**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DFS Checker (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, you are given a generic tree and the constraints on $n$ and $q$ are higher. You can make hacks only if both versions of the problem are solved.

You are given a rooted tree consisting of $n$ vertices. The vertices are numbered from $1$ to $n$, and the root is the vertex $1$. You are also given a permutation $p_1, p_2, \ldots, p_n$ of $[1,2,\ldots,n]$.

You need to answer $q$ queries. For each query, you are given two integers $x$, $y$; you need to swap $p_x$ and $p_y$ and determine if $p_1, p_2, \ldots, p_n$ is a valid DFS order$^\dagger$ of the given tree.

Please note that the swaps are persistent through queries.

$^\dagger$ A DFS order is found by calling the following $\texttt{dfs}$ function on the given tree.

              dfs_order = []          function dfs(v):         append v to the back of dfs_order         pick an arbitrary permutation s of children of v         for child in s:             dfs(child)     dfs(1)

Note that the DFS order is not unique.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $q$ ($2\le n\le 3\cdot 10^5$, $2\le q\le 10^5$) — the number of vertices in the tree and the number of queries.

The next line contains $n-1$ integers $a_2,a_3,\ldots,a_n$ ($1\le a_i<i$) — the parent of each vertex in the given tree.

The next line contains $n$ integers $p_1,p_2,\ldots,p_n$ ($1\le p_i\le n$, all $p_i$ are distinct) — the initial permutation $p$.

The next $q$ lines each contain two integers $x$, $y$ ($1\le x,y\le n,x\neq y$) — the positions of the elements to swap in the permutation.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$, and the sum of all $q$ does not exceed $10^5$.

Output: For each test case, print $q$ lines corresponding to the $q$ queries. For each query, output $\texttt{YES}$ if there is a DFS order that exactly equals the current permutation, and output $\texttt{NO}$ otherwise.

You can output $\texttt{Yes}$ and $\texttt{No}$ in any case (for example, strings $\texttt{yEs}$, $\texttt{yes}$, $\texttt{Yes}$, and $\texttt{YES}$ will be recognized as a positive response).

Examples:
input:
3
3 3
1 1
1 2 3
2 3
3 2
1 3
7 4
1 1 2 2 3 3
1 2 3 4 5 6 7
3 5
2 5
3 7
4 6
5 4
1 1 3 4
2 3 4 5 1
5 1
4 5
3 4
2 3
output:
YES
YES
NO
YES
NO
NO
YES
YES
NO
NO
YES

Note: In the first test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,3,2],[1,2,3],[3,2,1]$, respectively. The first two permutations are valid DFS orders; the third is not a DFS order.

In the second test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,2,5,4,3,6,7],[1,3,5,4,2,6,7],[1,3,7,4,2,6,5],[1,3,7,6,2,4,5]$, respectively.

Buggy Code:
#from math import ceil,log
#from collections import deque,Counter
#from bisect import bisect
#from heapq import heappush,heappop
#from random import randint

from sys import stdin,stdout
input=lambda:stdin.readline().rstrip()
print=lambda *x,sep=' ',end='\n':stdout.write(sep.join(map(str,x))+end)

def out(l):
    print('\n'.join(map(str,l)))

def yes():
    print('Yes')

def no():
    print('No')

def alice():
    print('Alice')

def bob():
    print('Bob')

def check(i):
    if p[i]==1:
        return 1 if i==1 else 0
    prev=p[i-1]
    pa=parent[p[i]]
    if pa==prev:
        return 1
    if tin[prev]!=tout[prev]:
        return 0
    if tin[pa]<=tin[prev]<=tout[pa]:
        return 1
    return 0

def dfs(par,grand_par=0):
    timer[-1]+=1
    tin[par]=tout[par]=timer[-1]
    for child in adj[par]:
        if child!=grand_par:
            dfs(child,par)
            tout[par]=tout[child]

def solve():
    global adj,tin,tout,p,parent,timer
    n,q=map(int,input().split())
    parent=[0,0]+list(map(int,input().split()))
    p=[0]+list(map(int,input().split()))
    query=[tuple(map(int,input().split())) for i in range(q)]
    adj=[[] for i in range(n+1)]
    tin=[0]*(n+1)
    tout=[0]*(n+1)
    timer=[0]
    for i,j in enumerate(parent):
        if j==0:
            continue
        adj[j].append(i)
    dfs(1,1)
    #print(tin,tout)
    cnt=0
    for i in range(1,n+1):
        cnt+=check(i)
    ans=[]
    for x,y in query:
        s=set()
        l=[x,y,x-1,x+1,y-1,y+1]
        for j in l:
            if 1<=j<=n:
                s.add(j)
        for j in s:
            cnt-=check(j)
        p[x],p[y]=p[y],p[x]
        for j in s:
            cnt+=check(j)
        ans.append('YES' if cnt==n else 'NO')
    out(ans)

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

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DFS Checker (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, you are given a generic tree and the constraints on $n$ and $q$ are higher. You can make hacks only if both versions of the problem are solved.

You are given a rooted tree consisting of $n$ vertices. The vertices are numbered from $1$ to $n$, and the root is the vertex $1$. You are also given a permutation $p_1, p_2, \ldots, p_n$ of $[1,2,\ldots,n]$.

You need to answer $q$ queries. For each query, you are given two integers $x$, $y$; you need to swap $p_x$ and $p_y$ and determine if $p_1, p_2, \ldots, p_n$ is a valid DFS order$^\dagger$ of the given tree.

Please note that the swaps are persistent through queries.

$^\dagger$ A DFS order is found by calling the following $\texttt{dfs}$ function on the given tree.

              dfs_order = []          function dfs(v):         append v to the back of dfs_order         pick an arbitrary permutation s of children of v         for child in s:             dfs(child)     dfs(1)

Note that the DFS order is not unique.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $q$ ($2\le n\le 3\cdot 10^5$, $2\le q\le 10^5$) — the number of vertices in the tree and the number of queries.

The next line contains $n-1$ integers $a_2,a_3,\ldots,a_n$ ($1\le a_i<i$) — the parent of each vertex in the given tree.

The next line contains $n$ integers $p_1,p_2,\ldots,p_n$ ($1\le p_i\le n$, all $p_i$ are distinct) — the initial permutation $p$.

The next $q$ lines each contain two integers $x$, $y$ ($1\le x,y\le n,x\neq y$) — the positions of the elements to swap in the permutation.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$, and the sum of all $q$ does not exceed $10^5$.

Output: For each test case, print $q$ lines corresponding to the $q$ queries. For each query, output $\texttt{YES}$ if there is a DFS order that exactly equals the current permutation, and output $\texttt{NO}$ otherwise.

You can output $\texttt{Yes}$ and $\texttt{No}$ in any case (for example, strings $\texttt{yEs}$, $\texttt{yes}$, $\texttt{Yes}$, and $\texttt{YES}$ will be recognized as a positive response).

Examples:
input:
3
3 3
1 1
1 2 3
2 3
3 2
1 3
7 4
1 1 2 2 3 3
1 2 3 4 5 6 7
3 5
2 5
3 7
4 6
5 4
1 1 3 4
2 3 4 5 1
5 1
4 5
3 4
2 3
output:
YES
YES
NO
YES
NO
NO
YES
YES
NO
NO
YES

Note: In the first test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,3,2],[1,2,3],[3,2,1]$, respectively. The first two permutations are valid DFS orders; the third is not a DFS order.

In the second test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,2,5,4,3,6,7],[1,3,5,4,2,6,7],[1,3,7,4,2,6,5],[1,3,7,6,2,4,5]$, respectively.

Buggy Code:
class TreeAncestor:
    def __init__(self, parent):
        n = len(parent) # 节点总数
        m = n.bit_length()  # 祖先表的列数
        g = [[] for _ in range(n + 1)]  # 邻接表，根节点编号为 1
        for i in range(2, n):  # 从 2 开始，因为节点 1 是根节点
            g[parent[i]].append(i)

        depth = [0] * (n + 1)  # 节点深度
        pa = [[-1] * m for _ in range(n + 1)]  # 祖先表

        def dfs(x: int, fa: int) -> None:
            pa[x][0] = fa
            for y in g[x]:
                if y != fa:
                    depth[y] = depth[x] + 1
                    dfs(y, x)

        dfs(1, -1)  # 从根节点 1 开始 DFS
        for i in range(m - 1):
            for x in range(1, n + 1):
                if (p := pa[x][i]) != -1:
                    pa[x][i + 1] = pa[p][i]

        self.depth = depth
        self.pa = pa

    def get_kth_ancestor(self, node: int, k: int) -> int:
        for i in range(k.bit_length()):
            if (k >> i) & 1:  # k 二进制从低到高第 i 位是 1
                node = self.pa[node][i]
                if node == -1:
                    return -1
        return node

    def get_lca(self, x: int, y: int) -> int:
        if self.depth[x] > self.depth[y]:
            x, y = y, x
        y = self.get_kth_ancestor(y, self.depth[y] - self.depth[x])
        if y == x:
            return x
        for i in range(len(self.pa[x]) - 1, -1, -1):
            px, py = self.pa[x][i], self.pa[y][i]
            if px != py:
                x, y = px, py  # 同时上跳 2**i 步
        return self.pa[x][0]

def check(a, b):
    # print(a, b)
    # print(tree.get_lca(a, b), parent[b])
    if a == 0:
        return b == 1
    return tree.get_lca(a, b) == parent[b]

for _ in range(int(input())):
    # n = int(input())
    n, q = list(map(int, input().split()))
    # nums = list(map(int, input().split()))
    # print(min(k, m) * min(k, n))

    parent = [1, 1] + list(map(int, input().split()))
    tree = TreeAncestor(parent)
    nums = [0] + list(map(int, input().split()))
    first = (n + 1) >> 1
    tf = [0] * (n+1)
    correct = 0
    for i in range(1, n+1):
        if check(nums[i-1], nums[i]):
            tf[i] = 1
            correct += 1

    for _ in range(q):

        a, b = list(map(int, input().split()))
        nums[a], nums[b] = nums[b], nums[a]
        t = [a, b, a+1, b+1]
        for node in t:
            if node <= n and tf[node] == 1:
                tf[node] = 0
                correct -= 1
        for i in t:
            if i <= n and tf[i] == 0 and check(nums[i-1], nums[i]):
                tf[i] = 1
                correct += 1
        # print(nums)
        # print(tf)

        if correct == n:
            print("YES")
        else:
            print("NO")

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DFS Checker (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, you are given a generic tree and the constraints on $n$ and $q$ are higher. You can make hacks only if both versions of the problem are solved.

You are given a rooted tree consisting of $n$ vertices. The vertices are numbered from $1$ to $n$, and the root is the vertex $1$. You are also given a permutation $p_1, p_2, \ldots, p_n$ of $[1,2,\ldots,n]$.

You need to answer $q$ queries. For each query, you are given two integers $x$, $y$; you need to swap $p_x$ and $p_y$ and determine if $p_1, p_2, \ldots, p_n$ is a valid DFS order$^\dagger$ of the given tree.

Please note that the swaps are persistent through queries.

$^\dagger$ A DFS order is found by calling the following $\texttt{dfs}$ function on the given tree.

              dfs_order = []          function dfs(v):         append v to the back of dfs_order         pick an arbitrary permutation s of children of v         for child in s:             dfs(child)     dfs(1)

Note that the DFS order is not unique.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $q$ ($2\le n\le 3\cdot 10^5$, $2\le q\le 10^5$) — the number of vertices in the tree and the number of queries.

The next line contains $n-1$ integers $a_2,a_3,\ldots,a_n$ ($1\le a_i<i$) — the parent of each vertex in the given tree.

The next line contains $n$ integers $p_1,p_2,\ldots,p_n$ ($1\le p_i\le n$, all $p_i$ are distinct) — the initial permutation $p$.

The next $q$ lines each contain two integers $x$, $y$ ($1\le x,y\le n,x\neq y$) — the positions of the elements to swap in the permutation.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$, and the sum of all $q$ does not exceed $10^5$.

Output: For each test case, print $q$ lines corresponding to the $q$ queries. For each query, output $\texttt{YES}$ if there is a DFS order that exactly equals the current permutation, and output $\texttt{NO}$ otherwise.

You can output $\texttt{Yes}$ and $\texttt{No}$ in any case (for example, strings $\texttt{yEs}$, $\texttt{yes}$, $\texttt{Yes}$, and $\texttt{YES}$ will be recognized as a positive response).

Examples:
input:
3
3 3
1 1
1 2 3
2 3
3 2
1 3
7 4
1 1 2 2 3 3
1 2 3 4 5 6 7
3 5
2 5
3 7
4 6
5 4
1 1 3 4
2 3 4 5 1
5 1
4 5
3 4
2 3
output:
YES
YES
NO
YES
NO
NO
YES
YES
NO
NO
YES

Note: In the first test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,3,2],[1,2,3],[3,2,1]$, respectively. The first two permutations are valid DFS orders; the third is not a DFS order.

In the second test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,2,5,4,3,6,7],[1,3,5,4,2,6,7],[1,3,7,4,2,6,5],[1,3,7,6,2,4,5]$, respectively.

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

public class D_2_DFS_Checker_Hard_Version {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

    static FastReader vt = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);

    static int[] par;
    static int[] per;
    static int[] pos;
    static ArrayList<ArrayList<Integer>> tree;
    static int[][] res;

    public static void main(String[] args) {
        // try {
        int testCases = vt.nextInt();
        while (testCases-- > 0) {
            solve();
        }
        out.flush();
        // } catch (Exception e) {
        // return;
        // }
    }

    static void solve() {
        int n = vt.nextInt();
        int q = vt.nextInt();

        par = new int[n + 1];
        per = new int[n + 1];
        pos = new int[n + 1];
        tree = new ArrayList<>();
        res = new int[n + 1][3];

        for (int i = 0; i <= n; ++i) {
            tree.add(new ArrayList<>());
        }

        for (int i = 2; i <= n; ++i) {
            par[i] = vt.nextInt();
            tree.get(par[i]).add(i);
        }

        for (int i = 1; i <= n; ++i) {
            per[i] = vt.nextInt();
            pos[per[i]] = i;
        }
        computeRes(1);
        for (int i = 0; i < q; ++i) {
            int x = vt.nextInt();
            int y = vt.nextInt();
            swap(x, y);
            x = per[x];
            y = per[y];
            int depthx = 0;
            int depthy = 0;
            int root = x;
            while (par[root] != 0) {
                root = par[root];
                ++depthx;
            }
            root = y;
            while (par[root] != 0) {
                root = par[root];
                ++depthy;
            }
            int rootx = x;
            int rooty = y;
            int t1 = 0;
            int t2 = 0;
            while (depthx >= 0 || depthy >= 0) {
                if (depthx > depthy) {
                    if (updateres(rootx) && t1 != 0) {
                        depthx = -1;
                    }
                    t1 = 1;
                    rootx = par[rootx];
                    depthx--;
                } else {
                    if (updateres(rooty) && t2 != 0) {
                        depthy = -1;
                    }
                    t2 = 1;
                    rooty = par[rooty];
                    --depthy;
                }
            }
            if (res[1][2] == 1) {
                out.println("YES");
            } else {
                out.println("NO");
            }
        }
    }

    static boolean updateres(int node) {
        if (node == 0) {
            return true;
        }
        if (tree.get(node).size() == 0) {
            res[node][0] = pos[node];
            res[node][1] = pos[node];
            res[node][2] = 1;
            return false;
        }
        int parent = node;
        int prevmin = res[parent][0];
        int prevmax = res[parent][1];
        int prev = res[parent][2];
        int minRange = 10000000;
        int maxRange = 0;
        int c = 1;
        ArrayList<int[]> ranges = new ArrayList<>();
        for (Integer child : tree.get(parent)) {
            ranges.add(new int[] { Math.min(res[child][0], pos[child]), Math.max(res[child][1], pos[child]) });
            minRange = Math.min(minRange, Math.min(res[child][0], pos[child]));
            maxRange = Math.max(maxRange, Math.max(res[child][1], pos[child]));
            c &= res[child][2];
        }
        res[parent][0] = minRange;
        res[parent][1] = maxRange;
        if (c != 1 || pos[parent] >= minRange) {
            res[parent][2] = 0;
            if (prevmin == minRange && prevmax == maxRange && prev == 0) {
                return true;
            }
            return false;
        }
        Collections.sort(ranges, (a, b) -> a[0] != b[0] ? a[0] - b[0] : a[1] - b[1]);
        boolean valid = true;
        for (int i = 1; i < ranges.size(); ++i) {
            if (ranges.get(i)[0] < ranges.get(i - 1)[1]) {
                valid = false;
                break;
            }
        }
        if (valid) {
            res[parent][2] = 1;
        } else {
            res[parent][2] = 0;
        }
        if (prevmin == minRange && prevmax == maxRange && prev == res[parent][2]) {
            return true;
        }
        return false;
    }

    static void computeRes(int node) {
        int minRange = 10000000;
        int maxRange = 0;
        boolean valid = true;
        ArrayList<int[]> ranges = new ArrayList<>();
        int c = 1;
        for (int child : tree.get(node)) {
            computeRes(child);
            ranges.add(new int[] { Math.min(res[child][0], pos[child]), Math.max(res[child][1], pos[child]) });
            minRange = Math.min(minRange, Math.min(res[child][0], pos[child]));
            maxRange = Math.max(maxRange, Math.max(res[child][1], pos[child]));
            c &= res[child][2];
        }
        if (tree.get(node).size() == 0) {
            res[node][0] = pos[node];
            res[node][1] = pos[node];
            res[node][2] = 1;
            return;
        }
        if (c != 1 || pos[node] >= minRange) {
            res[node][2] = 0;
            return;
        }
        Collections.sort(ranges, (a, b) -> a[0] != b[0] ? a[0] - b[0] : a[1] - b[1]);
        res[node][0] = minRange;
        res[node][1] = maxRange;
        for (int i = 1; i < ranges.size(); ++i) {
            if (ranges.get(i)[0] < ranges.get(i - 1)[1]) {
                valid = false;
                break;
            }
        }
        if (valid) {
            res[node][2] = 1;
        }
    }

    static void swap(int x, int y) {
        int temp = per[x];
        per[x] = per[y];
        per[y] = temp;
        pos[per[x]] = x;
        pos[per[y]] = y;
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DFS Checker (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, you are given a generic tree and the constraints on $n$ and $q$ are higher. You can make hacks only if both versions of the problem are solved.

You are given a rooted tree consisting of $n$ vertices. The vertices are numbered from $1$ to $n$, and the root is the vertex $1$. You are also given a permutation $p_1, p_2, \ldots, p_n$ of $[1,2,\ldots,n]$.

You need to answer $q$ queries. For each query, you are given two integers $x$, $y$; you need to swap $p_x$ and $p_y$ and determine if $p_1, p_2, \ldots, p_n$ is a valid DFS order$^\dagger$ of the given tree.

Please note that the swaps are persistent through queries.

$^\dagger$ A DFS order is found by calling the following $\texttt{dfs}$ function on the given tree.

              dfs_order = []          function dfs(v):         append v to the back of dfs_order         pick an arbitrary permutation s of children of v         for child in s:             dfs(child)     dfs(1)

Note that the DFS order is not unique.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $q$ ($2\le n\le 3\cdot 10^5$, $2\le q\le 10^5$) — the number of vertices in the tree and the number of queries.

The next line contains $n-1$ integers $a_2,a_3,\ldots,a_n$ ($1\le a_i<i$) — the parent of each vertex in the given tree.

The next line contains $n$ integers $p_1,p_2,\ldots,p_n$ ($1\le p_i\le n$, all $p_i$ are distinct) — the initial permutation $p$.

The next $q$ lines each contain two integers $x$, $y$ ($1\le x,y\le n,x\neq y$) — the positions of the elements to swap in the permutation.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$, and the sum of all $q$ does not exceed $10^5$.

Output: For each test case, print $q$ lines corresponding to the $q$ queries. For each query, output $\texttt{YES}$ if there is a DFS order that exactly equals the current permutation, and output $\texttt{NO}$ otherwise.

You can output $\texttt{Yes}$ and $\texttt{No}$ in any case (for example, strings $\texttt{yEs}$, $\texttt{yes}$, $\texttt{Yes}$, and $\texttt{YES}$ will be recognized as a positive response).

Examples:
input:
3
3 3
1 1
1 2 3
2 3
3 2
1 3
7 4
1 1 2 2 3 3
1 2 3 4 5 6 7
3 5
2 5
3 7
4 6
5 4
1 1 3 4
2 3 4 5 1
5 1
4 5
3 4
2 3
output:
YES
YES
NO
YES
NO
NO
YES
YES
NO
NO
YES

Note: In the first test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,3,2],[1,2,3],[3,2,1]$, respectively. The first two permutations are valid DFS orders; the third is not a DFS order.

In the second test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,2,5,4,3,6,7],[1,3,5,4,2,6,7],[1,3,7,4,2,6,5],[1,3,7,6,2,4,5]$, respectively.

Buggy Code:
import java.util.*;

public class Main {

    private static long[] a, p, pos, f;
    private static List<Set<Long>> s;
    private static List<List<Long>> adj;
    private static long bad;

    private static boolean calc(long i) {
        if (s.get((int)i).isEmpty()) return false;
        long x = pos[(int)i];
        long y = Collections.min(s.get((int)i));
        return y < x;
    }

    private static void del(long i, long x) {
        Set<Long> set = s.get((int)i);
        Long[] arr = set.toArray(new Long[0]);
        Arrays.sort(arr);

        long l = -1, r = -1;
        int j = 0;
        while (j < arr.length) {
            if (arr[j] == x) {
                if (j > 0) {
                    l = arr[j - 1];
                    if (l + f[(int)p[(int)l]] != x) bad--;
                }
                if (j < arr.length - 1) {
                    r = arr[j + 1];
                    if (x + f[(int)p[(int)x]] != r) bad--;
                }
                if (l != -1 && r != -1) {
                    if (l + f[(int)p[(int)l]] != r) bad++;
                }
                break;
            }
            j++;
        }
        set.remove(x);
    }

    private static void add(long i, long x) {
        Set<Long> set = s.get((int)i);
        set.add(x);
        Long[] arr = set.toArray(new Long[0]);
        Arrays.sort(arr);

        long l = -1, r = -1;
        int j = 0;
        while (j < arr.length) {
            if (arr[j] == x) {
                if (j > 0) {
                    l = arr[j - 1];
                    if (l + f[(int)p[(int)l]] != x) bad++;
                }
                if (j < arr.length - 1) {
                    r = arr[j + 1];
                    if (x + f[(int)p[(int)x]] != r) bad++;
                }
                if (l != -1 && r != -1) {
                    if (l + f[(int)p[(int)l]] != r) bad--;
                }
                break;
            }
            j++;
        }
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();

        while (t-- > 0) {
            int n = scanner.nextInt();
            int q = scanner.nextInt();

            a = new long[n];
            Arrays.fill(a, -1);

            int i = 1;
            while (i < n) {
                a[i] = scanner.nextLong() - 1;
                i++;
            }

            p = new long[n];
            i = 0;
            while (i < n) {
                p[i] = scanner.nextLong() - 1;
                i++;
            }

            pos = new long[n];
            i = 0;
            while (i < n) {
                pos[(int)p[i]] = i;
                i++;
            }

            long good = 0;
            s = new ArrayList<>(n);
            i = 0;
            while (i < n) {
                s.add(new HashSet<>());
                i++;
            }

            f = new long[n];
            adj = new ArrayList<>(n);
            i = 0;
            while (i < n) {
                adj.add(new ArrayList<>());
                i++;
            }

            i = n - 1;
            while (i >= 0) {
                if (i > 0) adj.get((int)a[i]).add((long)i);
                int j = 0;
                while (j < adj.get((int)i).size()) {
                    long k = adj.get((int)i).get(j);
                    f[(int)i] += f[(int)k];
                    j++;
                }
                f[(int)i]++;
                i--;
            }

            bad = 0;
            i = 0;
            while (i < n) {
                if (a[(int)p[i]] != -1) add(a[(int)p[i]], i);
                i++;
            }

            i = 0;
            while (i < n) {
                bad += (calc(i) ? 1 : 0);
                i++;
            }

            while (q-- > 0) {
                long x = scanner.nextLong() - 1;
                long y = scanner.nextLong() - 1;

                Set<Long> change = new HashSet<>();
                change.add(p[(int)x]);
                change.add(p[(int)y]);
                change.add(a[(int)p[(int)x]]);
                change.add(a[(int)p[(int)y]]);

                for (long z : change) {
                    if (z != -1) bad -= (calc(z) ? 1 : 0);
                }

                if (a[(int)p[(int)x]] != -1) del(a[(int)p[(int)x]], x);
                if (a[(int)p[(int)y]] != -1) del(a[(int)p[(int)y]], y);

                long temp = p[(int)x];
                p[(int)x] = p[(int)y];
                p[(int)y] = temp;

                long tempPos = pos[(int)p[(int)x]];
                pos[(int)p[(int)x]] = pos[(int)p[(int)y]];
                pos[(int)p[(int)y]] = tempPos;

                if (a[(int)p[(int)x]] != -1) add(a[(int)p[(int)x]], x);
                if (a[(int)p[(int)y]] != -1) add(a[(int)p[(int)y]], y);

                for (long z : change) {
                    if (z != -1) bad += (calc(z) ? 1 : 0);
                }

                System.out.println(bad == 0 ? "YES" : "NO");
            }
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given an integer array $a$ of size $n$.

Let's define the value of the array as its size minus the number of set bits in the bitwise OR of all elements of the array.

For example, for the array $[1, 0, 1, 2]$, the bitwise OR is $3$ (which contains $2$ set bits), and the value of the array is $4-2=2$.

Your task is to calculate the maximum possible value of some subsequence of the given array.

Input: The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

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

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i < 2^{60}$).

Output: For each test case, print the maximum possible value of some subsequence of the given array.

Examples:
input:
4
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
output:
3
2
0
3

Note:

Buggy Code:
#include<bits/stdc++.h>
#define endl '\n'
#define ll long long
#define pii pair<int, int>
const ll inf = 1145141919810;
using namespace std;
void slv(){
    ll n;
    cin >> n;
    vector<ll> a(n + 1);
    for(int i = 1; i <= n; ++i) cin >> a[i];

    sort(a.begin(), a.end());

    ll ans = 0;
    for(ll i = 1, j; i <= n; ++i) {
        for(j = 0; a[i] >= (1 << j); ++j);
        ans = max(ans, i - j);
    }
    cout << ans << endl;
}

int main(){
    ios::sync_with_stdio(0), cin.tie(0);
    int _ = 1;
    cin >> _;
    while(_--) slv();
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given an integer array $a$ of size $n$.

Let's define the value of the array as its size minus the number of set bits in the bitwise OR of all elements of the array.

For example, for the array $[1, 0, 1, 2]$, the bitwise OR is $3$ (which contains $2$ set bits), and the value of the array is $4-2=2$.

Your task is to calculate the maximum possible value of some subsequence of the given array.

Input: The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

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

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i < 2^{60}$).

Output: For each test case, print the maximum possible value of some subsequence of the given array.

Examples:
input:
4
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
output:
3
2
0
3

Note:

Buggy Code:
#include <iostream>
#include <math.h>
#include <algorithm>

#define int long long

using namespace std;

int a[101];

void solve()
{
	int n;

	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>a[i];

		if(a[i]) a[i]=(int)log2(a[i])+1;
	}

	sort(a+1,a+n+1);

	int maxi=0;
	for(int i=n;i>=1;i--){
		maxi=max(maxi,i-a[i]);
	}

	cout<<maxi<<'\n';

	return;
}

signed main()
{
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);

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

	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
4
1 2 4 8
```

### Explanation:
This test case is designed to expose the bug in the buggy code. Here’s why:

1. **Input Array**: The array `[1, 2, 4, 8]` consists of powers of 2. Each element has exactly one set bit in its binary representation.
2. **Bitwise OR**: The bitwise OR of all elements in the array is `1 | 2 | 4 | 8 = 15` (which is `1111` in binary, having 4 set bits).
3. **Value Calculation**: The value of the array is `4 - 4 = 0`.

However, the buggy code incorrectly calculates the value of each element using `log2(a[i]) + 1`, which gives the position of the highest set bit plus one. For the array `[1, 2, 4, 8]`, this results in `[1, 2, 3, 4]`. The code then sorts this array and tries to find the maximum value of `i - a[i]`, which will be `4 - 1 = 3`.

This test case will show that the buggy code outputs `3` instead of the correct answer `0`, thereby exposing the bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given an integer array $a$ of size $n$.

Let's define the value of the array as its size minus the number of set bits in the bitwise OR of all elements of the array.

For example, for the array $[1, 0, 1, 2]$, the bitwise OR is $3$ (which contains $2$ set bits), and the value of the array is $4-2=2$.

Your task is to calculate the maximum possible value of some subsequence of the given array.

Input: The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

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

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i < 2^{60}$).

Output: For each test case, print the maximum possible value of some subsequence of the given array.

Examples:
input:
4
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
output:
3
2
0
3

Note:

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

const ll inf = (1ll<<60);

void solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    map<ll, int> m;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        m[a[i]] = 0;
        for (int j = 0; (1<<j) <= a[i]; j++) {
            if ((a[i] >> j) & 1) {
                m[j + inf] = 0;
            }
        }
    }
    int k = 1;
    for (auto &[x, y] : m) {
        y = k++;
    }

    vector<vector<int>> adj(k);

    for (auto [x, y] : m) {
        if (x >= inf) continue;
        for (int j = 0; (1<<j) <= x; j++) {
            if ((x >> j) && 1) {
                adj[y].push_back(m[j + inf]);
                adj[m[j + inf]].push_back(y);
            }
        }
    }

    vector<ll> match(k, 0), dist(k, 0);

    auto bfs = [&]() -> bool {
        queue<int> q;

        for (int i = 1; i < k; i++) {
            if (!match[i]) {
                dist[i] = 0;
                q.push(i);
            } else {
                dist[i] = inf;
            }
        }

        dist[0] = inf;

        while (!q.empty()) {
            auto u = q.front();
            q.pop();
            if (dist[u] >= dist[0]) continue;
            for (auto v : adj[u]) {
                // if v is matched
                if (dist[match[v]] == inf) {
                    dist[match[v]] = dist[u] + 1;
                    q.push(match[v]);
                }
            }
        }

        return dist[0] != inf;
    };

    auto dfs = [&](int u, auto&& dfs) -> bool {
        if (u == 0) return true;
        for (auto v : adj[u]) {
            if (dist[match[v]] == dist[u] + 1 && dfs(match[v], dfs)) {
                match[u] = v;
                match[v] = u;
                return true;
            }
        }
        dist[u] = inf;
        return false;
    };

    auto hopcroft_karp = [&]() -> int {
        int cnt = 0;

        while (bfs()) {
            for (int i = 1; i < k; i++) {
                if (!match[i] && dfs(i, dfs)) cnt++;
            }
        }

        return cnt;
    };

    cout << max(0, n - hopcroft_karp()) << endl;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t = 1;
    cin >> t;
    while (t-->0) {
        solve();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
4
1 2 4 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given an integer array $a$ of size $n$.

Let's define the value of the array as its size minus the number of set bits in the bitwise OR of all elements of the array.

For example, for the array $[1, 0, 1, 2]$, the bitwise OR is $3$ (which contains $2$ set bits), and the value of the array is $4-2=2$.

Your task is to calculate the maximum possible value of some subsequence of the given array.

Input: The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

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

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i < 2^{60}$).

Output: For each test case, print the maximum possible value of some subsequence of the given array.

Examples:
input:
4
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
output:
3
2
0
3

Note:

Buggy Code:
import math
import io
import heapq
import bisect
from collections import defaultdict, deque, Counter
import re
import numpy as np
from itertools import permutations
import math

MOD = 1000000007
PI = 3.1415926535897932384626433832795
INF = 10**18

def gcd(a, b):
    if b == 0:
        return a
    return gcd(b, a % b)

def lcm(a, b):
    return (a // gcd(a, b)) * b

def yes():
    print("YES")

def no():
    print("NO")

def ok():
    print("OK")

mxN = int(2e5)

def calculateScore(arr):
    Or = 0
    for i in arr:
        Or |= i
    return len(arr) - bin(Or).count('1')

def solve(n, arr):
    max_score = 0
    for subset in range(1 << n):
        current_subset = [arr[i] for i in range(n) if subset & (1 << i)]
        max_score = max(max_score, calculateScore(current_subset))
    return max_score

def main():
    n = int(input())
    arr = list(map(int, input().split()))
    print(solve(n, arr))
    pass

if __name__ == '__main__':
    t = int(input())
    for _ in range(t):
        main()

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases or scenarios that might cause the code to fail. Here are some potential issues to consider:

1. **Empty Subsequence**: The code should handle the case where the best subsequence might be empty (i.e., the value of the array is 0).
2. **Large Bitwise OR**: The code should handle cases where the bitwise OR of the elements has a large number of set bits.
3. **Multiple Test Cases**: The code should handle multiple test cases correctly.
4. **Large Input Size**: The code should handle the maximum input size efficiently.

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

1. **Test Case 1**: An array with a single element that has all bits set.
2. **Test Case 2**: An array with elements that have a large number of set bits in their bitwise OR.
3. **Test Case 3**: An array with a mix of elements, some of which have overlapping set bits.

Here is the test input:

```plaintext
3
1
63
4
1 2 4 8
8
7 1 48 14 13 8 7 6
```

Explanation:
- **Test Case 1**: The array `[63]` has a single element with all bits set (in binary: `111111`), and the value of the array is `1 - 6 = -5`. This tests the handling of a single element with many set bits.
- **Test Case 2**: The array `[1, 2, 4, 8]` has elements with no overlapping set bits. The bitwise OR is `15` (in binary: `1111`), and the value of the array is `4 - 4 = 0`. This tests the handling of elements with no overlapping set bits.
- **Test Case 3**: The array `[7, 1, 48, 14, 13, 8, 7, 6]` has elements with overlapping set bits. The bitwise OR is `63` (in binary: `111111`), and the value of the array is `8 - 6 = 2`. This tests the handling of a mix of elements with overlapping set bits.

This test input should help expose any issues 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: Best Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given an integer array $a$ of size $n$.

Let's define the value of the array as its size minus the number of set bits in the bitwise OR of all elements of the array.

For example, for the array $[1, 0, 1, 2]$, the bitwise OR is $3$ (which contains $2$ set bits), and the value of the array is $4-2=2$.

Your task is to calculate the maximum possible value of some subsequence of the given array.

Input: The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

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

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i < 2^{60}$).

Output: For each test case, print the maximum possible value of some subsequence of the given array.

Examples:
input:
4
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
output:
3
2
0
3

Note:

Buggy Code:
import sys
sint = lambda: sys.stdin.readline().rstrip("\r\n")
nint = lambda: int(input())
mint = lambda: map(int, input().split())
lint = lambda: list(map(int, input().split()))

def solve(n, A):
    d = 0
    for mask in range(1 << n):
        curr = 0
        sz = 0
        for i in range(n):
            if mask & 1 << i :
                curr |= A[i]
                sz += 1

        d = max(d, sz - curr.bit_length())

    return d

if __name__ == "__main__":
    t = nint()
    for _ in range(t):
        n = nint()
        A = lint()
        ans = solve(n, A)
        print(ans)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on scenarios where the bitwise OR operation and the bit counting can lead to incorrect results. The bug in the code is that it calculates the value of the subsequence using `sz - curr.bit_length()`, which does not correctly account for the number of set bits in the bitwise OR of the subsequence.

### Steps to Generate the Test Input:

1. **Edge Case with All Zeroes:**
   - This should be handled correctly, but it's a good baseline.
   - Example: `3 0 0 0` (Expected output: 3)

2. **Single Element:**
   - A single element should return the value based on its bit length.
   - Example: `1 5` (Expected output: 0, since 5 in binary is 101, and 1 - 3 = -2, but the minimum value is 0)

3. **Multiple Elements with High Bit Length:**
   - Choose elements with high bit lengths to maximize the bitwise OR.
   - Example: `4 7 1 48 14` (Expected output: 2, since the bitwise OR is 63, which has 6 set bits, and 4 - 6 = -2, but the minimum value is 0)

4. **Elements with Overlapping Set Bits:**
   - Choose elements that have overlapping set bits to ensure the bitwise OR has fewer set bits than the sum of individual set bits.
   - Example: `3 1 2 3` (Expected output: 1, since the bitwise OR is 3, which has 2 set bits, and 3 - 2 = 1)

5. **Mixed Elements:**
   - Combine elements with different bit lengths to test the algorithm's ability to handle varied inputs.
   - Example: `8 7 1 48 14 13 8 7 6` (Expected output: 3, as given in the example)

### Chosen Test Input:
```plaintext
4
3 0 0 0
1 5
4 7 1 48 14
3 1 2 3
```

This test input covers a range of scenarios, including edge cases and combinations of elements with different bit lengths, to ensure that the buggy code's logic is thoroughly tested.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given an integer array $a$ of size $n$.

Let's define the value of the array as its size minus the number of set bits in the bitwise OR of all elements of the array.

For example, for the array $[1, 0, 1, 2]$, the bitwise OR is $3$ (which contains $2$ set bits), and the value of the array is $4-2=2$.

Your task is to calculate the maximum possible value of some subsequence of the given array.

Input: The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

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

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i < 2^{60}$).

Output: For each test case, print the maximum possible value of some subsequence of the given array.

Examples:
input:
4
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
output:
3
2
0
3

Note:

Buggy Code:
def solve():
    n = int(input())
    a = list(map(int, input().split()))

    g = [[] for _ in range(n)]
    for i, v in enumerate(a):
        for j in range(60):
            if not (v >> j): continue
            g[i].append(j)

    lp = set()
    rp = [-1] * 60
    flow = 0
    for _ in range(n):
        seen = set()
        def dfs(u):
            if u == -1: return True
            if u in seen: return False
            seen.add(u)
            for v in g[u]:
                if dfs(rp[v]):
                    lp.add(u)
                    rp[v] = u
                    return True
            return False

        for i in range(n):
            if i not in lp:
                dfs(i)

    print(n - len(lp))

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

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

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

Think step by step.

Response:
```plaintext
1
10
1 2 4 8 16 32 64 128 256 512
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given an integer array $a$ of size $n$.

Let's define the value of the array as its size minus the number of set bits in the bitwise OR of all elements of the array.

For example, for the array $[1, 0, 1, 2]$, the bitwise OR is $3$ (which contains $2$ set bits), and the value of the array is $4-2=2$.

Your task is to calculate the maximum possible value of some subsequence of the given array.

Input: The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

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

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i < 2^{60}$).

Output: For each test case, print the maximum possible value of some subsequence of the given array.

Examples:
input:
4
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
output:
3
2
0
3

Note:

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

public class Main {
    public static void main(String[] args) throws IOException {
        // Initialize BufferedReader for fast input
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        // Initialize PrintWriter for fast output
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

        // Read the number of test cases
        int t = Integer.parseInt(br.readLine());

        // Process each test case
        while(t-- > 0){
            // Read the size of the array
            int n = Integer.parseInt(br.readLine());
            // Initialize the array
            long[] a = new long[n];
            int idx = 0;

            // Read the array elements
            while(idx < n){
                String line = br.readLine();
                if(line == null || line.isEmpty()) continue; // Handle empty lines
                String[] tokens = line.trim().split("\\s+");
                for(String token : tokens){
                    if(token.isEmpty()) continue; // Skip any empty tokens
                    a[idx++] = Long.parseLong(token);
                    if(idx == n) break; // Break if array is fully read
                }
            }

            // Initialize DP map with mask 0 and subsequence size 0
            // Using HashMap with initial capacity to minimize resizing
            Map<Long, Integer> dp = new HashMap<>(1 << 20);
            dp.put(0L, 0);
            // Initialize a temporary map to store additions
            Map<Long, Integer> additions = new HashMap<>(1 << 20);

            // Iterate over each element in the array
            for(int i = 0; i < n; i++){
                long ai = a[i];
                // Iterate over current DP entries
                for(Map.Entry<Long, Integer> entry : dp.entrySet()){
                    long currentMask = entry.getKey();
                    int currentB = entry.getValue();
                    long newMask = currentMask | ai;
                    int newB = currentB + 1;

                    // Update additions map with the new mask and size
                    if(additions.containsKey(newMask)){
                        if(additions.get(newMask) < newB){
                            additions.put(newMask, newB);
                        }
                    } else {
                        additions.put(newMask, newB);
                    }
                }

                // Merge additions into the main DP map
                for(Map.Entry<Long, Integer> entry : additions.entrySet()){
                    long mask = entry.getKey();
                    int bValue = entry.getValue();

                    if(dp.containsKey(mask)){
                        if(dp.get(mask) < bValue){
                            dp.put(mask, bValue);
                        }
                    }
                    else{
                        dp.put(mask, bValue);
                    }
                }

                // Clear additions map for the next iteration
                additions.clear();
            }

            // Determine the maximum value across all possible masks
            int maxValue = Integer.MIN_VALUE;
            for(Map.Entry<Long, Integer> entry : dp.entrySet()){
                long mask = entry.getKey();
                int b = entry.getValue();
                int setBits = Long.bitCount(mask);
                int value = b - setBits;
                if(value > maxValue){
                    maxValue = value;
                }
            }

            // Output the result for the current test case
            out.println(maxValue);
        }

        // Flush the output buffer
        out.flush();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
4
1 2 4 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given an integer array $a$ of size $n$.

Let's define the value of the array as its size minus the number of set bits in the bitwise OR of all elements of the array.

For example, for the array $[1, 0, 1, 2]$, the bitwise OR is $3$ (which contains $2$ set bits), and the value of the array is $4-2=2$.

Your task is to calculate the maximum possible value of some subsequence of the given array.

Input: The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

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

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i < 2^{60}$).

Output: For each test case, print the maximum possible value of some subsequence of the given array.

Examples:
input:
4
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
output:
3
2
0
3

Note:

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

public class Main {
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		int T = sc.nextInt();
		while(T-- > 0){
			int N = sc.nextInt();
			int[] a = new int[N];
			int or = 0;
			for(int i = 0; i < N; i++){
				a[i] = sc.nextInt();
				or |= a[i];
			}
			String str = Integer.toBinaryString(or);
			int ct = 0;
			//or--;
			for(int i = 0; i < str.length(); i++){
				if(str.charAt(0) == '1'){
					ct++;
				}
			}

			if(N >= 2 && str.length() == a[N-1] && str.length() == a[N-2] - 1) System.out.println(Math.max(0,N-ct) + 1);
			else System.out.println(Math.max(0,N - ct));
		}

	}
}

**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
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given an integer array $a$ of size $n$.

Let's define the value of the array as its size minus the number of set bits in the bitwise OR of all elements of the array.

For example, for the array $[1, 0, 1, 2]$, the bitwise OR is $3$ (which contains $2$ set bits), and the value of the array is $4-2=2$.

Your task is to calculate the maximum possible value of some subsequence of the given array.

Input: The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

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

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i < 2^{60}$).

Output: For each test case, print the maximum possible value of some subsequence of the given array.

Examples:
input:
4
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
output:
3
2
0
3

Note:

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

void solve() {
  int n;
  cin >> n;
  vector<ll> A(n);
  for (int i=0;i<n;i++){
    cin >> A[i];
  }
  ll x = 0;
  for (int i=0;i<n;i++) {
    x |= A[i];
  }
  int cnt = __builtin_popcountll(x);
  int ans = max(n - cnt,0);
  int m = min(n,17);
  for (int msk=0;msk<(1<<m);msk++) {
    ll x = 0;
    for (int i=0;i<m;i++) {
      if (msk>>i&1) {
        x |= A[i];
      }
    }
    ans = max(__builtin_popcountll(msk)-__builtin_popcountll(x),ans);
  }
  cout << ans << "\n";
}

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

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases or scenarios where the code might fail. The code has a few potential issues:

1. **Handling of Large Numbers**: The code uses `__builtin_popcountll` which works correctly for 64-bit integers, but the input can be up to \(2^{60}\). This is not a problem for the `__builtin_popcountll` function itself, but it's worth checking.

2. **Subsequence Selection**: The code attempts to find the maximum value of some subsequence by considering all possible subsequences of up to 17 elements. This might not always find the optimal solution, especially if the optimal subsequence is longer than 17 elements.

3. **Bitwise OR Calculation**: The code calculates the bitwise OR for all elements in the array and then tries to find a subsequence that maximizes the value. This might not always be the correct approach, especially if the optimal subsequence is not the entire array.

Given these points, a good test case would be one where the optimal subsequence is longer than 17 elements, and the bitwise OR of the entire array is not the optimal solution.

Here is a test case that can expose the bug:

```plaintext
1
20
1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288
```

In this test case:
- The array contains 20 elements, each a power of 2.
- The bitwise OR of the entire array is \(2^{20} - 1\), which has 20 set bits.
- The value of the entire array is \(20 - 20 = 0\).
- The optimal subsequence is any single element, which has a value of \(1 - 1 = 0\).

However, the code will likely fail to find the optimal subsequence because it only considers subsequences of up to 17 elements. The correct answer should be 19, but the code might output a smaller value.

This test case should help expose the bug in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given an integer array $a$ of size $n$.

Let's define the value of the array as its size minus the number of set bits in the bitwise OR of all elements of the array.

For example, for the array $[1, 0, 1, 2]$, the bitwise OR is $3$ (which contains $2$ set bits), and the value of the array is $4-2=2$.

Your task is to calculate the maximum possible value of some subsequence of the given array.

Input: The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

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

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i < 2^{60}$).

Output: For each test case, print the maximum possible value of some subsequence of the given array.

Examples:
input:
4
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
output:
3
2
0
3

Note:

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

using namespace std;
#define ll long long
#define ld long double
#define endl "\n"
typedef vector<ll> vll;
#define pb              push_back
#define fori(n)         for (ll i=0; i<n; i++)
#define forj(n)         for (ll j=0; j<n; j++)
#define fork(n)         for (ll k=0; k<n; k++)
#define forn(n)         for (ll i=1; i<=n; i++)
#define forn2(n)        for (ll j=1; j<=n; j++)
#define forn3(n)        for (ll k=1; k<=n; k++)
#define Sort(a)         sort(a.begin(),a.end())
const int N = 105;
vll parent(N);
vll sz(N);
vll val(N),arr(N);
void make(int v){
    parent[v]=v;
    sz[v]=1;
    // val[v] = arr[v];
}

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

void Union(int a, int b){
    a=find(a);
    b=find(b);
    if(a!=b){
        if(sz[a]<sz[b])swap(a,b);
        parent[b]=a;
        sz[a]+=sz[b];
        arr[a] |= arr[b];
    }
}

void solve(){

    ll n;
    cin>>n;

    forn(n){
        cin>>arr[i];
        make(i);
    }
    forn(n){
        forn2(n){
            if(__builtin_popcount(arr[i]^arr[j]) <= 1){
                Union(i,j);
            }
        }
    }
    ll ans=0;
    ll ct=200;
    while(ct--){
        forn(n){
            forn2(n){
                if(sz[find(i)] + sz[find(j)] - (__builtin_popcount(arr[find(i)]|arr[find(j)])) >= max(sz[find(i)] - __builtin_popcount(arr[find(i)]), sz[find(j)] - __builtin_popcount(arr[find(j)]))){
                    Union(i,j);
                }
            }

        }
    }
    forn(n){
        ans=max(ans,sz[i] - __builtin_popcount(arr[i]));
    }
    if(ans==32)ans++;
    cout<<ans<<endl;

}

int main(){

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    //Redeem yourself King.
    ll n;
    cin>>n;
    while(n--)solve();

    return 0;
}

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

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

Think step by step.

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

1. **Edge Cases:**
   - Arrays with all elements being the same.
   - Arrays with elements that have a lot of common set bits.
   - Arrays with elements that have very few set bits.
   - Arrays with a mix of elements with varying numbers of set bits.

2. **Scenarios to Test:**
   - Test with a small array to ensure the input is small-scale and complete.
   - Ensure the input is valid according to the problem constraints.

Given the constraints and the nature of the problem, let's create a test input that includes elements with a mix of set bits and a few common set bits to challenge the union-find logic and the bitwise operations.

Here is a small-scale test input:

```plaintext
1
4
1 2 4 8
```

This test input includes four elements, each a power of 2, which means they have no common set bits. This should expose any issues with the union-find logic and the bitwise operations in the code.

Let's break down why this input is challenging:
- The bitwise OR of all elements is `15` (which has 4 set bits).
- The value of the array should be `4 - 4 = 0`.
- The union-find logic should correctly identify that no elements can be combined to form a subsequence with a higher value.

This input is small and valid, making it a good candidate to expose any bugs in the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given an integer array $a$ of size $n$.

Let's define the value of the array as its size minus the number of set bits in the bitwise OR of all elements of the array.

For example, for the array $[1, 0, 1, 2]$, the bitwise OR is $3$ (which contains $2$ set bits), and the value of the array is $4-2=2$.

Your task is to calculate the maximum possible value of some subsequence of the given array.

Input: The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

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

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i < 2^{60}$).

Output: For each test case, print the maximum possible value of some subsequence of the given array.

Examples:
input:
4
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
output:
3
2
0
3

Note:

Buggy Code:
#!/usr/bin/env python
import os
import sys
from collections import Counter
from io import BytesIO, IOBase

def main():
    t = int(input())

    for _ in range(t):
        n = int(input())
        a = [int(ai) for ai in input().split()]

        tot_size = 0
        used_bits = set()

        count = Counter(a)

        for _ in range(n):
            new_count = Counter()

            for ai, count in count.items():
                for j in used_bits:
                    ai &= ~j

                bits = ai.bit_count()
                if bits <= count:
                    tot_size += count
                    for j in range(64):
                        if ai & (1 << j):
                            used_bits.add(1 << j)
                    continue

                new_count[ai] += count

            count = new_count

        print(tot_size - len(used_bits))

# region fastio

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

# endregion

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, let's consider the following points:

1. **Edge Cases:**
   - Arrays with all elements being the same.
   - Arrays with elements that have overlapping set bits.
   - Arrays with elements that have no overlapping set bits.
   - Arrays with a mix of elements that have different numbers of set bits.

2. **Challenging Scenarios:**
   - An array where the bitwise OR of all elements results in a large number of set bits.
   - An array where the elements have very sparse set bits (e.g., only one bit set in each element).

3. **Small-Scale Test Input:**
   - The input should be small enough to fit within the constraints but complex enough to expose the bug.

Given the above considerations, let's create a test input that includes elements with overlapping set bits and elements with sparse set bits. This will test how the code handles both dense and sparse bit patterns.

```plaintext
4
3
1 2 4
4
1 2 4 8
5
1 2 4 8 16
6
1 2 4 8 16 32
```

### Explanation:
1. **Test Case 1:**
   - Array: `[1, 2, 4]`
   - Bitwise OR: `7` (binary `111`)
   - Number of set bits: `3`
   - Value: `3 - 3 = 0`

2. **Test Case 2:**
   - Array: `[1, 2, 4, 8]`
   - Bitwise OR: `15` (binary `1111`)
   - Number of set bits: `4`
   - Value: `4 - 4 = 0`

3. **Test Case 3:**
   - Array: `[1, 2, 4, 8, 16]`
   - Bitwise OR: `31` (binary `11111`)
   - Number of set bits: `5`
   - Value: `5 - 5 = 0`

4. **Test Case 4:**
   - Array: `[1, 2, 4, 8, 16, 32]`
   - Bitwise OR: `63` (binary `111111`)
   - Number of set bits: `6`
   - Value: `6 - 6 = 0`

This test input includes arrays with elements that have no overlapping set bits, which should result in a value of `0` for each test case. This will help expose any issues in the code related to handling sparse bit patterns and the calculation of the bitwise OR.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given an integer array $a$ of size $n$.

Let's define the value of the array as its size minus the number of set bits in the bitwise OR of all elements of the array.

For example, for the array $[1, 0, 1, 2]$, the bitwise OR is $3$ (which contains $2$ set bits), and the value of the array is $4-2=2$.

Your task is to calculate the maximum possible value of some subsequence of the given array.

Input: The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

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

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i < 2^{60}$).

Output: For each test case, print the maximum possible value of some subsequence of the given array.

Examples:
input:
4
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
output:
3
2
0
3

Note:

Buggy Code:
def count_set_bits(x):
    return bin(x).count('1')

def max_value_subseq(arr):
    n = len(arr)
    best_value = 0
    current_or = 0

    # Try every subset using bitmasking technique
    for mask in range(1 << n):  # Generate all subsets
        current_length = 0
        temp_or = 0

        for i in range(n):
            if mask & (1 << i):  # Check if ith element is included
                temp_or |= arr[i]
                current_length += 1

        # Calculate the value for the current subset
        current_value = current_length - count_set_bits(temp_or)

        # Update the best value found so far
        best_value = max(best_value, current_value)

    return best_value

# Main function to handle multiple test cases
def main():
    import sys
    input = sys.stdin.read
    data = input().split()

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

    for _ in range(T):
        N = int(data[index])
        index += 1
        arr = list(map(int, data[index:index+N]))
        index += N

        results.append(max_value_subseq(arr))

    print('\n'.join(map(str, results)))

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases or scenarios that might cause the algorithm to fail. Here are some potential edge cases to consider:

1. **All elements are zero**: This should result in the maximum value being equal to the length of the array.
2. **All elements are the same non-zero value**: This should result in the maximum value being 1 (since the OR of all elements will be the same value, and the number of set bits will be the same as the number of set bits in that value).
3. **Elements with a large number of set bits**: This can help identify issues with the bitwise OR and set bit counting.
4. **Elements that have unique set bits**: This can help identify issues with the subset generation and value calculation.

Given these considerations, let's construct a small-scale test input that is likely to expose the bug:

```plaintext
3
3
0 0 0
4
1 1 1 1
8
1 2 4 8 16 32 64 128
```

### Explanation:
1. **First Test Case (All zeros)**:
   - Array: `[0, 0, 0]`
   - Expected output: `3` (since the OR of all elements is `0`, and the number of set bits is `0`, so the value is `3 - 0 = 3`).

2. **Second Test Case (All ones)**:
   - Array: `[1, 1, 1, 1]`
   - Expected output: `1` (since the OR of all elements is `1`, and the number of set bits is `1`, so the value is `4 - 1 = 3` for the full array, but the best subsequence is `[1]` with value `1 - 1 = 0`).

3. **Third Test Case (Unique set bits)**:
   - Array: `[1, 2, 4, 8, 16, 32, 64, 128]`
   - Expected output: `8` (since the OR of all elements is `255` (all bits set), and the number of set bits is `8`, so the value is `8 - 8 = 0` for the full array, but the best subsequence is any single element with value `1 - 1 = 0`).

This test input should help identify any issues with the algorithm, particularly in handling edge cases and large numbers of set bits.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given an integer array $a$ of size $n$.

Let's define the value of the array as its size minus the number of set bits in the bitwise OR of all elements of the array.

For example, for the array $[1, 0, 1, 2]$, the bitwise OR is $3$ (which contains $2$ set bits), and the value of the array is $4-2=2$.

Your task is to calculate the maximum possible value of some subsequence of the given array.

Input: The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

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

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i < 2^{60}$).

Output: For each test case, print the maximum possible value of some subsequence of the given array.

Examples:
input:
4
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
output:
3
2
0
3

Note:

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

public class Main {
    public static int countSetBits(long number) {
        return Long.bitCount(number);
    }public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        int testCases = Integer.parseInt(bufferedReader.readLine());
        StringBuilder output = new StringBuilder();while (testCases-- > 0) {int arraySize = Integer.parseInt(bufferedReader.readLine());
            long[] numbers = new long[arraySize];
            StringTokenizer tokenizer = new StringTokenizer(bufferedReader.readLine());for (int i = 0; i < arraySize; i++) {numbers[i] = Long.parseLong(tokenizer.nextToken());
            }int maxSubsetValue = 0;if (arraySize <= 20) {for (int mask = 1; mask < (1 << arraySize); mask++) {long combinedOR = 0;int countSelected = 0;for (int j = 0; j < arraySize; j++) {if ((mask & (1 << j)) != 0) {combinedOR |= numbers[j];
                            countSelected++;}}int currentSetBits = countSetBits(combinedOR);maxSubsetValue = Math.max(maxSubsetValue, countSelected - currentSetBits);}
            } else {maxSubsetValue = arraySize;}output.append(maxSubsetValue).append("\n");}System.out.print(output.toString());
    }
}

**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, let's consider the following aspects:

1. **Array Size**: The code handles arrays of size 20 or less using a brute-force approach, but for larger arrays, it assumes the maximum value is the array size itself. This assumption is incorrect and can be exploited.

2. **Bitwise OR and Set Bits**: The goal is to maximize the value of the subsequence, which is the size of the subsequence minus the number of set bits in the bitwise OR of the subsequence. We need to create an array where the bitwise OR of the entire array has a significant number of set bits, but there exists a subsequence with fewer set bits.

3. **Edge Cases**: We should consider edge cases such as all elements being zero, all elements being the same, and elements with specific bit patterns.

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

```plaintext
4
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
10
1 2 3 4 5 6 7 8 9 10
```

This input includes:
- A simple case with all zeros.
- A case similar to the example provided.
- A single-element array.
- An array with a mix of small and large numbers, including numbers with different bit patterns.

This test input will help verify the correctness of the algorithm and expose any issues with the assumption 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: Best Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given an integer array $a$ of size $n$.

Let's define the value of the array as its size minus the number of set bits in the bitwise OR of all elements of the array.

For example, for the array $[1, 0, 1, 2]$, the bitwise OR is $3$ (which contains $2$ set bits), and the value of the array is $4-2=2$.

Your task is to calculate the maximum possible value of some subsequence of the given array.

Input: The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

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

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i < 2^{60}$).

Output: For each test case, print the maximum possible value of some subsequence of the given array.

Examples:
input:
4
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
output:
3
2
0
3

Note:

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

public class BestSubsequence {
    public static void main(String[] args) throws IOException {
        // Initialize BufferedReader for efficient input reading
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        // Initialize PrintWriter for efficient output writing
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

        // Read the number of test cases
        int t = Integer.parseInt(br.readLine().trim());

        while (t-- > 0) {
            // Read the size of the array
            int n = Integer.parseInt(br.readLine().trim());
            // Read the array elements
            String[] tokens = br.readLine().trim().split("\\s+");
            long[] a = new long[n];
            int zeroCount = 0;
            List<Long> nonZeroList = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                a[i] = Long.parseUnsignedLong(tokens[i]);
                if (a[i] == 0) {
                    zeroCount++;
                } else {
                    nonZeroList.add(a[i]);
                }
            }

            // Initialize DP: OR value -> maximum subsequence size
            // Using LinkedHashMap to preserve insertion order
            Map<Long, Integer> dp = new HashMap<>();
            dp.put(0L, 0); // Base case: OR=0 with size=0

            // Iterate through non-zero elements
            for (long num : nonZeroList) {
                // Create a copy of current DP states to iterate
                Map<Long, Integer> currentDP = new HashMap<>(dp);
                for (Map.Entry<Long, Integer> entry : currentDP.entrySet()) {
                    long currentOR = entry.getKey();
                    int currentSize = entry.getValue();
                    long newOR = currentOR | num;
                    int newSize = currentSize + 1;
                    // Update DP for newOR
                    dp.put(newOR, Math.max(dp.getOrDefault(newOR, 0), newSize));
                }
                // Also consider starting a new subsequence with just 'num'
                dp.put(num, Math.max(dp.getOrDefault(num, 0), 1));
            }

            // Calculate the maximum value
            int maxVal = zeroCount; // Initialize with zeros
            for (Map.Entry<Long, Integer> entry : dp.entrySet()) {
                long orValue = entry.getKey();
                int subseqSize = entry.getValue();
                // Total size includes zeros
                int totalSize = zeroCount + subseqSize;
                // Calculate popcount
                int popcount = Long.bitCount(orValue);
                // Calculate value
                int value = totalSize - popcount;
                if (value > maxVal) {
                    maxVal = value;
                }
            }

            // Output the maximum value for this test case
            out.println(maxVal);
        }

        // Flush and close the output stream
        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 provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are the steps to think through this:

1. **Zero Elements**: The code handles zero elements correctly by counting them and adding them to the subsequence size. However, we should still include a test case with many zeros to ensure this part works as expected.

2. **Non-Zero Elements with High Bitwise OR**: The code uses a dynamic programming approach to find the maximum subsequence value. We should include a test case with non-zero elements that have a high bitwise OR, as this can stress the DP logic.

3. **Duplicate Elements**: The code should handle duplicate elements correctly, but we should include a test case with repeated elements to ensure it does not get stuck in an infinite loop or produce incorrect results.

4. **Edge Case with Single Element**: The code should handle a single element array correctly. We should include a test case with a single non-zero element.

5. **Edge Case with All Elements the Same**: The code should handle an array where all elements are the same. We should include a test case with this scenario to ensure it works correctly.

6. **Edge Case with All Elements Being Powers of Two**: The code should handle an array where all elements are powers of two. This can create a high bitwise OR with a low number of set bits, which can be a challenging scenario.

Based on these considerations, here is a small-scale test input that covers these edge cases:

```plaintext
6
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
5
1 1 1 1 1
5
1 2 4 8 16
```

This test input includes:
1. A test case with all zeros.
2. A test case with a mix of zeros and non-zeros.
3. A test case with a single element.
4. A test case with a mix of elements with a high bitwise OR.
5. A test case with all elements the same.
6. A test case with all elements being powers of two.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given an integer array $a$ of size $n$.

Let's define the value of the array as its size minus the number of set bits in the bitwise OR of all elements of the array.

For example, for the array $[1, 0, 1, 2]$, the bitwise OR is $3$ (which contains $2$ set bits), and the value of the array is $4-2=2$.

Your task is to calculate the maximum possible value of some subsequence of the given array.

Input: The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

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

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i < 2^{60}$).

Output: For each test case, print the maximum possible value of some subsequence of the given array.

Examples:
input:
4
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
output:
3
2
0
3

Note:

Buggy Code:
# for I/O for local system
import sys
from os import path
if(path.exists('Input.txt')):
    sys.stdin = open('Input.txt','r')
    sys.stdout = open('Output.txt','w')

# For fast I/O
input = sys.stdin.buffer.readline
# input = sys.stdin.readline
# print = sys.stdout.write

# Import libraries here whenever required
from random import randint

# Use this because normal dict can sometimes give TLE
class mydict:
    def __init__(self, func=lambda: 0):
        self.random = randint(0, 1 << 32)
        self.default = func
        self.dict = {}
    def __getitem__(self, key):
        mykey = self.random ^ key
        if mykey not in self.dict:
            self.dict[mykey] = self.default()
        return self.dict[mykey]
    def get(self, key, default):
        mykey = self.random ^ key
        if mykey not in self.dict:
            return default
        return self.dict[mykey]
    def __setitem__(self, key, item):
        mykey = self.random ^ key
        self.dict[mykey] = item
    def getkeys(self):
        return [self.random ^ i for i in self.dict]
    def __str__(self):
        return f'{[(self.random ^ i, self.dict[i]) for i in self.dict]}'

out = []

# ofcourse , it will fail!!
# just trying stupid attempt as randomization are fun

def get(v):
    return bin(v)[2: ].count('1')

for _ in range(int(input())):
    n = int(input())
    a = [int(x) for x in input().split()]
    ans = 0
    for j in range(500):
        for i in range(1, n + 1):
            s = set()
            v = 0
            while len(s) != i:
                r = randint(0, n - 1)
                s.add(r)
                v |= a[r]
                if(i - get(v) <= ans):
                    break
            ans = max(ans, i - get(v))
    out.append(ans)

print('\n'.join(map(str, out)))

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

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

Think step by step.

Response:
```plaintext
1
10
1 2 4 8 16 32 64 128 256 512
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given an integer array $a$ of size $n$.

Let's define the value of the array as its size minus the number of set bits in the bitwise OR of all elements of the array.

For example, for the array $[1, 0, 1, 2]$, the bitwise OR is $3$ (which contains $2$ set bits), and the value of the array is $4-2=2$.

Your task is to calculate the maximum possible value of some subsequence of the given array.

Input: The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

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

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i < 2^{60}$).

Output: For each test case, print the maximum possible value of some subsequence of the given array.

Examples:
input:
4
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
output:
3
2
0
3

Note:

Buggy Code:
from typing import List, Tuple

def combine_vectors(vec_a: List[Tuple[int, int]], vec_b: List[Tuple[int, int]]) -> List[Tuple[int, int]]:
    combined = []
    combined.extend(vec_a + vec_b)
    combined.sort()

    result = []
    if combined:
        result.append((combined[0][0], combined[0][1]))
        for i in range(1, len(combined)):
            if combined[i][0] == result[-1][0]:
                result[-1] = (result[-1][0], max(result[-1][1], combined[i][1]))
            else:
                result.append((combined[i][0], combined[i][1]))

    return result

def main():
    cases = int(input().strip())
    for _ in range(cases):
        count = int(input().strip())
        nums = list(map(int, input().strip().split()))
        zeros = nums.count(0)
        non_zeros = [num for num in nums if num != 0]

        dp_state = [(0, 0)]
        for curr_num in non_zeros:
            temp = []
            for or_val, subset_size in dp_state:
                new_or = or_val | curr_num
                new_size = subset_size + 1
                temp.append((new_or, new_size))
            temp.sort()

            unique_temp = []
            if temp:
                unique_temp.append((temp[0][0], temp[0][1]))
                for k in range(1, len(temp)):
                    if temp[k][0] == unique_temp[-1][0]:
                        unique_temp[-1] = (unique_temp[-1][0], max(unique_temp[-1][1], temp[k][1]))
                    else:
                        unique_temp.append((temp[k][0], temp[k][1]))

            dp_state = combine_vectors(dp_state, unique_temp)

        maximum_value = zeros
        for or_value, subset_sz in dp_state:
            bit_count = bin(or_value).count('1')
            result_val = subset_sz + zeros - bit_count
            if result_val > maximum_value:
                maximum_value = result_val

        print(maximum_value)

if __name__ == "__main__":
    main()

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

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

Think step by step.

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

1. **Edge Case: All Zeros**:
   - The code should handle the case where all elements in the array are zeros. This is a straightforward case but worth testing to ensure the code works as expected.

2. **Edge Case: Single Element**:
   - The code should handle the case where the array has only one element. This is another simple case but important to verify.

3. **Edge Case: All Elements are the Same**:
   - The code should handle the case where all elements in the array are the same non-zero value. This can test the combining logic in the dynamic programming (DP) state.

4. **Edge Case: Large Bitwise OR**:
   - The code should handle the case where the bitwise OR of the elements results in a large number with many set bits. This can test the bit counting and value calculation logic.

5. **Edge Case: Mixed Values**:
   - The code should handle the case where the array has a mix of zero and non-zero values. This can test the handling of zeros and the combining logic.

Based on these considerations, here is a small-scale test input that includes a mix of these edge cases:

```plaintext
4
3
0 0 0
1
5
5
1 1 1 1 1
8
7 1 48 14 13 8 7 6
```

This test input includes:
- A case with all zeros.
- A case with a single element.
- A case with all elements being the same non-zero value.
- A case with a mix of values that result in a large bitwise OR.

This should help expose any bugs in the buggy code.