    startTime = clock();
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    // init(mod);
    int t=1;
    cin>>t;
    for(int i=1;i<=t;i++)
    {
        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 focus on edge cases and scenarios that maximize the failure probability. Here are the steps to think through the problem:

1. **Tree Structure**: The code deals with a tree structure, so we should consider different tree topologies such as a linear chain, a star, or a balanced tree.
2. **Attack Points**: The attack points of the monsters should be carefully chosen to create scenarios where the code might fail.
3. **Edge Cases**: Consider edge cases such as a tree with a single vertex, a tree with all vertices having the same attack points, or a tree where the optimal strategy is not immediately obvious.

Let's create a test input that combines these elements:

### Test Input
```plaintext
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
```

### Explanation
1. **Single Vertex Tree**:
   - This is a simple case to ensure the code handles a tree with a single vertex correctly.
   - Input: `1\n1000000000000`
   - Expected Output: `1000000000000`

2. **Star Tree**:
   - A star tree where one central vertex is connected to all other vertices.
   - Input: `5\n47 15 32 29 23\n1 2\n1 3\n2 4\n2 5`
   - Expected Output: `193`
   - Explanation: In the first round, kill the central vertex and two of its children. In the second round, kill the remaining two children.

3. **Balanced Tree**:
   - A more complex tree structure with multiple levels.
   - Input: `7\n8 10 2 3 5 7 4\n1 2\n1 4\n3 2\n5 3\n6 2\n7 5`
   - Expected Output: `57`
   - Explanation: In the first round, kill the vertices 1, 3, 6, and 7. In the second round, kill the vertices 2, 4, and 5.

### Test Input in Format
```plaintext
3
1
1000000000000
5
47 15 32 29 23
1 2
1 3
2 4
2 5
7
8 10 2 3 5 7 4
1 2
1 4
3 2
5 3
6 2
7 5
```

This test input covers a range of scenarios including a single vertex, a star tree, and a balanced tree, which should help in exposing any bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bridge Renovation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, Monocarp started working as a director of a park located near his house. The park is quite large, so it even has a small river splitting it into several zones. Several bridges are built across this river. Three of these bridges are especially old and need to be repaired.

All three bridges have the same length but differ in width. Their widths are $$$18$$$, $$$21$$$ and $$$25$$$ units, respectively.

During the park renovation process, Monocarp has to replace the old planks that served as the surface of the bridges with the new ones.

Planks are sold with a standard length of $$$60$$$ units. Monocarp already knows that he needs $$$n$$$ planks for each bridge. But since the bridges have different widths, he needs $$$n$$$ planks of length $$$18$$$ for the first bridge, $$$n$$$ planks of length $$$21$$$ for the second one, and $$$n$$$ planks of length $$$25$$$ for the last one.

Workers in charge of renovation have no problem with cutting planks into parts but refuse to join planks, since it creates weak spots and looks ugly.

Monocarp wants to buy as few planks as possible but struggles to calculate the required number of planks. Can you help him?

Input: The first and only line contains a single integer $$$n$$$ ($$$1 \le n \le 1000$$$) — the number of planks required for each of the three bridges.

Output: Print a single integer — the minimum number of planks of standard length ($$$60$$$ units) Monocarp needs to cover all three bridges if the planks can be cut into parts.

Examples:
input:
1
output:
2

input:
3
output:
4

input:
1000
output:
1167

Note: In the first example, it is possible to cut one plank of length $$$60$$$ into three planks with lengths $$$25$$$, $$$18$$$ and $$$17$$$, and cut another plank of length $$$60$$$ into two planks with lengths $$$39$$$ and $$$21$$$. That way, Monocarp will have all the required planks.

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

using namespace std;

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

    // Total number of planks needed
    int total_planks = 0;

    // Number of planks of each width needed
    int planks_18 = n;
    int planks_21 = n;
    int planks_25 = n;

    while (planks_18 > 0 || planks_21 > 0 || planks_25 > 0) {
        // Try to fit the largest possible combination of planks into a 60-unit plank
        if (planks_25 > 0 && planks_18 > 0) {
            // Fit 25 + 18 + 17 (remaining 17 units)
            planks_25--;
            planks_18--;
            total_planks++;
        } else if (planks_25 > 0 && planks_21 > 0) {
            // Fit 25 + 21 + 14 (remaining 14 units)
            planks_25--;
            planks_21--;
            total_planks++;
        } else if (planks_21 > 0 && planks_18 > 0) {
            // Fit 21 + 18 + 21 (remaining 21 units)
            planks_21--;
            planks_18--;
            total_planks++;
        } else if (planks_25 > 0) {
            // Fit 25 + 25 + 10 (remaining 10 units)
            planks_25 -= 2;
            total_planks++;
        } else if (planks_21 > 0) {
            // Fit 21 + 21 + 18 (remaining 18 units)
            planks_21 -= 2;
            planks_18--;
            total_planks++;
        } else if (planks_18 > 0) {
            // Fit 18 + 18 + 18 + 6 (remaining 6 units)
            planks_18 -= 3;
            total_planks++;
        }
    }

    cout << total_planks << endl;

    return 0;
}

// 5iXeDOjYQZR3SAC7VlL

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

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

Think step by step.

Response:
```plaintext
2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bridge Renovation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, Monocarp started working as a director of a park located near his house. The park is quite large, so it even has a small river splitting it into several zones. Several bridges are built across this river. Three of these bridges are especially old and need to be repaired.

All three bridges have the same length but differ in width. Their widths are $$$18$$$, $$$21$$$ and $$$25$$$ units, respectively.

During the park renovation process, Monocarp has to replace the old planks that served as the surface of the bridges with the new ones.

Planks are sold with a standard length of $$$60$$$ units. Monocarp already knows that he needs $$$n$$$ planks for each bridge. But since the bridges have different widths, he needs $$$n$$$ planks of length $$$18$$$ for the first bridge, $$$n$$$ planks of length $$$21$$$ for the second one, and $$$n$$$ planks of length $$$25$$$ for the last one.

Workers in charge of renovation have no problem with cutting planks into parts but refuse to join planks, since it creates weak spots and looks ugly.

Monocarp wants to buy as few planks as possible but struggles to calculate the required number of planks. Can you help him?

Input: The first and only line contains a single integer $$$n$$$ ($$$1 \le n \le 1000$$$) — the number of planks required for each of the three bridges.

Output: Print a single integer — the minimum number of planks of standard length ($$$60$$$ units) Monocarp needs to cover all three bridges if the planks can be cut into parts.

Examples:
input:
1
output:
2

input:
3
output:
4

input:
1000
output:
1167

Note: In the first example, it is possible to cut one plank of length $$$60$$$ into three planks with lengths $$$25$$$, $$$18$$$ and $$$17$$$, and cut another plank of length $$$60$$$ into two planks with lengths $$$39$$$ and $$$21$$$. That way, Monocarp will have all the required planks.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define rp(n) for (int i = 0; i < (n); i++)
#define rp2(n, j) for (int j = 0; j < (n); j++)
#define ll long long int
#define set(a,n,k) for(int lambda=0;lambda<n;lambda++){a[lambda]=k;}
#define out(a,n,k) for(int lambda=0;lambda<n;lambda++){cout<<a[lambda]<<k;}
#define ain(a,n) for(int lambda=0;lambda<n;lambda++){cin>>a[lambda];}
#define v1(type,n,name) vector<type> name(n); for(int lam=0; lam<n; lam++) cin >> name[lam];
#define v2(txype,x,y) type x,y;cin>>x>>y;
#define v3(type,x,y,z) type x,y,z;cin>>x>>y>>z;
#define incsort(name) std::sort(name.begin(),name.end());
#define decsort(name) {incsort(name);std::reverse(name.begin(),name.end());}
#define remax(maxo,new) maxo=max(maxo,new);
#define printvec(name) for(auto lamb : name){cout<<lamb<<" ";}
#define doublevec(type) std::vector<std::vector<type>>
int factorial(int n) {
    int result = 1;
    for (int i = 1; i <= n; ++i) {
        result *= i;
    }
    return result;
}

int n;
void solve() {

    cin >> n;
    cout<< n + (n+5/6);
}

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr); std::cout.tie(nullptr);
    // freopen("moocast.in", "r", stdin);
    // freopen("moocast.out", "w", stdout);
    int t =1 ;
    rp(t)solve();
    return 0;
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided code, let's analyze the code and the problem statement.

### Problem Analysis
- The task is to determine the minimum number of planks of length 60 units needed to cover the three bridges with widths 18, 21, and 25 units.
- Each bridge requires `n` planks of the respective widths.
- The planks can be cut but not joined.

### Buggy Code Analysis
The buggy code is:
```cpp
cout<< n + (n+5/6);
```
This expression is incorrect because:
- The expression `5/6` evaluates to `0` in integer division.
- The correct formula should be `n + (n + 5) / 6`.

### Correct Formula
The correct formula to calculate the minimum number of planks needed is:
- `n` planks for the first bridge (18 units each)
- `n` planks for the second bridge (21 units each)
- `n` planks for the third bridge (25 units each)
- The total length required is `18n + 21n + 25n = 64n` units.
- Since each plank is 60 units, the minimum number of planks required is `64n / 60`, rounded up to the nearest integer.

The correct formula in C++ is:
```cpp
(n * 64 + 59) / 60
```

### Test Input
To expose the bug, we need a test case where the incorrect formula `n + (n + 5/6)` produces a different result from the correct formula `(n * 64 + 59) / 60`.

Let's choose `n = 5`:
- Incorrect formula: `5 + (5 + 5/6) = 5 + (5 + 0) = 10`
- Correct formula: `(5 * 64 + 59) / 60 = (320 + 59) / 60 = 379 / 60 = 7`

### Test Input
```plaintext
5
```

This test input will show that the buggy code outputs `10` while the correct output should be `7`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bridge Renovation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, Monocarp started working as a director of a park located near his house. The park is quite large, so it even has a small river splitting it into several zones. Several bridges are built across this river. Three of these bridges are especially old and need to be repaired.

All three bridges have the same length but differ in width. Their widths are $$$18$$$, $$$21$$$ and $$$25$$$ units, respectively.

During the park renovation process, Monocarp has to replace the old planks that served as the surface of the bridges with the new ones.

Planks are sold with a standard length of $$$60$$$ units. Monocarp already knows that he needs $$$n$$$ planks for each bridge. But since the bridges have different widths, he needs $$$n$$$ planks of length $$$18$$$ for the first bridge, $$$n$$$ planks of length $$$21$$$ for the second one, and $$$n$$$ planks of length $$$25$$$ for the last one.

Workers in charge of renovation have no problem with cutting planks into parts but refuse to join planks, since it creates weak spots and looks ugly.

Monocarp wants to buy as few planks as possible but struggles to calculate the required number of planks. Can you help him?

Input: The first and only line contains a single integer $$$n$$$ ($$$1 \le n \le 1000$$$) — the number of planks required for each of the three bridges.

Output: Print a single integer — the minimum number of planks of standard length ($$$60$$$ units) Monocarp needs to cover all three bridges if the planks can be cut into parts.

Examples:
input:
1
output:
2

input:
3
output:
4

input:
1000
output:
1167

Note: In the first example, it is possible to cut one plank of length $$$60$$$ into three planks with lengths $$$25$$$, $$$18$$$ and $$$17$$$, and cut another plank of length $$$60$$$ into two planks with lengths $$$39$$$ and $$$21$$$. That way, Monocarp will have all the required planks.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
void prime(int n)
{int i,j;
    int prime[n];
   memset(prime,0,sizeof(prime));
    for(i=2;i<=sqrt(n);i++){

        if(prime[i]==0){
            for(j=i*i;j<=n;j+=i){
                prime[j] = 1;
            }

        }

    }
    for(i=2;i<=n;i++){
        if(prime[i]==0){
            cout<<i<<endl;
        }
    }

}
long long int power(int n){
    int i,j=1;
    for(i=0;i<n;i++){
        j*=10;
    }
    return j;

}
int compare(const void *x_void,const void *y_void){

    int x = *(int *)x_void;
    int y = *(int *)y_void;
    return x-y;
}
int compare1(const void *x_void,const void *y_void){

    int x = *(int *)x_void;
    int y = *(int *)y_void;
    return y-x;
}
long long int binary_search(long long int arr[],long long int n,long long int k){
long long int s = 0;
long long int e = n;

for(long long int i=0;s<=e;i++){
    long long int mid = (s+e)/2 ;
    if(arr[mid]==k){
        return mid;
    }
    else if(k<arr[mid]){
            e = mid-1;
    }
    else {s = mid+1;}

}
return -1;
}
void mergee(int a[],int low,int mid,int high){
int n = high-low+1;
int *b = new int[n];
int left = low;
int right = mid+1;
int bidx = 0;
while(left<=mid  && right<=high){
    if(a[left]<=a[right]){
        b[bidx++] = a[left++];
    }
    else{b[bidx++] = a[right++];}
}
while(left<=mid){
    b[bidx++] = a[left++];
}
while(right<=high){
    b[bidx++] = a[right++];
}
for(int k=0;k<n;k++){
    a[low+k] = b[k];
}
}
void mergesort(int a[],int low,int high){
if(low<high){
    int mid = (low+high)/2;
   mergesort(a,low,mid);
   mergesort(a,mid+1,high);
   mergee(a,low,mid,high);
   }

}
int gcd(int a,int b){
    if(b==0){
        return 1;
    }
    else{return gcd(b,a%b);}
}
long long int fun(long long int a){
    long long int c=1;
    for(long long i=0;i<a;i++){
        c*=-1;
    }
    return c;

}
 long long int primal(long long int n){
int i;
for(i=2;i<=sqrt(n);i++){
    if(n%i==0){
        return -1;
    }

}
return 1;
}
long long int findsum(vector<int>v,long long int n){

    long long int sum = 0;
    for(int i =0;i<=n;i++){
        sum+=v[i];
    }
    return sum;

   }
vector<long long int> binary_searchh(long long int low,long long int high,long long int n,long long int k){
    vector<long long int>v;
  long long int rv = 0;
    while(low<=high){
        long long int mid = (low+high)/2;
        if((mid*n)>=k ){
                rv=mid;
                v.push_back(rv);
            high = mid-1;

        }
        else if((mid*n)<k){
            low = mid+1;
        }
    }
    return v;
}
void permutation(int ind,vector<int>v,set<vector<int>>&permit,int n,vector<int>emp){
        if(ind==n){
            return ;
        }
    if(ind==n-1){
        swap(v[0],v[ind]);
    }
    else{swap(v[ind],v[ind+1]);}
    permutation(ind+1,v,permit,n,emp);
    permit.insert(v);
     if(ind==n-1){
        swap(v[0],v[ind]);
    }
    else{swap(v[ind],v[ind+1]);}
    permutation(ind+1,v,permit,n,emp);
   // permit.push_back(v);
}
void sum_sequence(int ind,long long int sum,vector<int>v,vector<int>arr,vector<long long int>&deck,int n){
    if(ind==n){
        if(sum==0){
                int c=0;
            for(int i=0;i<arr.size();i++){
                if(arr[i]==1){
                    c+=1;
                }
            }
            deck.push_back(c);

        }
        return ;
    }
    if(v[ind]<=sum){
        arr.push_back(v[ind]);
    sum_sequence(ind,sum-v[ind],v,arr,deck,n);
    arr.pop_back();
    }
     sum_sequence(ind+1,sum,v,arr,deck,n);
}
void vector_box(int ind,vector<int>v,vector<vector<int>>&ans,vector<int>test,int n){

}
long long int  dp2(int ind,int m,int a,int b,vector<int>&dp){
if(ind<=0){
    return 0;
}
if(dp[ind]!=-1){
    return dp[ind];
}
long long int left = dp2(ind-1,m,a,b,dp)+a;
long long int right = INT_MAX;
    right = dp2(ind-m,m,a,b,dp)+b;

return dp[ind] = min(left,right);
}
bool sequence_sum(int ind,vector<int>v,int sum,vector<vector<int>>&dp){
if(sum==0){
    return true;
}
if(ind==0){
    return (v[ind]==sum);
}
if(dp[ind][sum]!=-1){
    return dp[ind][sum];
}
bool not_pick = sequence_sum(ind-1,v,sum-v[ind],dp);
bool pick  =false;
if(v[ind]<=sum){
bool pick = sequence_sum(ind-1,v,sum-v[ind],dp);
}

return dp[ind][sum] = not_pick | pick;
}
int find_prefixes(int ind,vector<int>v,int sum){
if(ind==0){
    if(sum%v[ind]==0){
        return sum/v[ind];
    }
    else{
        return INT_MAX;
    }
}
int not_take = find_prefixes(ind-1,v,sum);
int take = INT_MAX;
if(v[ind]<=sum){
    take = 1+find_prefixes(ind,v,sum-v[ind]);
}
return min(not_take,take);
}
int  binary(long long int n){
long long int first=1;
long long int last=n;
while(first<=last){
    long long int mid =(first+last)/2;
    if((mid*mid*mid)<n){
        return mid;
    }
    else{last=mid-1;}
}
return -1;
}
int find_val(long long int n,long long int m){
long long int strt=1;
long long int last=m;
long long int val=0;
while(strt<=last){
    long long int mid=(strt+last)/2;
        if(mid*n>=m && ((mid*n)%m==0 || (mid*n)>n)){
            last=mid-1;
            val=mid;
        }
        else{
            strt=mid+1;
        }
}
return val;
}
int idx(vector<long long int>v,long long int l){
    long long int idx;
    long long int low =0;
    long long int high=v.size()-1;
    while(low<=high){
        long long int mid=(low+high)/2;
        if(v[mid]>l){
            high=mid-1;
        }
        else if(v[mid]<l){
            low=mid+1;
        }
        else if(v[mid]==l){
            return mid;
        }
    }

    return -1;
}
int binary_searchh(vector<int>v){
int i = 1;
int j = v.size()-2;
int maximum=0;
while(i<=j){
    int mid = (i+j)/2;

}
   return maximum;
}
string swapper(string &s,int present,int next){
        char s1 = s[present];
        s[present] =s[next];
        s[next] = s1;
        return s;

}
void swapfunction(set<string>&v,int present,int next,int size,string permuter){
        if(next==size || present==size){
            return ;
        }
        swapper(permuter,present,next);
        v.insert(permuter);
        swapper(permuter,present,next);
        swapfunction(v,present,next+1,size,permuter);
        swapper(permuter,present,next);
        swapfunction(v,present+1,0,size,permuter);
}
long long int binary_search_robinHood(vector<long long int>v,int size,double avg,long long int sum){
    int left = 0;
    int right = size;
    if(size%2==0){
    while(left<=right){
            int mid = (right+left)/2;
            if(v[mid]<avg){
                return 0;
            }
            else if(v[mid]==avg){
                return 1;
            }
            else if(v[mid]>avg){
                    return 1+((v[mid]*((size+1)*2))-sum);
            }
            return -1;
    }
    }
    else{
     while(left<=right){
            int mid = (right+left)/2;
            if(v[mid+1]<avg){
                return 0;
            }
            else if(v[mid+1]==avg){
                return 1;
            }
            else if(v[mid+1]>avg){
                    return 1+((v[mid+1]*((size+1)*2))-sum);
            }
            return -1;
    }
    }

}
int findmin(vector<int>v,int n,int strt_idx,int num_idx){
    int final_idx = v.size();
    int ret=-1;
    if(strt_idx==0){
        final_idx = num_idx;
    }
    for(int i=strt_idx;i<final_idx;i++){
            if(v[i]<n){
                ret = i;
                return ret;
            }
    }
    return ret;
}
int print_bst(vector<int>&v,int i){
    int ret_val=-1,mid_val=-1;
    int left = 0;
    int right = v.size()-1;
    while(left<=right){
        int mid = (left+right)/2;
        if(v[mid]>=i){
            ret_val = 1;
            mid_val = mid;
            right =mid-1;
        }
        else if(v[mid]<i){
            left = mid+1;
        }
    }
    if(mid_val!=-1){
        v[mid_val]=-1;
        sort(v.begin(),v.end(),greater<int>());
        v.pop_back();
        sort(v.begin(),v.end());
    }
    return ret_val;
}
bool finddivisor(int n){
    int low = 1;
    int high = sqrt(n);
    int c=1;
    while(low<=high){
        int mid = (low+high)/2;
        if(n%mid==0){
            c+=1;
        }
          high=mid-1;
    }
    return c==3;
}
int binary(vector<long long int>v,long long int find){
    long long int low =0;
    long long int  high =v.size()-1;
    while(low<=high){
        long long int mid = (low+high)/2;
        if(v[mid]==find){
            return 1;
        }
        else if(v[mid]<find){
            low = mid+1;
        }
        else{
            high = mid-1;
        }
    }
return 0;
}
int checkTPrime(long long int number){
    int divisors=0;
        for(int i=2;i<number;i++){
            if(number%i==0){
                divisors+=1;
            }
            if(divisors==2){
                return divisors;
            }
        }
    return divisors;
}
int correctorrder(vector<int>v,int n){
    int size = v.size();
    int c=0;
    for(int i=n;i<size-1;i++){
        if(v[i]>v[i+1]){
            c+=1;
            swap(v[i],v[i+1]);
        }
        else{
            break;
        }
    }
    return c;
}
int findpower(int n){
    int m=1;;
    for(int i=0;i<n;i++){
            m*=10;
    }

    return m;
}
int checkifprime(int n){
    int root = sqrt(n);
    for(int i=2;i<=root;i++){
        if(n%i==0){
            return 1;
        }
    }
    return -1;
}

int main(){
long long int n1,n0,digits,rem,q,r,g,e,m=0,x=0,y,s,i,j,c=0,d=0,n,a,b,maximum=0,t,f=0,k,l,sum;
cin>>n;
sum=n;
if(n<3){
    cout<<n+1;
    return 0;
}
c=0;
while(n>0){
    c+=1;
    n-=3;
}
cout<<sum+ceil(c/2);
return 0;
}

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

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

Think step by step.

Response:
```plaintext
2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bridge Renovation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, Monocarp started working as a director of a park located near his house. The park is quite large, so it even has a small river splitting it into several zones. Several bridges are built across this river. Three of these bridges are especially old and need to be repaired.

All three bridges have the same length but differ in width. Their widths are $$$18$$$, $$$21$$$ and $$$25$$$ units, respectively.

During the park renovation process, Monocarp has to replace the old planks that served as the surface of the bridges with the new ones.

Planks are sold with a standard length of $$$60$$$ units. Monocarp already knows that he needs $$$n$$$ planks for each bridge. But since the bridges have different widths, he needs $$$n$$$ planks of length $$$18$$$ for the first bridge, $$$n$$$ planks of length $$$21$$$ for the second one, and $$$n$$$ planks of length $$$25$$$ for the last one.

Workers in charge of renovation have no problem with cutting planks into parts but refuse to join planks, since it creates weak spots and looks ugly.

Monocarp wants to buy as few planks as possible but struggles to calculate the required number of planks. Can you help him?

Input: The first and only line contains a single integer $$$n$$$ ($$$1 \le n \le 1000$$$) — the number of planks required for each of the three bridges.

Output: Print a single integer — the minimum number of planks of standard length ($$$60$$$ units) Monocarp needs to cover all three bridges if the planks can be cut into parts.

Examples:
input:
1
output:
2

input:
3
output:
4

input:
1000
output:
1167

Note: In the first example, it is possible to cut one plank of length $$$60$$$ into three planks with lengths $$$25$$$, $$$18$$$ and $$$17$$$, and cut another plank of length $$$60$$$ into two planks with lengths $$$39$$$ and $$$21$$$. That way, Monocarp will have all the required planks.

Buggy Code:
n = int(input())
a = 0
c = 0
from math import ceil

l = 0
if n == 5:
    l = 1
    print(6)
if n % 2 == 0:
    a += ceil(n / 2)
    p = ceil(n / 2)
    f = ceil(n / 6)
if n % 2 != 0 and n != 1:
    c = 2
    n = (n - 1)
    f = ceil((n - 3) / 6)

elif n == 1 and l==0:
    print(2)
else:

    print(ceil(n + f + c))

**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
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bridge Renovation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, Monocarp started working as a director of a park located near his house. The park is quite large, so it even has a small river splitting it into several zones. Several bridges are built across this river. Three of these bridges are especially old and need to be repaired.

All three bridges have the same length but differ in width. Their widths are $$$18$$$, $$$21$$$ and $$$25$$$ units, respectively.

During the park renovation process, Monocarp has to replace the old planks that served as the surface of the bridges with the new ones.

Planks are sold with a standard length of $$$60$$$ units. Monocarp already knows that he needs $$$n$$$ planks for each bridge. But since the bridges have different widths, he needs $$$n$$$ planks of length $$$18$$$ for the first bridge, $$$n$$$ planks of length $$$21$$$ for the second one, and $$$n$$$ planks of length $$$25$$$ for the last one.

Workers in charge of renovation have no problem with cutting planks into parts but refuse to join planks, since it creates weak spots and looks ugly.

Monocarp wants to buy as few planks as possible but struggles to calculate the required number of planks. Can you help him?

Input: The first and only line contains a single integer $$$n$$$ ($$$1 \le n \le 1000$$$) — the number of planks required for each of the three bridges.

Output: Print a single integer — the minimum number of planks of standard length ($$$60$$$ units) Monocarp needs to cover all three bridges if the planks can be cut into parts.

Examples:
input:
1
output:
2

input:
3
output:
4

input:
1000
output:
1167

Note: In the first example, it is possible to cut one plank of length $$$60$$$ into three planks with lengths $$$25$$$, $$$18$$$ and $$$17$$$, and cut another plank of length $$$60$$$ into two planks with lengths $$$39$$$ and $$$21$$$. That way, Monocarp will have all the required planks.

Buggy Code:
n=int(input())
doski=60
b=1
for i in range(n):
    if doski >= 21:
        doski=doski-21
    elif doski <21:
        doski=60
        b+=1
        doski = doski - 25
    if doski >= 18:
        doski=doski-18
    elif doski <18:
        doski=60
        b+=1
        doski = doski - 25
    if doski >= 25:
        doski = doski - 25
    elif doski < 25:
        doski = 60
        b += 1
        doski = doski - 25
print(b)

**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 points:

1. **Edge Cases**: The code should handle cases where the remaining length of the plank is just enough to cover one of the bridge widths.
2. **Optimal Usage**: The code should ensure that the planks are used as efficiently as possible, minimizing the number of planks needed.
3. **Multiple Iterations**: The code should handle multiple iterations of plank usage without getting stuck in an inefficient loop.

Let's analyze the provided buggy code. The main issue is that the code does not always use the planks optimally. For example, it tries to use the 21-unit width first, then the 18-unit width, and finally the 25-unit width. This can lead to suboptimal usage of planks.

To expose this bug, we can use a test case where the optimal solution requires a different order of plank usage. Specifically, we can use a case where the 25-unit width should be used first to minimize the number of planks needed.

### Challenging Test Input
Let's consider the case where `n = 5`. This will force the code to handle multiple iterations and show the inefficiency in plank usage.

### Explanation
- For `n = 5`, the optimal solution would be to use the 25-unit width first to minimize the number of planks needed.
- If the code always tries to use the 21-unit width first, it will end up using more planks than necessary.

### Test Input
```plaintext
5
```

This input will force the code to handle multiple iterations and show the inefficiency in plank usage, exposing the bug in the algorithm.