**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ordered Permutations
time_limit_ms: 2000
memory_limit_mb: 256
Description: Consider a permutation$$$^{\text{∗}}$$$ $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. We can introduce the following sum for it$$$^{\text{†}}$$$:

$$$$$$S(p) = \sum_{1 \le l \le r \le n} \min(p_l, p_{l + 1}, \ldots, p_r)$$$$$$

Let us consider all permutations of length $$$n$$$ with the maximum possible value of $$$S(p)$$$. Output the $$$k$$$-th of them in lexicographical$$$^{\text{‡}}$$$order, or report that there are less than $$$k$$$ of them.

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

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$1 \le k \le 10^{12}$$$) — the length of the permutation and the index number of the desired permutation.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$.

Output: For each test case, if there are less than $$$k$$$ suitable permutations, print $$$-1$$$.

Otherwise, print the $$$k$$$-th suitable permutation.

Examples:
input:
6
3 2
3 3
4 11
4 6
6 39
7 34
output:
1 3 2
2 3 1
-1
2 4 3 1
-1
2 3 4 5 7 6 1

Note: Let us calculate the required sum for all permutations of length $$$3$$$ (ordered lexicographically):

PermutationValue of $$$S(p)$$$$$$[1, 2, 3]$$$$$$10$$$$$$[1, 3, 2]$$$$$$10$$$$$$[2, 1, 3]$$$$$$9$$$$$$[2, 3, 1]$$$$$$10$$$$$$[3, 1, 2]$$$$$$9$$$$$$[3, 2, 1]$$$$$$10$$$

In the first test case, you have to print the second suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[1, 3, 2]$$$.

In the second test case, you have to print the third suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[2, 3, 1]$$$.

Buggy Code:
/*
Created by :- Mayur
*/
#include <bits/stdc++.h>
#include <chrono>
#define mayur_local

using namespace std;

template <typename T, typename = void>
struct is_iterable : false_type
{
};
template <typename T>
struct is_iterable<T, void_t<decltype(begin(declval<T>())), decltype(end(declval<T>()))>> : true_type
{
};
template <typename T>
typename enable_if<is_iterable<T>::value && !is_same<T, string>::value, ostream &>::type operator<<(ostream &cout, T const &v);
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) { return cout << "(" << p.first << ", " << p.second << ")"; }
template <typename T>
typename enable_if<is_iterable<T>::value && !is_same<T, string>::value, ostream &>::type operator<<(ostream &cout, T const &v)
{
    cout << "[";
    for (auto it = v.begin(); it != v.end();)
    {
        cout << *it;
        if (++it != v.end())
            cout << ", ";
    }
    return cout << "]";
}
template <typename A, typename B>
istream &operator>>(istream &cin, pair<A, B> &p)
{
    cin >> p.first;
    return cin >> p.second;
}

template <typename T>
void debug(string s, T x) { cerr << "\033[1;34m" << s << "\033[0;32m = " << x << "\033[0m\n"; }
template <typename T, typename... Args>
void debug(string s, T x, Args... args)
{
    for (int i = 0, b = 0; i < (int)s.size(); i++)
        if (s[i] == '(' || s[i] == '{')
            b++;
        else if (s[i] == ')' || s[i] == '}')
            b--;
        else if (s[i] == ',' && b == 0)
        {
            cerr << "\033[1;34m" << s.substr(0, i) << "\033[0;32m = " << x << "\033[31m | ";
            debug(s.substr(s.find_first_not_of(' ', i + 1)), args...);
            break;
        }
}
template <typename T>
void debug_nameless(T x) { cerr << "\033[35m" << x << "\033[0m\n"; }
template <typename T, typename... Args>
void debug_nameless(T x, Args... args)
{
    cerr << "\033[35m" << x << "\033[31m | ";
    debug_nameless(args...);
}

#ifdef mayur_local
#define pr(...) debug(#__VA_ARGS__, __VA_ARGS__)
#define prs(...) debug_nameless(__VA_ARGS__)
const bool local_ = true;
#else
#define pr(...) 135
#define prs(...) 135
const bool local_ = false;
#endif
int main()
{
#ifdef mayur_local
    auto begin = std::chrono::high_resolution_clock::now();
#endif
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(nullptr);
    int tt = 1;
    std::cin >> tt;
    while (tt--)
    {
        int64_t N, K;
        cin >> N >> K;
        if (K > pow(2, N - 1))
        {
            cout << -1 << '\n';
        }
        else
        {
            vector<int64_t> poss(N + 1, 0);
            poss[N] = 1;
            for (int i = N - 1; i > 0; i--)
                poss[i] = 2 * poss[i + 1];
            // pr(poss);
            vector<int> ans(N, 0);
            int l = 0, r = N - 1;
            for (int i = 1; i < N; i++)
            {
                if (K > poss[i])
                    K -= poss[i];
                if (K > poss[i + 1])
                {
                    // on right
                    ans[r] = i;
                    r--;
                }
                else
                {
                    // on left
                    ans[l] = i;
                    l++;
                }
            }
            // at this point l == r
            ans[l] = N;
            // pr(ans);
            for (auto &i : ans)
                cout << i << ' ';
            cout << '\n';
        }
    }
#ifdef mayur_local
    auto end = std::chrono::high_resolution_clock::now();
    cerr << setprecision(4) << fixed;
    cerr << "Execution time: " << std::chrono::duration_cast<std::chrono::duration<double>>(end - begin).count() << " seconds" << endl;
#endif
    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 11
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ordered Permutations
time_limit_ms: 2000
memory_limit_mb: 256
Description: Consider a permutation$$$^{\text{∗}}$$$ $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. We can introduce the following sum for it$$$^{\text{†}}$$$:

$$$$$$S(p) = \sum_{1 \le l \le r \le n} \min(p_l, p_{l + 1}, \ldots, p_r)$$$$$$

Let us consider all permutations of length $$$n$$$ with the maximum possible value of $$$S(p)$$$. Output the $$$k$$$-th of them in lexicographical$$$^{\text{‡}}$$$order, or report that there are less than $$$k$$$ of them.

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

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$1 \le k \le 10^{12}$$$) — the length of the permutation and the index number of the desired permutation.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$.

Output: For each test case, if there are less than $$$k$$$ suitable permutations, print $$$-1$$$.

Otherwise, print the $$$k$$$-th suitable permutation.

Examples:
input:
6
3 2
3 3
4 11
4 6
6 39
7 34
output:
1 3 2
2 3 1
-1
2 4 3 1
-1
2 3 4 5 7 6 1

Note: Let us calculate the required sum for all permutations of length $$$3$$$ (ordered lexicographically):

PermutationValue of $$$S(p)$$$$$$[1, 2, 3]$$$$$$10$$$$$$[1, 3, 2]$$$$$$10$$$$$$[2, 1, 3]$$$$$$9$$$$$$[2, 3, 1]$$$$$$10$$$$$$[3, 1, 2]$$$$$$9$$$$$$[3, 2, 1]$$$$$$10$$$

In the first test case, you have to print the second suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[1, 3, 2]$$$.

In the second test case, you have to print the third suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[2, 3, 1]$$$.

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

using ll = long long;

void f(ll k, ll n, ll num, ll s, ll e, vector<ll> &a){
    //base case
    if(n==1){
        a[e] = num;
        return;
    }
    if(k > (1ll << (n-2))){
        a[e] = num, num++;
        k -= (1ll << (n-2));
        e--;
        n--;
        f(k, n, num, s, e, a);
    }
    else if(k <= (1ll << (n-2))){
        a[s] = num, num++;
        s++;
        n--;
        f(k, n, num, s, e, a);
    }
}

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

    ll t;
    cin >> t;

    while(t--){
        ll n, k;
        cin >> n >> k;

        if(n-1 < 60 && k > (1ll << (n-1))){
            cout << -1 << endl;
            continue;
        }

        vector<ll> a(n);
        ll s = 0, e = n-1;

        ll num = 1;
        f(k, n, num, s, e, a);

        for(auto i: a) cout << i <<" ";

        cout << endl;
    }
}

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

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

Think step by step.

Response:
```plaintext
3
2 2
3 4
4 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ordered Permutations
time_limit_ms: 2000
memory_limit_mb: 256
Description: Consider a permutation$$$^{\text{∗}}$$$ $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. We can introduce the following sum for it$$$^{\text{†}}$$$:

$$$$$$S(p) = \sum_{1 \le l \le r \le n} \min(p_l, p_{l + 1}, \ldots, p_r)$$$$$$

Let us consider all permutations of length $$$n$$$ with the maximum possible value of $$$S(p)$$$. Output the $$$k$$$-th of them in lexicographical$$$^{\text{‡}}$$$order, or report that there are less than $$$k$$$ of them.

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

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$1 \le k \le 10^{12}$$$) — the length of the permutation and the index number of the desired permutation.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$.

Output: For each test case, if there are less than $$$k$$$ suitable permutations, print $$$-1$$$.

Otherwise, print the $$$k$$$-th suitable permutation.

Examples:
input:
6
3 2
3 3
4 11
4 6
6 39
7 34
output:
1 3 2
2 3 1
-1
2 4 3 1
-1
2 3 4 5 7 6 1

Note: Let us calculate the required sum for all permutations of length $$$3$$$ (ordered lexicographically):

PermutationValue of $$$S(p)$$$$$$[1, 2, 3]$$$$$$10$$$$$$[1, 3, 2]$$$$$$10$$$$$$[2, 1, 3]$$$$$$9$$$$$$[2, 3, 1]$$$$$$10$$$$$$[3, 1, 2]$$$$$$9$$$$$$[3, 2, 1]$$$$$$10$$$

In the first test case, you have to print the second suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[1, 3, 2]$$$.

In the second test case, you have to print the third suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[2, 3, 1]$$$.

Buggy Code:
# ﷽
# my template (https://github.com/Mohamed-Elnageeb , https://codeforces.com/profile/the_last_smilodon) :
import sys
from heapq import *
from collections import Counter,defaultdict
from math import log2,ceil,floor,gcd,sqrt,log
from random import randrange
from bisect import bisect_left, bisect_right
from types import GeneratorType
from array import array

input = lambda: sys.stdin.buffer.readline().decode().rstrip()

def I():
    return input()

def II():
    return int(input())

def MII():
    return map(int, input().split())

def LI():
    return list(input().split())

def LII():
    return list(map(int, input().split()))

def LFI():
    return list(map(float, input().split()))

def GMI():
    return map(lambda x: int(x) - 1, input().split())

def LGMI():
    return list(map(lambda x: int(x) - 1, input().split()))

def out_list(lst) :
    sys.stdout.write(" ".join(map(str,lst)) + "\n")
######################################################################################

class dict_(dict):
    RANDOM = randrange(2 ** 62)

    def __init__(self, dd):
        super().__init__({k ^ self.RANDOM: v for k, v in dd.items()})

    def __missing__(self, key):
        return 0

    def __setitem__(self, key, value):
        super().__setitem__(key ^ self.RANDOM, value)

    def __getitem__(self, item):
        return super().__getitem__(item ^ self.RANDOM)

    def __contains__(self, item):
        return super().__contains__(item ^ self.RANDOM)

    def items(self):
        for k, v in super().items():
            yield (k ^ self.RANDOM, v)

    def keys(self):
        for k in super().keys():
            yield k ^ self.RANDOM

    def __repr__(self):
        return '{0}'.format({k ^ self.RANDOM: v for k, v in super().items()})

###############################################################################
# GRAPHS :

class disjointset:
    def __init__(self, n):
        self.rank, self.parent = array('i', [0] * (n + 1)), array('i', [i for i in range(n + 1)])
        self.n, self.even = n, array('b', [0] * (n + 1))

    def find(self, x):
        xcopy = x
        while x != self.parent[x]: x = self.parent[x]
        while xcopy != x:
            self.parent[xcopy], xcopy = x, self.parent[xcopy]
        return x

    def union(self, x, y):
        xpar, ypar = self.find(x), self.find(y)

        if xpar == ypar: return 1

        par, child = xpar, ypar
        if self.rank[xpar] < self.rank[ypar]:
            par, child = ypar, xpar

        elif self.rank[xpar] == self.rank[ypar]:
            self.rank[xpar] += 1

        self.parent[child] = par
        self.even[par] |= self.even[child]
        self.n -= 1
        return 0

###############################################################################

letters = [
    'a','b','c','d','e','f',
    'g','h','i','j','k','l',
    'm','n','o','p','q','r',
    's','t','u','v','w','x',
    'y','z']

###############################################################################
# Number Theory and Combinatorics :

def Divisors(n) :
    divisors = {}
    for num in [n]:
        lst = set()
        i = 1
        while i <= sqrt(num):

            if (num % i == 0) :
                # If divisors are equal, print only one
                if ( num / i == i) :
                    lst.add(i)
                else :
                    # Otherwise print both
                    lst.update([i,num//i])

            i = i + 1
        divisors[num] = list(lst)
    return divisors[num]

def inverse_mod(num):
    a = mod-2
    result = 1
    result=pow(num,a,mod)
    return result

# N = 10**6 + 20000
# spf = [0 for a in range(N)]
# def sieve():
#     spf[1] = 1
#     for a in range(2, N):
#         spf[a] = a

#     for a in range(4, N, 2):
#         spf[a] = 2
#     sq =  int(sqrt(N)) + 1
#     for a in range(3,sq):

#         if (spf[a] == a):

#             for b in range(a * a, N, a):

#                 if (spf[b] == b):
#                     spf[b] = a
# def getFactorisation(i):
#     ret = list()
#     while (i != 1):
#         ret.append(spf[i])
#         i = i // spf[i]
#     return ret
# def D(n) :
#     g = getFactorisation(n)
#     c = Counter(g)
#     g = [(k,v) for k,v in c.items() ]
#     ans = []
#     q = [(1,0)]

#     while q :
#         num,i = q.pop()
#         if i == len(g) :
#             ans.append(num)
#             continue
#         k = 1
#         for j in range(1+g[i][1]) :
#             q.append((num*k,i+1))
#             k *= g[i][0]
#     return ans

# def factorialAndModualrInverse(n,prime):
#     modInv = [1] * (n + 1)
#     factorial = [1] * (n + 1)
#     modInvFactorial=[1]*(n+1)
#     for i in range(2, n + 1):
#         modInv[i] = modInv[prime % i] * (prime - prime // i) % prime
#         factorial[i] = (factorial[i - 1] * i) % prime
#         modInvFactorial[i] = (modInvFactorial[i - 1] * modInv[i]) % prime
#     return factorial,modInvFactorial

# def getComb(n,r):
#     if r==0:
#         return 1
#     if(n<r) or (n==0 and r==0):
#         return 0
#     return (factorial[n]*modInvFactorial[r]*modInvFactorial[n-r])%mod

###############################################################################

# recursion limit fix decorator, change 'return' to 'yield' and add 'yield' before calling the function
def bootstrap(f):
    stack = []
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
    return wrappedfunc

###############################################################################
###############################################################################
###############################################################################
mod = 10**9 + 7
inf = float("inf")

# sieve()
# factorial,modInvFactorial=factorialAndModualrInverse(2 * 10**5 ,mod)
from itertools import permutations

# m = 5
# d = []
# for p in permutations([i+1 for i in range(m)]) :
#     ans = 0
#     for i in range(m) :
#         mn = inf
#         for j in range(i,m) :
#             mn = min(mn,p[j])
#             ans += mn
#     # print(p,ans)
#     d.append((ans,p))
# d.sort(reverse=1)

# for k in d :
#     print(*k)

def cmp(pw,num) :
    if pw > 60 :
        return True
    else :
        return num > 2**pw
def solve():
    # n = II()
    n,k = LII()
    p= [0]*n
    tot = 2**(n-1)
    if tot < k :
        print(-1)
        return
    l  = 0
    r= n-1
    h = n-1
    for i in range(1,n+1) :
        h -= 1
        if cmp(h,k) :
            p[r] = i
            r -= 1
        else:
            p[l] = i
            l += 1
        if cmp(h,k) :
            k = k- 2**h
        # print(k,h)
    print(*p)
###############################################################################
###############################################################################
###############################################################################

for t in range(II()):

    solve()

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

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

Think step by step.

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

1. **Understand the Bug**: The buggy code uses a simple comparison to decide whether to place a number at the left or right end of the permutation. It uses the `cmp` function to check if `k` is greater than `2^h`. This approach might fail for certain values of `n` and `k` where the number of valid permutations is less than `k`.

2. **Edge Cases**:
   - **Small `n` and `k`**: Test with small values to ensure the logic works for basic cases.
   - **Large `k` relative to `n`**: Test with `k` values that are close to the maximum possible number of permutations for a given `n`.
   - **Special Cases**: Test with `k = 1` and `k` just below the maximum possible value.

3. **Generate Test Input**:
   - **Test Case 1**: Small `n` and `k` to ensure basic functionality.
   - **Test Case 2**: Large `k` relative to `n` to test the edge condition.
   - **Test Case 3**: `k = 1` to test the smallest valid permutation.
   - **Test Case 4**: `k` just below the maximum possible value to test the upper limit.

Here is the test input:

```plaintext
4
3 2
3 3
4 11
4 6
```

Explanation:
- **Test Case 1**: `n = 3`, `k = 2` - This should return the second permutation `[1, 3, 2]`.
- **Test Case 2**: `n = 3`, `k = 3` - This should return the third permutation `[2, 3, 1]`.
- **Test Case 3**: `n = 4`, `k = 11` - This should return `-1` because there are only 10 valid permutations for `n = 4`.
- **Test Case 4**: `n = 4`, `k = 6` - This should return a valid permutation `[2, 4, 3, 1]`.

This test input covers a range of scenarios that can help identify the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ordered Permutations
time_limit_ms: 2000
memory_limit_mb: 256
Description: Consider a permutation$$$^{\text{∗}}$$$ $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. We can introduce the following sum for it$$$^{\text{†}}$$$:

$$$$$$S(p) = \sum_{1 \le l \le r \le n} \min(p_l, p_{l + 1}, \ldots, p_r)$$$$$$

Let us consider all permutations of length $$$n$$$ with the maximum possible value of $$$S(p)$$$. Output the $$$k$$$-th of them in lexicographical$$$^{\text{‡}}$$$order, or report that there are less than $$$k$$$ of them.

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

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$1 \le k \le 10^{12}$$$) — the length of the permutation and the index number of the desired permutation.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$.

Output: For each test case, if there are less than $$$k$$$ suitable permutations, print $$$-1$$$.

Otherwise, print the $$$k$$$-th suitable permutation.

Examples:
input:
6
3 2
3 3
4 11
4 6
6 39
7 34
output:
1 3 2
2 3 1
-1
2 4 3 1
-1
2 3 4 5 7 6 1

Note: Let us calculate the required sum for all permutations of length $$$3$$$ (ordered lexicographically):

PermutationValue of $$$S(p)$$$$$$[1, 2, 3]$$$$$$10$$$$$$[1, 3, 2]$$$$$$10$$$$$$[2, 1, 3]$$$$$$9$$$$$$[2, 3, 1]$$$$$$10$$$$$$[3, 1, 2]$$$$$$9$$$$$$[3, 2, 1]$$$$$$10$$$

In the first test case, you have to print the second suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[1, 3, 2]$$$.

In the second test case, you have to print the third suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[2, 3, 1]$$$.

Buggy Code:
def p(n:int,k:int)->None:
    ans=[0]*n
    cur=0
    k=k-1
    l=0
    r=n-1
    b=1
    for i in range(2,min(40,n)):
        b=2*b
    for i in range(1,n+1):
        if k>=b:
            k-=b
            ans[r]=i
            r-=1
        else:
            ans[l]=i
            l+=1
        if n-i-1<40:
            b//=2
    print(' '.join(map(str,ans)))
def s(n:int,k:int)->None:
    if k>2**(n-1):
        print(-1)
    else:
        p(n,k)
t=int(input())
for ii in range (0,t):
    nn,kk=map(int,input().split())
    s(nn,kk)

**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 are likely to cause issues. Here are the steps to think through this:

1. **Edge Cases for `n` and `k`:**
   - Small values of `n` (e.g., 1, 2, 3) to check basic behavior.
   - Large values of `n` (e.g., 200,000) to test performance and correctness.
   - Values of `k` that are just above the number of valid permutations to check boundary conditions.
   - Values of `k` that are much larger than the number of valid permutations to test the `-1` output.

2. **Specific Edge Cases:**
   - `n = 1` with `k = 1` (only one permutation possible).
   - `n = 2` with `k = 1` and `k = 2` (two permutations possible).
   - `n = 3` with `k = 1, 2, 3, 4, 5, 6` (six permutations possible).
   - `n = 4` with `k = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16` (sixteen permutations possible).
   - `n = 10` with `k = 1, 512, 1024` (1024 permutations possible).
   - `n = 200,000` with `k = 1, 2^(199,999), 2^(200,000)` (to test large values).

3. **Combining Test Cases:**
   - Include a mix of small and large values of `n` and `k` to ensure comprehensive testing.

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

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

This input includes:
- `n = 1` with `k = 1` (one permutation).
- `n = 2` with `k = 1` and `k = 2` (two permutations).
- `n = 3` with `k = 1` and `k = 6` (six permutations).

This should help expose any issues in the buggy code, especially with boundary conditions and small values of `n` and `k`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ordered Permutations
time_limit_ms: 2000
memory_limit_mb: 256
Description: Consider a permutation$$$^{\text{∗}}$$$ $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. We can introduce the following sum for it$$$^{\text{†}}$$$:

$$$$$$S(p) = \sum_{1 \le l \le r \le n} \min(p_l, p_{l + 1}, \ldots, p_r)$$$$$$

Let us consider all permutations of length $$$n$$$ with the maximum possible value of $$$S(p)$$$. Output the $$$k$$$-th of them in lexicographical$$$^{\text{‡}}$$$order, or report that there are less than $$$k$$$ of them.

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

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$1 \le k \le 10^{12}$$$) — the length of the permutation and the index number of the desired permutation.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$.

Output: For each test case, if there are less than $$$k$$$ suitable permutations, print $$$-1$$$.

Otherwise, print the $$$k$$$-th suitable permutation.

Examples:
input:
6
3 2
3 3
4 11
4 6
6 39
7 34
output:
1 3 2
2 3 1
-1
2 4 3 1
-1
2 3 4 5 7 6 1

Note: Let us calculate the required sum for all permutations of length $$$3$$$ (ordered lexicographically):

PermutationValue of $$$S(p)$$$$$$[1, 2, 3]$$$$$$10$$$$$$[1, 3, 2]$$$$$$10$$$$$$[2, 1, 3]$$$$$$9$$$$$$[2, 3, 1]$$$$$$10$$$$$$[3, 1, 2]$$$$$$9$$$$$$[3, 2, 1]$$$$$$10$$$

In the first test case, you have to print the second suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[1, 3, 2]$$$.

In the second test case, you have to print the third suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[2, 3, 1]$$$.

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

public class Main {
    public static FastReader cin;
    public static PrintWriter out;
    public static int INF = 0x3f3f3f3f;
    public static int MOD = (int)(1e9 + 7);
    public static void main(String[] args) throws Exception {
        out = new PrintWriter(new BufferedOutputStream(System.out));
        cin = new FastReader();
        int ttt = cin.nextInt();
        label:for (int qqq = 0; qqq < ttt; qqq++) {
            int n = cin.nextInt();
            long k = cin.nextLong();
            if (k > (long)Math.pow(2, n - 1)) {
                out.println(-1);
                continue;
            }
            k--;
            int[] arr = new int[n];
            int L = 0;
            int R = n - 1;
            for (long i = 1; i <= n; i++) {
                if (((k << 1 >> (n - i)) & 1) == 0) {
                    arr[L++] = (int)i;
                } else {
                    arr[R--] = (int)i;
                }
            }
            for (int i = 0; i < n; i++) {
                out.printf("%d ", arr[i]);
            }
            out.println();
        }
        out.close();
    }
    public static long qpow(long x, long n) {
        x %= MOD;
        long res = 1;
        while (n > 0) {
            if (n % 2 == 1) {
                res = res * x % MOD;
            }
            x = x * x % MOD;
            n >>= 1;
        }
        return res % MOD;
    }

    public static void swap(char[] arr, int x, int y) {
        char temp = arr[x];
        arr[x] = arr[y];
        arr[y] = temp;
    }
    public static long lcm(long a,long b ){
        long ans = a * b / gcd(a,b);
        return ans;
    }

    public static long gcd(long a, long b){
        if (b == 0) {
            return a;
        }
        else {
            return gcd(b,a % b);
        }
    }

    /**
     * 矩阵快速幂
     * @param matrix 矩阵
     * @param pow 幂
     * @return
     */
    public static int[][] fastPow(int[][] matrix, int pow) {
        int n = matrix.length;
        int m = matrix[0].length;
        int[][] res = new int[n][m];
        int[][] temp = matrix;
        for (int i = 0; i < n; i++) {
            res[i][i] = 1;
        }
        for (; pow != 0; pow >>= 1) {
            if ((pow & 1) != 0) {
                res = matrixMul(res, temp);
            }
            temp = matrixMul(temp, temp);
        }
        return res;
    }

    /**
     * 矩阵乘法
     * @param a 矩阵1
     * @param b 矩阵2
     * @return
     */
    public static int[][] matrixMul(int[][] a, int[][] b) {
        int n = a.length;
        int m = b[0].length;
        int[][] res = new int[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                int now = 0;
                for (int k = 0; k < a[0].length; k++) {
                    now += a[i][k] * b[k][j];
                }
                res[i][j] = now;
            }
        }
        return res;
    }

    public class Matrix{
        // arr[行][列]
        int[][] arr;
        public Matrix(int[][] arr) {
            this.arr = arr;
        }

        public int[][] getArr() {
            return arr;
        }

        public void setD(int[][] arr) {
            this.arr = arr;
        }
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer str;

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

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

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

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

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

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

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. The code provided has a few potential weaknesses:

1. **Handling of Large `k` Values**: The code uses a simple check to determine if `k` is greater than \(2^{n-1}\), which might not be the correct condition for all cases.
2. **Bit Manipulation**: The code uses bit manipulation to determine the positions of elements in the permutation, which might not handle all edge cases correctly.
3. **Lexicographical Order**: The code assumes a specific way to generate permutations, which might not align with the lexicographical order of permutations with the maximum `S(p)`.

To create a test input that exposes these issues, we can consider the following scenarios:

- **Small `n` with large `k`**: This will test the condition `if (k > (long)Math.pow(2, n - 1))`.
- **Boundary values of `n` and `k`**: This will test the edge cases where `n` and `k` are at their limits.
- **Specific values of `k` that should result in `-1`**: This will test the correctness of the condition for invalid `k`.

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

```plaintext
6
3 2
3 3
4 11
4 6
6 39
7 34
```

This test input includes:
- `3 2`: A valid case with a small `n` and a small `k`.
- `3 3`: Another valid case with a small `n` and a small `k`.
- `4 11`: A case where `k` is larger than the number of valid permutations.
- `4 6`: A valid case with a small `n` and a medium `k`.
- `6 39`: A case where `k` is larger than the number of valid permutations.
- `7 34`: A case with a medium `n` and a medium `k`.

These cases should help expose any issues in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ordered Permutations
time_limit_ms: 2000
memory_limit_mb: 256
Description: Consider a permutation$$$^{\text{∗}}$$$ $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. We can introduce the following sum for it$$$^{\text{†}}$$$:

$$$$$$S(p) = \sum_{1 \le l \le r \le n} \min(p_l, p_{l + 1}, \ldots, p_r)$$$$$$

Let us consider all permutations of length $$$n$$$ with the maximum possible value of $$$S(p)$$$. Output the $$$k$$$-th of them in lexicographical$$$^{\text{‡}}$$$order, or report that there are less than $$$k$$$ of them.

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

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$1 \le k \le 10^{12}$$$) — the length of the permutation and the index number of the desired permutation.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$.

Output: For each test case, if there are less than $$$k$$$ suitable permutations, print $$$-1$$$.

Otherwise, print the $$$k$$$-th suitable permutation.

Examples:
input:
6
3 2
3 3
4 11
4 6
6 39
7 34
output:
1 3 2
2 3 1
-1
2 4 3 1
-1
2 3 4 5 7 6 1

Note: Let us calculate the required sum for all permutations of length $$$3$$$ (ordered lexicographically):

PermutationValue of $$$S(p)$$$$$$[1, 2, 3]$$$$$$10$$$$$$[1, 3, 2]$$$$$$10$$$$$$[2, 1, 3]$$$$$$9$$$$$$[2, 3, 1]$$$$$$10$$$$$$[3, 1, 2]$$$$$$9$$$$$$[3, 2, 1]$$$$$$10$$$

In the first test case, you have to print the second suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[1, 3, 2]$$$.

In the second test case, you have to print the third suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[2, 3, 1]$$$.

Buggy Code:
import java.util.*;

public class Main {

  public static void help(int n,long k){
    long x = (long)Math.pow(2,n-1);

    if(k>x){
      System.out.println(-1);
      return;
    }

    int[] arr = new int[n];
    int i=0;
    int j=arr.length-1;
    int a=1;
    while(x>0){
      if(k<=x/2){
        arr[i] = a;
        i++;
      }
      else{
        arr[j] = a;
        j--;
      }

      // System.out.println(k+" "+x);
      a++;
      x/=2;
      if(k>x){
        k-=x;
      }

    }

    for(i=0; i<n ; i++){
      System.out.print(arr[i]+" ");
    }
    System.out.println();
  }
    public static void main(String[] args) {
      Scanner sc = new Scanner(System.in);

      int t = sc.nextInt();

      while(t>0){
        int n = sc.nextInt();
        long k = sc.nextLong();
        help(n,k);
        t--;
      }
  }
}

**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 consider edge cases and scenarios where the code might fail. The buggy code has a few potential issues:

1. **Incorrect handling of the maximum number of permutations**: The code uses a fixed formula to determine the maximum number of permutations, which might not be accurate for all cases.
2. **Incorrect placement of elements**: The code places elements in the array based on a binary decision (if `k <= x/2` or `k > x/2`), which might not correctly handle all permutations.

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

1. **Test Case 1**: `n = 3, k = 4`
   - This case should return `-1` because there are only 3 suitable permutations for `n = 3`.

2. **Test Case 2**: `n = 4, k = 10`
   - This case should return `-1` because there are only 8 suitable permutations for `n = 4`.

3. **Test Case 3**: `n = 5, k = 16`
   - This case should return `-1` because there are only 16 suitable permutations for `n = 5`.

4. **Test Case 4**: `n = 6, k = 32`
   - This case should return `-1` because there are only 32 suitable permutations for `n = 6`.

Here is the test input:

```plaintext
4
3 4
4 10
5 16
6 32
```

This input will help identify if the code correctly handles the maximum number of permutations and returns `-1` when there are fewer suitable permutations than the requested `k`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ordered Permutations
time_limit_ms: 2000
memory_limit_mb: 256
Description: Consider a permutation$$$^{\text{∗}}$$$ $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. We can introduce the following sum for it$$$^{\text{†}}$$$:

$$$$$$S(p) = \sum_{1 \le l \le r \le n} \min(p_l, p_{l + 1}, \ldots, p_r)$$$$$$

Let us consider all permutations of length $$$n$$$ with the maximum possible value of $$$S(p)$$$. Output the $$$k$$$-th of them in lexicographical$$$^{\text{‡}}$$$order, or report that there are less than $$$k$$$ of them.

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

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$1 \le k \le 10^{12}$$$) — the length of the permutation and the index number of the desired permutation.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$.

Output: For each test case, if there are less than $$$k$$$ suitable permutations, print $$$-1$$$.

Otherwise, print the $$$k$$$-th suitable permutation.

Examples:
input:
6
3 2
3 3
4 11
4 6
6 39
7 34
output:
1 3 2
2 3 1
-1
2 4 3 1
-1
2 3 4 5 7 6 1

Note: Let us calculate the required sum for all permutations of length $$$3$$$ (ordered lexicographically):

PermutationValue of $$$S(p)$$$$$$[1, 2, 3]$$$$$$10$$$$$$[1, 3, 2]$$$$$$10$$$$$$[2, 1, 3]$$$$$$9$$$$$$[2, 3, 1]$$$$$$10$$$$$$[3, 1, 2]$$$$$$9$$$$$$[3, 2, 1]$$$$$$10$$$

In the first test case, you have to print the second suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[1, 3, 2]$$$.

In the second test case, you have to print the third suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[2, 3, 1]$$$.

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds; // find_by_order, order_of_key
#define fr(i, a, b) for (int i = (a); i < (int)(b); ++i)
#define frr(i, a, b) for (int i = (a); i > (int)(b); --i)
#define ina(x, n)               \
    for (int i = 0; i < n; ++i) \
        cin >> x[i];
#define in(n)    \
    long long n; \
    cin >> n;
#define int long long
#define double long double
#define pb push_back
#define vi vector<int>
#define mii map<int, int>
#define vvi vector<vector<int>>
#define vpi vector<pair<int, int>>
#define mdi map<double, int>
#define mci map<char, int>
#define mide map<int, deque<int>>
#define pi pair<int, int>
#define si set<int>
#define oyes cout << "YES" << endl;
#define ono cout << "NO" << endl;
#define oyess cout << "Yes" << endl;
#define onoo cout << "No" << endl;
#define ve1 cout << "-1" << endl;
#define ff first
#define ss second
#define all(x) (x).begin(), (x).end()
#define rev(x) reverse((x).begin(), (x).end())

void solve()
{
    in(n);
    in(k);
    if (n <= 63 && (1LL << (n - 1)) < k)
    {
        ve1;
        return;
    }
    vi ans(n);
    int ind1 = 0, ind2 = n - 1;
    si st;
    fr(i, 1, n + 1) st.insert(i);
    fr(i, 0, n)
    {
        if (k <= (1LL << (n - i - 2)))
        {
            ans[ind1] = *st.begin();
            ind1++;
            st.erase(st.begin());
        }
        else
        {
            ans[ind2] = *st.begin();
            ind2--;
            st.erase(st.begin());
            k -= (1LL << (n - i - 2));
        }
    }
    fr(i, 0, n)
    {
        if (ans[i] == 0)
            ans[i] = *st.begin();
    }
    fr(i, 0, n) cout << ans[i] << " ";
    cout << endl;
}

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    in(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
6
3 2
3 3
4 11
4 6
6 39
7 34
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ordered Permutations
time_limit_ms: 2000
memory_limit_mb: 256
Description: Consider a permutation$$$^{\text{∗}}$$$ $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. We can introduce the following sum for it$$$^{\text{†}}$$$:

$$$$$$S(p) = \sum_{1 \le l \le r \le n} \min(p_l, p_{l + 1}, \ldots, p_r)$$$$$$

Let us consider all permutations of length $$$n$$$ with the maximum possible value of $$$S(p)$$$. Output the $$$k$$$-th of them in lexicographical$$$^{\text{‡}}$$$order, or report that there are less than $$$k$$$ of them.

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

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$1 \le k \le 10^{12}$$$) — the length of the permutation and the index number of the desired permutation.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$.

Output: For each test case, if there are less than $$$k$$$ suitable permutations, print $$$-1$$$.

Otherwise, print the $$$k$$$-th suitable permutation.

Examples:
input:
6
3 2
3 3
4 11
4 6
6 39
7 34
output:
1 3 2
2 3 1
-1
2 4 3 1
-1
2 3 4 5 7 6 1

Note: Let us calculate the required sum for all permutations of length $$$3$$$ (ordered lexicographically):

PermutationValue of $$$S(p)$$$$$$[1, 2, 3]$$$$$$10$$$$$$[1, 3, 2]$$$$$$10$$$$$$[2, 1, 3]$$$$$$9$$$$$$[2, 3, 1]$$$$$$10$$$$$$[3, 1, 2]$$$$$$9$$$$$$[3, 2, 1]$$$$$$10$$$

In the first test case, you have to print the second suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[1, 3, 2]$$$.

In the second test case, you have to print the third suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[2, 3, 1]$$$.

Buggy Code:
// Problem: C. Ordered Permutations
// Contest: Codeforces - Codeforces Round 992 (Div. 2)
// URL: https://codeforces.com/contest/2040/problem/C
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
#include <cmath>
#define int long long int
#define pb push_back
/*---------------------------------------------------------------------------------------------------------------------------------*/
/*ASCII small case-> a-97 to z-122
upper case-> A-65 to Z-90*/
#define nl "\n"
#define mp map<int, int>
#define all(v) v.begin(), v.end()
#define F first
#define S second
#define ump unordered_map<long long, int, custom_hash>
#define mstpi multiset<pair<int, int> >
#define mst multiset<int>
#define vec vector<int>
#define pi pair<int, int>
#define ld long double
#define vep vector<pair<int, int>>
#define ul unsigned long long
#define loop1 for (int i = 0; i < n; i++)
#define REP(i, a, b) for (int i = a; i <= b; i++)
#define NO cout << "NO" << endl
#define YES cout << "YES" << endl
using namespace std;
using namespace __gnu_pbds;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>ordered_set; //find_by_order, order_of_key, greater_equal<int> -> making it desc order
/*---------------------------------------------------------------------------------------------------------------------------------*/
const double PI = 3.1415926535;
const int inf = 1e18;
const int mod = inf;
/*---------------------------------------------------------------------------------------------------------------------------------*/
int modmul(int a, int b, int m)
{
    a %= m;
    b %= m;
    return (a * b) % m;
}
int modexp(int n, int x)
{
    int ans = 1;
    while (n > 0)
    {
        if (n & 1)
        {
            ans = (ans * x) % mod;
        }
        n >>= 1;
        x = (x * x) % mod;
    }
    return ans;
}
/*---------------------------------------------------------------------------------------------------------------------------------*/
string add(string a, string b)
{
    string ans = "";
    int carry = 0;
    while (!a.empty() || !b.empty() || carry > 0)
    {
        int digitA = a.empty() ? 0 : a.back() - '0';
        int digitB = b.empty() ? 0 : b.back() - '0';
        int sum = digitA + digitB + carry;
        ans = to_string(sum % 10) + ans;
        carry = sum / 10;
        if (!a.empty())
            a.pop_back();
        if (!b.empty())
            b.pop_back();
    }
    return ans;
}
string ministrnum(string a, string b)
{
    if (a.length() > b.length())
        return b;
    else if (b.length() > a.length())
        return a;
    else
    {
        int n = a.length();
        for (int i = 0; i < n; i++)
        {
            if (a[i] > b[i])
                return b;
            else if (a[i] < b[i])
                return a;
        }
    }
    return a;
}
void sol(int n, int k, int i, int l, int r, vec &v) {
	if(l > r) return;
	int tot = n > 50 ? LLONG_MAX:modexp(n-1, 2);
	if(k > tot/2) {
		v[r--] = i;
	}
	else {
		v[l++] = i;
	}
	if(k > tot/2) k -= tot/2;
	sol(n-1, k, i+1, l, r, v);
}
void solve()
{
	int n, k;
	cin>>n>> k;
	if(n < 40 && modexp(n-1, 2) < k) {
		cout<<-1<<endl;
		return;
	}
	vector<int> v(n);
	sol(n, k, 1, 0, n-1, v);
	for(auto x: v) cout<<x<<" ";
	cout<<endl;
}
int32_t main()
{
    ios_base ::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--)
        solve();
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are the steps to consider:

1. **Small n and k values**: Test the smallest possible values to ensure the code handles basic cases correctly.
2. **Large n and k values**: Test the largest possible values within the constraints to check for overflow and performance issues.
3. **Boundary values**: Test values at the boundaries of the input constraints.
4. **Special cases**: Test specific values that might trigger edge cases in the algorithm, such as when `k` is just above or below the maximum possible value for a given `n`.

Given the constraints:
- \(1 \le t \le 10^4\)
- \(1 \le n \le 2 \cdot 10^5\)
- \(1 \le k \le 10^{12}\)

Let's create a small-scale test input that includes a mix of these cases:

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

This input includes:
- The smallest possible values for `n` and `k` (1 1)
- A few small values of `n` with `k` set to 1, which should be the lexicographically smallest permutation for each `n`.

This input is designed to check if the code correctly handles the smallest and simplest cases, which are often where bugs can be most easily identified. If the code fails on any of these simple cases, it will be clear that there is a fundamental issue.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ordered Permutations
time_limit_ms: 2000
memory_limit_mb: 256
Description: Consider a permutation$$$^{\text{∗}}$$$ $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. We can introduce the following sum for it$$$^{\text{†}}$$$:

$$$$$$S(p) = \sum_{1 \le l \le r \le n} \min(p_l, p_{l + 1}, \ldots, p_r)$$$$$$

Let us consider all permutations of length $$$n$$$ with the maximum possible value of $$$S(p)$$$. Output the $$$k$$$-th of them in lexicographical$$$^{\text{‡}}$$$order, or report that there are less than $$$k$$$ of them.

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

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$1 \le k \le 10^{12}$$$) — the length of the permutation and the index number of the desired permutation.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$.

Output: For each test case, if there are less than $$$k$$$ suitable permutations, print $$$-1$$$.

Otherwise, print the $$$k$$$-th suitable permutation.

Examples:
input:
6
3 2
3 3
4 11
4 6
6 39
7 34
output:
1 3 2
2 3 1
-1
2 4 3 1
-1
2 3 4 5 7 6 1

Note: Let us calculate the required sum for all permutations of length $$$3$$$ (ordered lexicographically):

PermutationValue of $$$S(p)$$$$$$[1, 2, 3]$$$$$$10$$$$$$[1, 3, 2]$$$$$$10$$$$$$[2, 1, 3]$$$$$$9$$$$$$[2, 3, 1]$$$$$$10$$$$$$[3, 1, 2]$$$$$$9$$$$$$[3, 2, 1]$$$$$$10$$$

In the first test case, you have to print the second suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[1, 3, 2]$$$.

In the second test case, you have to print the third suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[2, 3, 1]$$$.

Buggy Code:
def calculate_bits(n, k):
    """
    Helper function to calculate the bits and positions for the result.
    """
    L = []
    remaining = k - 1
    bits = [1] * 64  # Simulate a 64-bit set
    borrow = 0
    i = 0

    while i < n - 1:
        a = bits[i]
        b = remaining & 1
        remaining >>= 1
        res = a - b - borrow
        if res < 0:
            res += 2
            borrow = 1
        else:
            borrow = 0
        bits[i] = res
        if i >= 50 and borrow == 0 and remaining == 0:
            break
        i += 1

    if borrow != 0:
        i = 50
        while i < n - 1:
            if bits[i] == 1:
                bits[i] = 0
                borrow = 0
                break
            i += 1

    j = 1
    while j <= n - 1:
        pos = n - 1 - j
        if bits[pos] == 1:
            L.append(j)
        j += 1

    return L

def solve(n, k):
    """
    Main function to solve the problem.
    """
    if n == 1:
        return [-1] if k != 1 else [1]

    limit = (1 << (n - 1)) if n - 1 < 50 else int(1e13)

    if (n - 1 < 50 and k > (1 << (n - 1))) or (n - 1 >= 50 and k > int(1e13) + 1):
        return [-1]

    L = calculate_bits(n, k)
    result = L + [n]

    inL = {x: True for x in L}
    index = n - 1
    while index >= 1:
        if index not in inL:
            result.append(index)
        index -= 1

    return result

def main():

    t = int(input().strip())
    for _ in range(t):
        try:
            n, k = map(int, input().strip().split())
            ans = solve(n, k)
            print(" ".join(map(str, ans)))
        except ValueError:
            print("Invalid input for n and k.")

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 provided code, we need to consider edge cases and scenarios that are likely to cause issues. The code involves calculating permutations and handling large numbers, so we should focus on:

1. **Small values of \( n \) and \( k \)**: These can help in verifying basic functionality.
2. **Large values of \( n \) and \( k \)**: These can test the limits of the algorithm and expose issues with large number handling.
3. **Boundary values**: Values that are close to the limits of the constraints.

Let's construct a test input that includes these scenarios:

```plaintext
6
1 1
3 2
3 3
4 6
4 11
6 39
```

### Explanation:
1. **Test Case 1: \( n = 1 \), \( k = 1 \)**
   - This tests the base case where \( n = 1 \). The expected output is `[1]`.

2. **Test Case 2: \( n = 3 \), \( k = 2 \)**
   - This tests a small permutation length with a valid \( k \). The expected output is `[1 3 2]`.

3. **Test Case 3: \( n = 3 \), \( k = 3 \)**
   - This tests a small permutation length with a valid \( k \). The expected output is `[2 3 1]`.

4. **Test Case 4: \( n = 4 \), \( k = 6 \)**
   - This tests a slightly larger permutation length with a valid \( k \). The expected output is `[2 4 3 1]`.

5. **Test Case 5: \( n = 4 \), \( k = 11 \)**
   - This tests a slightly larger permutation length with an invalid \( k \). The expected output is `[-1]`.

6. **Test Case 6: \( n = 6 \), \( k = 39 \)**
   - This tests a larger permutation length with a valid \( k \). The expected output is `[-1]` (since there are fewer than 39 valid permutations).

This test input is designed to cover a range of scenarios, including small and large values, and boundary conditions, to help identify any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ordered Permutations
time_limit_ms: 2000
memory_limit_mb: 256
Description: Consider a permutation$$$^{\text{∗}}$$$ $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. We can introduce the following sum for it$$$^{\text{†}}$$$:

$$$$$$S(p) = \sum_{1 \le l \le r \le n} \min(p_l, p_{l + 1}, \ldots, p_r)$$$$$$

Let us consider all permutations of length $$$n$$$ with the maximum possible value of $$$S(p)$$$. Output the $$$k$$$-th of them in lexicographical$$$^{\text{‡}}$$$order, or report that there are less than $$$k$$$ of them.

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

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$1 \le k \le 10^{12}$$$) — the length of the permutation and the index number of the desired permutation.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$.

Output: For each test case, if there are less than $$$k$$$ suitable permutations, print $$$-1$$$.

Otherwise, print the $$$k$$$-th suitable permutation.

Examples:
input:
6
3 2
3 3
4 11
4 6
6 39
7 34
output:
1 3 2
2 3 1
-1
2 4 3 1
-1
2 3 4 5 7 6 1

Note: Let us calculate the required sum for all permutations of length $$$3$$$ (ordered lexicographically):

PermutationValue of $$$S(p)$$$$$$[1, 2, 3]$$$$$$10$$$$$$[1, 3, 2]$$$$$$10$$$$$$[2, 1, 3]$$$$$$9$$$$$$[2, 3, 1]$$$$$$10$$$$$$[3, 1, 2]$$$$$$9$$$$$$[3, 2, 1]$$$$$$10$$$

In the first test case, you have to print the second suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[1, 3, 2]$$$.

In the second test case, you have to print the third suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[2, 3, 1]$$$.

Buggy Code:
import sys; input=sys.stdin.readline
for i in range(int(input())):
  n,k=map(int,input().split())
  if k<=1<<(n-1):
    out=[0]*n
    pl,pr=0,n-1
    cnt=n-2
    k-=1
    for j in range(1,n):
      if cnt>40:
        out[pr]=j
        pr-=1
      else:
        if not(k&(1<<(cnt))):
          out[pl]=j
          pl+=1
        else:
          out[pr]=j
          pr-=1
      cnt-=1
    out[pl]=n
    print(*out)
  else:
    print(-1)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios where the code might fail. The bug in the code is related to the logic for determining the positions of elements in the permutation, especially when the value of `k` is close to the maximum possible value for a given `n`.

Here are the steps to identify a challenging test input:

1. **Edge Cases for `k`:** The code might fail when `k` is very close to the maximum number of valid permutations. For `n`, the maximum number of valid permutations is `2^(n-1)`. We should test values of `k` that are just below this maximum.

2. **Edge Cases for `n`:** The code might also fail for small values of `n` or very large values of `n` within the given constraints.

3. **Special Values for `n` and `k`:** Test with values where `n` is a power of 2, as these can often reveal issues in bitwise operations.

4. **Multiple Test Cases:** To ensure the test input is complete and valid, we can include multiple test cases, each designed to expose different aspects of the bug.

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

```plaintext
4
3 3
4 7
5 15
6 31
```

Explanation:
- **Test Case 1 (3, 3):** This is a simple case where `k` is just below the maximum number of valid permutations for `n = 3` (which is `4`).
- **Test Case 2 (4, 7):** This is a case where `k` is just below the maximum number of valid permutations for `n = 4` (which is `8`).
- **Test Case 3 (5, 15):** This is a case where `k` is just below the maximum number of valid permutations for `n = 5` (which is `16`).
- **Test Case 4 (6, 31):** This is a case where `k` is just below the maximum number of valid permutations for `n = 6` (which is `32`).

These test cases are designed to push the code to its limits and expose any issues in the logic for determining the positions of elements in the permutation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ordered Permutations
time_limit_ms: 2000
memory_limit_mb: 256
Description: Consider a permutation$$$^{\text{∗}}$$$ $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. We can introduce the following sum for it$$$^{\text{†}}$$$:

$$$$$$S(p) = \sum_{1 \le l \le r \le n} \min(p_l, p_{l + 1}, \ldots, p_r)$$$$$$

Let us consider all permutations of length $$$n$$$ with the maximum possible value of $$$S(p)$$$. Output the $$$k$$$-th of them in lexicographical$$$^{\text{‡}}$$$order, or report that there are less than $$$k$$$ of them.

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

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$1 \le k \le 10^{12}$$$) — the length of the permutation and the index number of the desired permutation.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$.

Output: For each test case, if there are less than $$$k$$$ suitable permutations, print $$$-1$$$.

Otherwise, print the $$$k$$$-th suitable permutation.

Examples:
input:
6
3 2
3 3
4 11
4 6
6 39
7 34
output:
1 3 2
2 3 1
-1
2 4 3 1
-1
2 3 4 5 7 6 1

Note: Let us calculate the required sum for all permutations of length $$$3$$$ (ordered lexicographically):

PermutationValue of $$$S(p)$$$$$$[1, 2, 3]$$$$$$10$$$$$$[1, 3, 2]$$$$$$10$$$$$$[2, 1, 3]$$$$$$9$$$$$$[2, 3, 1]$$$$$$10$$$$$$[3, 1, 2]$$$$$$9$$$$$$[3, 2, 1]$$$$$$10$$$

In the first test case, you have to print the second suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[1, 3, 2]$$$.

In the second test case, you have to print the third suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[2, 3, 1]$$$.

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

public class Main {
    static String vv = "";
    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int t =sc.nextInt();
        while (t-->0){
            int n = sc.nextInt();
            long k = sc.nextLong();
            int [] res = new int[n];
            int first = 0, last=n-1;
            int start = 1;

            boolean f = false;
            while (n>63){
                f = true;
                res[first]=start;
                start++;
                first++;
                n--;
            }
            long sum = (long) Math.pow(2,n-1);

            if(f || k<=sum) {
                for (int i = start; i <= n+start-1; i++) {
                    long x = (long) Math.pow(2, n - 1 - i);
                    if (k > x) {
                        res[last] = i;
                        last--;
                        k -= x;
                    } else {
                        res[first] = i;
                        first++;
                    }
                }
                for (int z : res)
                    pw.print(z + " ");
                pw.println();
            }
            else
                pw.println("-1");

        }

        pw.flush();

    }

    public static void rec(int [] b,int i){
        if (i==b.length){
            boolean f = true;
            for (int z =0;z<b.length-2;z++){
                if (b[z]>b[z+1] && b[z+2]>b[z+1]){
                    f = false;
                    break;
                }
            }
            if (f) {
                for (int z : b)
                    System.out.print(z + " ");
                System.out.println();
            }
            return;
        }
        for(int j = i;j<b.length;j++){
            int temp = b[i];
            b[i]= b[j];
            b[j]=temp;
            rec(b,i+1);
            temp = b[i];
            b[i]= b[j];
            b[j]=temp;
        }

    }

    public static int lcs(String s1,String s2 , int i , int j,String res){
        if (i == s1.length() || j == s2.length())
            return 0;
        if (s1.charAt(i)==s2.charAt(j)) {
            if (res.length()+1>vv.length())
                vv = res+s1.charAt(i);
            return 1 + lcs(s1, s2, i + 1, j + 1, res + s1.charAt(i));
        }
        return Math.max(lcs(s1, s2, i+1, j,res),lcs(s1, s2, i, j+1,res));
    }

    public static long dp(long [][] arr,long [] [] memo,int i,int j, int max ){
        if(i==arr.length)
            return 0;
        if (memo[i][j]==-1) {
            memo[i][j] = dp(arr, memo, i + 1,j,max);
            if (j+ arr[i][1] <= max)
                memo[i][j]= Math.max(arr[i][0] + dp(arr, memo,i+1,(int) (j+arr[i][1]),max ),memo[i][j]);
        }
        return memo[i][j];
    }

    public static long getMaxArea(long hist[]) {
        // your code here
        Stack<Integer> st = new Stack<>();
        long [] res = new long [hist.length];
        for(int i =0;i<hist.length;i++){
            if(st.isEmpty())
                st.push(i);
            else{
                while(!st.isEmpty() && hist[st.peek()]>hist[i]){
                    res[st.peek()]=i-st.peek()-1;
                    st.pop();
                }
                st.push(i);
            }
        }
        while(!st.isEmpty()){
            res[(int)st.peek()]= hist.length-st.peek()-1;
            st.pop();
        }
        for(int i =(int)(hist.length-1);i>=0;i--){
            if(st.isEmpty())
                st.push(i);
            else{
                while(!st.isEmpty() && hist[st.peek()]>hist[i]){
                    res[st.peek()]+=st.peek()-i-1;
                    st.pop();
                }
                st.push(i);
            }
        }
        while(!st.isEmpty()){
            res[st.peek()]+=st.peek();
            st.pop();
        }
        long max = 0;
        for(int i =0;i<res.length;i++)
            max = Math.max(max,(res[i]+1)*hist[i]);
        return max;

    }

        public static String removeKdigits(String num, int k) {
            Stack<String> st = new Stack<>();
            st.push(num.charAt(0)+"");
            for(int i =1;i<num.length();i++){
                while(!st.isEmpty() && k>0 && num.charAt(i)-'0'<st.peek().charAt(0)-'0'){
                    st.pop();
                    k--;
                }
                st.push(num.charAt(i)+"");
            }
            String res ="";
            while(k-->0)
                st.pop();
            while(!st.isEmpty())
                res = st.pop()+res;
            while(res.length()>0 &&  res.charAt(0)=='0') {
                res = res.substring(1);
                if(res.length()==0){
                    res ="0";
                    break;
                }
            }
            if(res.length()==0)
                res ="0";
            return res;
        }

    //-------------------------------------------------------
    static int[] vis;
    static ArrayList<Integer>[] adj;
    static int s,d;
    static int zz = 0;
    static int dfs(ArrayList<HashMap>[] arr,int node,int max){
        for (HashMap z : arr[node]){

        }

        return max;
    }
    static boolean bfs(){
        Queue<Integer> q = new LinkedList<>();
        q.add(s);
        vis[s]=1;
        while(!q.isEmpty()){
            int cur = q.poll();
            if(cur==d)return true;
            for(int e:adj[cur]){
                if(vis[e]==1)continue;
                vis[e]= 1;
                q.add(e);
            }
        }
        return false;
    }

    public static boolean isPrime(int num) {
        if (num <= 1) return false;
        if (num <= 3) return true;
        if (num % 2 == 0 || num % 3 == 0) return false;
        for (int i = 5; i * i <= num; i += 6) {
            if (num % i == 0 || num % (i + 2) == 0) return false;
        }
        return true;
    }

    public static boolean knapsack(int [] arr,int target,int i,Stack<Integer> x, String s){
        //basecase
        if (i==arr.length){
            int sum =0;
            Stack<Integer> temp =(Stack<Integer>)x.clone();
            if (!s.isEmpty()) {
                for (int ff = s.length()-1; ff >= 0; ff--) {
                    int g = temp.pop();
                    if (!temp.empty() && ff == s.length()-1)
                        sum =  g + temp.pop();
                    else
                        sum+=g;

                }
            }
            if (sum==target)
                return true;
            return false;
        }

        //recursive call

        int z = x.pop() * arr[i];
        x.push(z);
        boolean times = knapsack(arr,target,i+1,x,s);
        if (times)
            return true;
        if (!x.empty())
            x.pop();
        x.push(z/arr[i]);
        x.push(arr[i]);
        boolean plus = knapsack(arr,target,i+1,x,s+"+");
        if (plus)
            return true;
        arr[i]=-arr[i];
        int sd = -x.pop();
        x.push(sd);
        boolean minus = knapsack(arr,target,i+1,x,s+"+");
        arr[i]=-arr[i];
        sd = -x.pop();
        x.push(sd);
        x.pop();
        if (minus)
            return true;
        return false;
    }
    static class Pair implements Comparable<Pair>{
        int x, y;
        int [] z;
        Pair(int x, int y){
            this.x = x;
            this.y = y;
        }
        Pair(int x , int [] z){
            this.x =x;
            this.z=z;
        }
        public int compareTo(Pair p){
            return this.x - p.x;
        }
    }
    static class Scanner {

        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(String file) throws IOException {
            br = new BufferedReader(new FileReader(file));
        }

        public Scanner(FileReader r) {
            br = new BufferedReader(r);
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public String readAllLines(BufferedReader reader) throws IOException {
            StringBuilder content = new StringBuilder();
            String line;
            while ((line = reader.readLine()) != null) {
                content.append(line);
                content.append(System.lineSeparator());
            }
            return content.toString();
        }

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

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

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec)
                        f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

        public long[] nextlongArray(int n) throws IOException {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public Long[] nextLongArray(int n) throws IOException {
            Long[] a = new Long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public int[] nextIntArray(int n) throws IOException {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public Integer[] nextIntegerArray(int n) throws IOException {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public boolean ready() throws IOException {
            return br.ready();
        }

    }
}

**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
6
3 2
3 3
4 11
4 6
6 39
7 34
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ordered Permutations
time_limit_ms: 2000
memory_limit_mb: 256
Description: Consider a permutation$$$^{\text{∗}}$$$ $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. We can introduce the following sum for it$$$^{\text{†}}$$$:

$$$$$$S(p) = \sum_{1 \le l \le r \le n} \min(p_l, p_{l + 1}, \ldots, p_r)$$$$$$

Let us consider all permutations of length $$$n$$$ with the maximum possible value of $$$S(p)$$$. Output the $$$k$$$-th of them in lexicographical$$$^{\text{‡}}$$$order, or report that there are less than $$$k$$$ of them.

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

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$1 \le k \le 10^{12}$$$) — the length of the permutation and the index number of the desired permutation.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$.

Output: For each test case, if there are less than $$$k$$$ suitable permutations, print $$$-1$$$.

Otherwise, print the $$$k$$$-th suitable permutation.

Examples:
input:
6
3 2
3 3
4 11
4 6
6 39
7 34
output:
1 3 2
2 3 1
-1
2 4 3 1
-1
2 3 4 5 7 6 1

Note: Let us calculate the required sum for all permutations of length $$$3$$$ (ordered lexicographically):

PermutationValue of $$$S(p)$$$$$$[1, 2, 3]$$$$$$10$$$$$$[1, 3, 2]$$$$$$10$$$$$$[2, 1, 3]$$$$$$9$$$$$$[2, 3, 1]$$$$$$10$$$$$$[3, 1, 2]$$$$$$9$$$$$$[3, 2, 1]$$$$$$10$$$

In the first test case, you have to print the second suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[1, 3, 2]$$$.

In the second test case, you have to print the third suitable permutation of length $$$3$$$. Looking at the table, we see that it is the permutation $$$[2, 3, 1]$$$.

Buggy Code:
import java.util.*;

public class codeforces {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        while (t-- > 0) {
            long n = in.nextLong(), k = in.nextLong();
            long cal = (1l << (n - 1l));
            if (k > cal) {
                System.out.println(-1);
            } else {
                long[] arr = new long[(int) n];
                int i = 0, j = (int) n - 1;
                for (long a = 1; a <= n; a++) {
                    if (k > ((cal) / 2l)) {
                        arr[j--] = a;
                        k -= cal / 2l;
                    } else {
                        arr[i++] = a;
                    }
                    cal /= 2l;
                }
                print(arr);
            }
        }
    }

    static void print(long[] arr){
        for (long i : arr) {
            System.out.print(i+" ");
        }
        System.out.println();
    }

    static void tryIt() {
        int x = 6;
        while (x-- > 0) {
            int[] arr = doIt(x);
            int k = (int) factorialTillN((long) (arr.length));
            int maxSum = sum(arr);
            int i = 1, numbersOf = 0;
            ArrayList<Integer> l = new ArrayList<>();
            while (k-- > 0) {
                int sum = sum(arr);
                if (sum == maxSum) {
                    l.add(i);
                    numbersOf++;
                }
                if (sum == maxSum)
                    System.out.println(Arrays.toString(arr) + " sum = " + sum);
                nextPermutation(arr);
                i++;
            }
            iii(l);
            System.out.println(l);
            // System.out.println("numbres = " + numbersOf);
        }
    }

    static void iii(ArrayList<Integer> l) {
        for (Integer integer : l) {
            System.out.print((integer / 2) + " ");
        }
        System.out.println();
    }

    static int[] doIt(int n) {
        int[] arr = new int[n];
        for (int i = 1; i <= n; i++) {
            arr[i - 1] = i;
        }
        return arr;
    }

    static long factorialTillN(long n) {
        long pow = 1;
        long[] arr = new long[(int) n];
        for (int i = 1; i <= n; i++) {
            pow *= i;
            // arr[i] = pow;
        }
        return pow;
    }

    static int sum(int[] arr) {
        int n = arr.length, ans = 0;
        for (int i = 0; i < n; i++) {
            int min = arr[i];
            for (int j = i; j < n; j++) {
                min = Math.min(min, arr[j]);
                ans += min;
            }
        }
        return ans;
    }

    static void nextPermutation(int[] arr) {
        int x = arr.length - 1;
        for (int i = arr.length - 1; i > 0; i--) {
            if (arr[i] > arr[i - 1]) {
                int ind = BS(arr, i, arr.length - 1, arr[i - 1]);
                int swap = arr[i - 1];
                arr[i - 1] = arr[ind];
                arr[ind] = swap;
                reeverse(arr, i, arr.length - 1);
                break;
            }
        }
    }

    static int BS(int[] arr, int i, int j, int key) {
        int start = i, end = j, mid = 0, ans = -1;
        while (start <= end) {
            mid = start + (end - start) / 2;
            if (arr[mid] < key) {
                end = mid - 1;
            } else {
                ans = mid;
                start = mid + 1;
            }
        }
        return ans;
    }

    static void reeverse(int[] arr, int i, int j) {
        while (i < j) {
            int swap = arr[i];
            arr[i] = arr[j];
            arr[j] = swap;
            i++;
            j--;
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that can cause the code to fail. The code is supposed to find the k-th permutation with the maximum possible value of \( S(p) \) in lexicographical order. Here are the steps to identify such a test input:

1. **Edge Cases for \( n \) and \( k \):**
   - \( n = 1 \): The only permutation is [1].
   - \( n = 2 \): Permutations are [1, 2] and [2, 1].
   - \( n = 3 \): Permutations are [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1].

2. **Large \( k \) Values:**
   - For \( n = 4 \), there are 16 permutations, and \( k \) can be up to 16.
   - For \( n = 5 \), there are 32 permutations, and \( k \) can be up to 32.

3. **Specific \( k \) Values:**
   - \( k = 1 \): The first permutation in lexicographical order.
   - \( k = 2 \): The second permutation in lexicographical order.
   - \( k = 10 \): A middle permutation in lexicographical order.
   - \( k = 15 \): A large \( k \) value for \( n = 4 \).

4. **Test Input:**
   - We will include multiple test cases with different values of \( n \) and \( k \) to cover various scenarios.

Here is the test input:

```plaintext
5
1 1
2 2
3 2
4 10
5 32
```

This test input includes:
- \( n = 1 \) and \( k = 1 \) (edge case with only one permutation).
- \( n = 2 \) and \( k = 2 \) (edge case with two permutations).
- \( n = 3 \) and \( k = 2 \) (middle case with six permutations).
- \( n = 4 \) and \( k = 10 \) (middle case with 16 permutations).
- \( n = 5 \) and \( k = 32 \) (large \( k \) value with 32 permutations).

This input should help in identifying any issues in the buggy code, particularly with the logic for generating the k-th permutation and handling large values of \( k \).


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Billetes MX (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, it is guaranteed that $q \leq 10^5$. You can make hacks only if both versions of the problem are solved.

An integer grid $A$ with $p$ rows and $q$ columns is called beautiful if:

  * All elements of the grid are integers between $0$ and $2^{30}-1$, and    * For any subgrid, the XOR of the values at the corners is equal to $0$. Formally, for any four integers $i_1$, $i_2$, $j_1$, $j_2$ ($1 \le i_1 < i_2 \le p$; $1 \le j_1 < j_2 \le q$), $A_{i_1, j_1} \oplus A_{i_1, j_2} \oplus A_{i_2, j_1} \oplus A_{i_2, j_2} = 0$, where $\oplus$ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

There is a partially filled integer grid $G$ with $n$ rows and $m$ columns where only $k$ cells are filled. Polycarp wants to know how many ways he can assign integers to the unfilled cells so that the grid is beautiful.

However, Monocarp thinks that this problem is too easy. Therefore, he will perform $q$ updates on the grid. In each update, he will choose an unfilled cell and assign an integer to it. Note that these updates are persistent. That is, changes made to the grid will apply when processing future updates.

For each of the $q + 1$ states of the grid, the initial state and after each of the $q$ queries, determine the number of ways Polycarp can assign integers to the unfilled cells so that the grid is beautiful. Since this number can be very large, you are only required to output their values modulo $10^9+7$.

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

The first line of each test case contains four integers $n$, $m$, $k$ and $q$ ($2 \le n, m \le 10^5$; $0 \le k, q \leq 10^5$) — the number of rows, the number of columns, the number of fixed cells, and the number of updates.

The following $k$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) indicating that $G_{r, c}$ is assigned the integer $v$.

The following $q$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) indicating that $G_{r, c}$ is assigned the integer $v$.

It is guaranteed that the pairs $(r,c)$ over all assignments are distinct.

It is guaranteed that the sum of $n$, $m$, $k$ and $q$ over all test cases does not exceed $10^5$ respectively.

Output: For each test case, output $q + 1$ lines. The $i$-th line of output should contain the answer of the $i$-th state of the grid modulo $10^9 + 7$.

Examples:
input:
3
3 3 8 1
2 1 6
3 2 12
1 2 6
2 2 0
1 3 10
1 1 0
2 3 12
3 1 10
3 3 1
2 5 2 0
1 1 10
1 2 30
2 5 0 2
1 1 10
1 2 30
output:
1
0
489373567
651321892
769740174
489373567

Note: In the first test case of the example, we initially have the following grid:

$0$| $6$| $10$   ---|---|---   $6$| $0$| $12$   $10$| $12$| $?$      It can be proven that the only valid value for tile $(3, 3)$ is $0$, so the first answer is $1$. For the second query, the grid does not satisfy the condition, and thus the answer is $0$.

Buggy Code:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <set>
#include <vector>
#include <queue>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <cassert>
#include <functional>
#include <stack>
#include <random>
#define all(arr) arr.begin(), arr.end()
#define ll long long
#define MS(arr, val) memset(arr, val, sizeof(arr))
#define vi vector<int>
#define pb emplace_back
#define PI pair<int, int>
#define PL pair<ll, ll>
#define lll __int128
#define ull unsigned long long
#define Kases int T; rd(T); rep(kase, 1, T)
#define rep(i, a, b) for(int i = (a), __ ## i = (b); i <= __ ## i; ++i)
#define per(i, b, a) for(int i = (b), __ ## i = (a); i >= __ ## i; --i)

using namespace std;

// static char buf[1000000],*p1=buf,*p2=buf,obuf[1000000],*p3=obuf;
// #define getchar() p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++
// #define putchar(x) (p3-obuf<1000000)?(*p3++=x):(fwrite(obuf,p3-obuf,1,stdout),p3=obuf,*p3++=x)
inline void rd(char* x){scanf("%s", x);}
inline void rd(double &x){scanf("%lf", &x);}
// inline void rd(int &x){scanf("%d", &x);}
template<typename T> inline void rd(T &x){
    x = 0; bool is = 0; char ch = getchar();
    while(!isdigit(ch)) is |= (ch == '-'), ch = getchar();
    while(isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
    is && (x = -x);
}
// template<typename T>void write(T x){
//     x < 0 && (x = -x, putchar('-')); if(x >= 10) write(x / 10); putchar(x % 10 + '0');
// }
template<typename T, typename ...U> inline void rd(T &x, U &...y){rd(x), rd(y...);}

// std::random_device rnd;
// std::mt19937 rng(rnd());
// int gcd(int a, int b){return b == 0 ? a : gcd(b, a % b);}
const int maxn = 214514;
const ll mod = 1e9 + 7;
int par[maxn], f[maxn], rf[maxn];
int fr(int u)
{
    if(par[u] == u) return u;
    int rt = fr(par[u]);
    f[u] ^= f[rt]; par[u] = rt;
    return rt;
}
ll pw[maxn];
int main()
{
    pw[0] = 1, pw[1] = (1 << 30) % mod;
    rep(i, 2, maxn - 1) pw[i] = (pw[i - 1] * pw[1]) % mod;
    Kases
    {
        int n, m, k, q; rd(n, m, k, q);
        int can_11 = -1, can = 1;
        rep(i, 1, n + m) par[i] = i, f[i] = 0, rf[i] = -1;
        int cnt = (n - 1) + (m - 1);

        auto upd = [&](int x, int y, int val)
        {
            if(can == 0) return;
            if(x == 1 && y == 1)
            {
                if(can_11 != -1 && can_11 != val) can = 0;
                can_11 = val;
            }
            else if(x == 1 || y == 1)
            {
                int u = (x == 1 ? y : m + x);
                int rt = fr(u);
                if(rf[rt] == -1) --cnt, rf[rt] = (f[u] ^ val);
                else if(rf[rt] ^ f[u] ^ val) can = 0;
            }
            else
            {
                int u = y, v = m + x;
                int rtu = fr(u), rtv = fr(v);
                if(rf[rtv] != -1) swap(rtu, rtv), swap(u, v);
                if(rtu == rtv)
                {
                    if(f[u] ^ f[v] ^ val) can = 0;
                    return;
                }
                if(rf[rtv] == -1) --cnt;
                if(rf[rtu] != -1 && rf[rtv] != -1)
                {
                    if(rf[rtu] ^ rf[rtv] ^ f[u] ^ f[v] ^ val)
                    {
                        can = 0;
                        return;
                    }
                }

                f[rtv] = (val ^ f[u] ^ f[v]); par[rtv] = rtu;
            }
        };

        rep(i, 1, k)
        {
            int x, y, v; rd(x, y, v); upd(x, y, v);
        }
        assert(cnt >= 0);
        printf("%lld\n", can == 0 ? 0 : pw[cnt + (can_11 == -1)]);
        while(q--)
        {
            int x, y, v; rd(x, y, v); upd(x, y, v);
            assert(cnt >= 0);
            printf("%lld\n", can == 0 ? 0 : pw[cnt + (can_11 == -1)]);
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes bugs in the provided code, we need to focus on edge cases and scenarios that are likely to cause issues. Here are the steps to create such a test input:

1. **Initial Grid Configuration**: Start with a small grid and a few fixed cells to ensure the grid is not trivial.
2. **Updates**: Apply updates that challenge the logic, especially focusing on the XOR condition and the handling of root nodes in the disjoint set union (DSU) structure.
3. **Edge Cases**: Consider cases where the grid is initially valid but becomes invalid after an update, or where the grid is initially invalid but becomes valid after an update.

Let's create a test case based on these considerations:

### Test Case
```plaintext
1
3 3 2 2
1 1 0
2 2 0
2 3 1
3 1 1
```

### Explanation
1. **Initial Grid**:
   - The grid is 3x3.
   - Two cells are initially fixed: \( G_{1,1} = 0 \) and \( G_{2,2} = 0 \).

2. **Updates**:
   - First update: \( G_{2,3} = 1 \)
   - Second update: \( G_{3,1} = 1 \)

### Expected Behavior
- **Initial State**: The grid is valid, and there are multiple ways to fill the remaining cells to keep it valid. The number of ways is \( 2^{30} \) (since there are no other constraints).
- **First Update**: After setting \( G_{2,3} = 1 \), the grid should still be valid, but the number of ways to fill the remaining cells might change.
- **Second Update**: After setting \( G_{3,1} = 1 \), the grid should still be valid, but the number of ways to fill the remaining cells might change again.

### Expected Output
```plaintext
1073741824
1073741824
1073741824
```

### Why This Test Case is Challenging
- **Initial Validity**: The initial grid is valid, and the number of ways to fill the remaining cells is large.
- **Updates**: The updates are designed to challenge the logic of the XOR condition and the handling of root nodes in the DSU structure.
- **Edge Cases**: The updates might cause the grid to become invalid if the XOR condition is not correctly handled, but in this case, the grid remains valid, which can help expose issues in the code's ability to maintain the validity of the grid.

This test case should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Billetes MX (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, it is guaranteed that $q \leq 10^5$. You can make hacks only if both versions of the problem are solved.

An integer grid $A$ with $p$ rows and $q$ columns is called beautiful if:

  * All elements of the grid are integers between $0$ and $2^{30}-1$, and    * For any subgrid, the XOR of the values at the corners is equal to $0$. Formally, for any four integers $i_1$, $i_2$, $j_1$, $j_2$ ($1 \le i_1 < i_2 \le p$; $1 \le j_1 < j_2 \le q$), $A_{i_1, j_1} \oplus A_{i_1, j_2} \oplus A_{i_2, j_1} \oplus A_{i_2, j_2} = 0$, where $\oplus$ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

There is a partially filled integer grid $G$ with $n$ rows and $m$ columns where only $k$ cells are filled. Polycarp wants to know how many ways he can assign integers to the unfilled cells so that the grid is beautiful.

However, Monocarp thinks that this problem is too easy. Therefore, he will perform $q$ updates on the grid. In each update, he will choose an unfilled cell and assign an integer to it. Note that these updates are persistent. That is, changes made to the grid will apply when processing future updates.

For each of the $q + 1$ states of the grid, the initial state and after each of the $q$ queries, determine the number of ways Polycarp can assign integers to the unfilled cells so that the grid is beautiful. Since this number can be very large, you are only required to output their values modulo $10^9+7$.

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

The first line of each test case contains four integers $n$, $m$, $k$ and $q$ ($2 \le n, m \le 10^5$; $0 \le k, q \leq 10^5$) — the number of rows, the number of columns, the number of fixed cells, and the number of updates.

The following $k$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) indicating that $G_{r, c}$ is assigned the integer $v$.

The following $q$ lines contain three integers $r$, $c$ and $v$ ($1 \le r \le n, 1 \le c \le m$; $0 \le v < 2^{30}$) indicating that $G_{r, c}$ is assigned the integer $v$.

It is guaranteed that the pairs $(r,c)$ over all assignments are distinct.

It is guaranteed that the sum of $n$, $m$, $k$ and $q$ over all test cases does not exceed $10^5$ respectively.

Output: For each test case, output $q + 1$ lines. The $i$-th line of output should contain the answer of the $i$-th state of the grid modulo $10^9 + 7$.

Examples:
input:
3
3 3 8 1
2 1 6
3 2 12
1 2 6
2 2 0
1 3 10
1 1 0
2 3 12
3 1 10
3 3 1
2 5 2 0
1 1 10
1 2 30
2 5 0 2
1 1 10
1 2 30
output:
1
0
489373567
651321892
769740174
489373567

Note: In the first test case of the example, we initially have the following grid:

$0$| $6$| $10$   ---|---|---   $6$| $0$| $12$   $10$| $12$| $?$      It can be proven that the only valid value for tile $(3, 3)$ is $0$, so the first answer is $1$. For the second query, the grid does not satisfy the condition, and thus the answer is $0$.

Buggy Code:
#include<bits/stdc++.h>
#ifdef DEBUG_LOCAL
#include <mydebug/debug.h>
#endif
using ll = long long;
const int N = 2e5+5,mod = 1e9 + 7,inv = (1e9 + 8) / 2;
using namespace std;
int T,n,m,k,q,x,y,z,fg,pw[N * 2],res; vector<pair<int,int>> g[N];
struct dsu{
    int tg[N],w[N],fa[N],as,fg,siz[N];
    int fd(int x){
        if(x == fa[x]) return x;
        int t = fa[x]; fa[x] = fd(fa[x]),w[x] ^= w[t];
        return fa[x];
    }void mg(int u,int v,int c){
        int ou = fd(u),ov = fd(v);
        if(siz[ou] < siz[ov]) swap(ou,ov),swap(u,v);
        if(ou == ov){
            if(w[u] ^ w[v] ^ c) fg = 1;
            return ;
        }siz[ou] += siz[ov];
        if(tg[ou] == -1) as --;
        if(tg[ov] == -1) as --;
        fa[ov] = ou,w[ov] = w[v] ^ w[u] ^ c;
        if(tg[ov] >= 0) tg[ov] ^= w[ov];
        tg[ou] = (tg[ou] == -2 ? -2 : (tg[ou] == -1 ? tg[ov] : (tg[ov] == -1 || tg[ou] == tg[ov] ? tg[ou] : -2)));
        if(tg[ou] == -1) as ++;
        if(tg[ou] == -2) fg = 1;
    }void init(){
        as = 0,fg = 0;
        for(int i = 1;i <= n + m;i ++) fa[i] = i,siz[i] = 1,w[i] = 0,tg[i] = -1;
        for(int i = 2;i <= n + m;i ++) if(i != m + 1) as ++;
    }void tt(int x,int y){
        int ox = fd(x); y ^= w[x];
        if(tg[ox] == -1) as --;
        tg[ox] = (tg[ox] == -2 ? -2 : (tg[ox] == -1 ? y : (y == -1 || y == tg[ox] ? tg[ox] : -2)));
        if(tg[ox] == -2) fg = 1;
    }
}d;
inline int calc(int x,int y,int z){
    int sm = 0;
    if(fg || d.fg) return 0;
    if(x == 1 && y == 1) res --;
    else if(y == 1) d.tt(m + x,z);
    else if(x == 1) d.tt(y,z);
    else d.mg(x + m,y,z);
    if(fg || d.fg) return 0;
    return pw[d.as + res];
}void los(){
    cin >> n >> m >> k >> q;  int as = 1; pw[0] = 1;
    for(int i = 1;i <= n + m;i ++) pw[i] = (1ll << 30) * pw[i - 1] % mod;
    d.init(),fg = 0,res = 1;
    for(int i = 1;i < n + m;i ++) as = 1ll * as * (1ll << 30) % mod;
    for(int i = 1;i <= k;i ++) if(cin >> x >> y >> z,as) as = calc(x,y,z);
    cout << as << "\n";
    for(int i = 1;i <= q;i ++) cin >> x >> y >> z,as = as ? calc(x,y,z) : 0,cout << as << "\n";
}int main(){
   ios::sync_with_stdio(0),cin.tie(0);
    for(cin >> T;T --;) los();
}

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