**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun
time_limit_ms: 2000
memory_limit_mb: 256
Description: Counting is Fun!

— satyam343

Given two integers $n$ and $x$, find the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Note that order matters (e.g. ($1, 1, 2$) and ($1, 2, 1$) are treated as different) and $a$, $b$, $c$ must be strictly greater than $0$.

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

Each test case contains two integers $n$ and $x$ ($1 \leq n,x \leq 10^6$).

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

Output: Output a single integer — the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Examples:
input:
4
7 4
10 5
7 1000
900000 400000
output:
4
10
7
1768016938

Note: In the first test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 2, 1$), and ($2, 1, 1$).

In the second test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 1, 3$), ($1, 2, 1$), ($1, 2, 2$), ($1, 3, 1$), ($2, 1, 1$), ($2, 1, 2$), ($2, 2, 1$), and ($3, 1, 1$).

Buggy Code:
### ATG
# for _ in range(int(input())):
#     n,m = map(int,input().split())
#     a = list(map(int,input().split()))
#     n=int(input())
#     a=input()

def los(n):
    a = input()
    return list(a[:n])

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

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

def muls():
    return input().split()

def liststring(a,n):
    return list(map(int,a[:n]))

def bina(a):
    return bin(a)[2:]

# from math import log2
# from collections import defaultdict
# import sys
# input = sys.stdin.readline
# from collections import deque
#dequeue : append, appendleft, pop, popleft
# from queue import Queue
#Queue : put, get
# mo = 10**9+7

# loi : list of integer
# los : list of charachter
# muli : multiple integer input
# muls : multiple string input
# for _ in range(int(input())):

# import math
for _ in range(int(input())):

    n,x = muli()
    ans = 0
    if x>n:
        print(n)
    else:

        for a in range(1,min(n,x)):
            b = 1
            while True:
                # print(a,b)
                if a*b<n and a+b<x:
                    ans += int(min(x-a-b,(n-a*b)/(a+b)))
                else:
                    break
                b += 1

        print(ans)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun
time_limit_ms: 2000
memory_limit_mb: 256
Description: Counting is Fun!

— satyam343

Given two integers $n$ and $x$, find the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Note that order matters (e.g. ($1, 1, 2$) and ($1, 2, 1$) are treated as different) and $a$, $b$, $c$ must be strictly greater than $0$.

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

Each test case contains two integers $n$ and $x$ ($1 \leq n,x \leq 10^6$).

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

Output: Output a single integer — the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Examples:
input:
4
7 4
10 5
7 1000
900000 400000
output:
4
10
7
1768016938

Note: In the first test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 2, 1$), and ($2, 1, 1$).

In the second test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 1, 3$), ($1, 2, 1$), ($1, 2, 2$), ($1, 3, 1$), ($2, 1, 1$), ($2, 1, 2$), ($2, 2, 1$), and ($3, 1, 1$).

Buggy Code:
import java.util.*;
    import java.io.*;
        public class Main {
            static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
            static StringTokenizer st;
            public static void main(String[] args) throws IOException {
                int t = readInt();
                for(int ab = 0; ab < t; ab++){
                    int n = readInt();
                    int x = readInt();
                    long ans = 0;
                    for(int i = 1; i <= 1000; i++) {
                        for(int j = i; j <= 1000; j++) {
                            for(int k = j; k <= 1000000; k++){
                                if(i+j+k <= x && i*j + i*k + j*k <= n){
                                    if(i == j && j == k)ans+=1;
                                    if(i == j && j != k)ans+=3;
                                    if(i != j && j == k)ans+=3;
                                    if(i != j && i == k)ans+=3;
                                    if(i != j && j != k)ans+=6;
                                } else break;
                            }
                        }
                    }
                    System.out.println(ans);
                }
            }
            static String next() throws IOException {
                while (st == null || !st.hasMoreTokens()) {
                    st = new StringTokenizer(br.readLine().trim());
                }
                return st.nextToken();
            }
            static int readInt() throws IOException {
                return Integer.parseInt(next());
            }
            static long readLong () throws IOException {
                return Long.parseLong(next());
            }
            static double readDouble () throws IOException {
                return Double.parseDouble(next());
            }
            static String readLine() throws IOException {
                return br.readLine().trim();
            }
        }

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The buggy code has a few key issues:

1. **Limited Range**: The code limits the values of `i`, `j`, and `k` to a maximum of 1000, which can miss valid triplets for larger values of `n` and `x`.
2. **Incorrect Triplet Counting**: The code incorrectly handles the counting of permutations of triplets, especially when `i`, `j`, and `k` are not all distinct.

To expose these issues, we can create a test case where the values of `n` and `x` are large enough to require triplets with values greater than 1000, and where the triplets have a mix of distinct and non-distinct values.

Let's create a test input with a single test case where `n` and `x` are large enough to ensure that the code's limitations are evident.

```plaintext
1
1000000 1000000
```

This test case will force the code to consider triplets with values greater than 1000, which the current implementation will not handle correctly. Additionally, the large values of `n` and `x` will ensure that there are many valid triplets, making it more likely to expose the incorrect counting of permutations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun
time_limit_ms: 2000
memory_limit_mb: 256
Description: Counting is Fun!

— satyam343

Given two integers $n$ and $x$, find the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Note that order matters (e.g. ($1, 1, 2$) and ($1, 2, 1$) are treated as different) and $a$, $b$, $c$ must be strictly greater than $0$.

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

Each test case contains two integers $n$ and $x$ ($1 \leq n,x \leq 10^6$).

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

Output: Output a single integer — the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Examples:
input:
4
7 4
10 5
7 1000
900000 400000
output:
4
10
7
1768016938

Note: In the first test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 2, 1$), and ($2, 1, 1$).

In the second test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 1, 3$), ($1, 2, 1$), ($1, 2, 2$), ($1, 3, 1$), ($2, 1, 1$), ($2, 1, 2$), ($2, 2, 1$), and ($3, 1, 1$).

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

public class Main4 {

    public static void solve() throws IOException{
        int n = in.nextInt();
        int x = in.nextInt();
        long ans = 0;
        for (int i = 1; i <= x; i++) {
            if (3 * i > x || (long) i * i + (long) i * i + (long) i * i > n) break;
            ans++;
            for (int j = i + 1; j <= x; j++) {
                if (i + i + j > x || (long) i * i + (long) i * j + (long) i * j > n) break;
                ans += 3;
                if (i + j + j > x || (long) i * j + (long) i * j + (long) j * j > n) continue;
                ans += 3;
                int l = j + 1, r = x;
                int p = -1;
                while (l <= r) {
                    int mid = l + r >> 1;
                    if ((long) i * j + (long) i * mid + (long) j * mid <= n) {
                        p = mid;
                        l = mid + 1;
                    }else {
                        r = mid - 1;
                    }
                }
                if (p == -1) continue;
                ans += 6L * (p - j);
            }
        }
        out.println(ans);
    }

    static boolean MULTI_CASE = true;

    public static void main(String[] args) throws IOException {
        if (MULTI_CASE) {
            int T = in.nextInt();
            for (int i = 0; i < T; ++i) {
                solve();
            }
        } else {
            solve();
        }
        out.close();
    }

    static InputReader in = new InputReader();
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

    static class InputReader {
        private StringTokenizer st;
        private BufferedReader bf;

        public InputReader() {
            bf = new BufferedReader(new InputStreamReader(System.in));
            st = null;
        }

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

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

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

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

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

/*
 */

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun
time_limit_ms: 2000
memory_limit_mb: 256
Description: Counting is Fun!

— satyam343

Given two integers $n$ and $x$, find the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Note that order matters (e.g. ($1, 1, 2$) and ($1, 2, 1$) are treated as different) and $a$, $b$, $c$ must be strictly greater than $0$.

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

Each test case contains two integers $n$ and $x$ ($1 \leq n,x \leq 10^6$).

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

Output: Output a single integer — the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Examples:
input:
4
7 4
10 5
7 1000
900000 400000
output:
4
10
7
1768016938

Note: In the first test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 2, 1$), and ($2, 1, 1$).

In the second test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 1, 3$), ($1, 2, 1$), ($1, 2, 2$), ($1, 3, 1$), ($2, 1, 1$), ($2, 1, 2$), ($2, 2, 1$), and ($3, 1, 1$).

Buggy Code:
#include <algorithm>
#include <cassert>
#include <cctype>
#include <cmath>
#include <cstring>
#include <iostream>
#include <vector>

#if __cplusplus >= 202002l
    #include <format>
#endif

using ull = unsigned long long;
using ll = long long;
using pii = std::pair<int, int>;
using std::cin;

// clang-format off
ll read()
{
    ll x=0,f=1;char ch=cin.get();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=cin.get();}
    while(isdigit(ch)){x=x*10+ch-'0';ch=cin.get();}
    return x*f;
}
template<typename T>
void read(std::vector<T> &vr,int n){for(int i=1;i<=n;++i)vr.push_back(read());}
template<typename T>
void read(std::vector<T> &vr,int a,int b){for(int i=a;i<=b;++i)vr[i]=read();}

// clang-format on
ll mysqrt(ll x)
{
    ll l = 0, r = std::sqrt(x) + 100;
    while (l + 1 != r)
    {
        ll mid = (l + r) / 2;
        if (mid * mid <= x)
            l = mid;
        else
            r = mid;
    }
    return l;
}
void solve()
{
    ll n = read(), x = read();

    ll ans1 = std::min(x / 3, mysqrt(n / 3));
    ll ans2 = 0;
    for (ll a = 1; a <= std::min(n, x) / 2; ++a)
    {
        ll b = std::min((n - a * a) / (2 * a), x - 2 * a);
        if (b <= 0)
            break;
        if (a <= b)
            ans2 = ans2 + (b - 1) * 3;
        else
            ans2 = ans2 + 3 * b;
    }
    ll ans3 = 0;
    for (ll i = 1; i < x / 3; ++i)
        for (ll j = i + 1; j <= (x - i) / 2; ++j)
        {
            ll c = std::min(x - i - j, (n - i * j) / (i + j));
            if (c <= j)
                break;
            ans3 = ans3 + (c - j) * 6;
        }
    std::cout << ans1 + ans2 + ans3 << '\n';
}

int main()
{
#ifdef Fresca
    freopen("in.in", "r", stdin);
    // freopen("out.out", "w", stdout);
#endif
    cin.tie(nullptr)->sync_with_stdio(false);

    int T = read();
    for(int i=1;i<=T;++i)
    {
        if(i!=3726)
            solve();
        else
        {
            int n=read(),x=read();
            std::cout<<std::format("{}-{}\n",n,x);
        }
    }
    return 0;
}

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

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

Think step by step.

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

1. **Edge Cases**: Test cases with the smallest and largest possible values for `n` and `x`.
2. **Boundary Conditions**: Test cases where `n` and `x` are close to their limits.
3. **Special Cases**: Test cases where the constraints `ab + ac + bc ≤ n` and `a + b + c ≤ x` are tight.
4. **Symmetry and Order**: Test cases with values that highlight the importance of order in the triplets.

Given the constraints:
- \(1 \leq t \leq 10^4\)
- \(1 \leq n, x \leq 10^6\)

We will create a small-scale test input with a few test cases that cover these aspects.

### Test Input
```plaintext
4
1 1
1000000 1000000
3 3
10 6
```

### Explanation
1. **Test Case 1: \(1 1\)**
   - This is the smallest possible value for both `n` and `x`.
   - The only possible triplet is \((1, 1, 1)\), but it does not satisfy \(ab + ac + bc \le n\) since \(1*1 + 1*1 + 1*1 = 3 > 1\).

2. **Test Case 2: \(1000000 1000000\)**
   - This is the largest possible value for both `n` and `x`.
   - This test case will check the performance and correctness of the algorithm for large inputs.

3. **Test Case 3: \(3 3\)**
   - This test case has both `n` and `x` equal to 3.
   - The possible triplets are \((1, 1, 1)\), which satisfies both conditions.

4. **Test Case 4: \(10 6\)**
   - This test case has `n` and `x` such that the constraints are tight.
   - The possible triplets are \((1, 1, 1)\), \((1, 1, 2)\), \((1, 2, 1)\), \((2, 1, 1)\), \((1, 1, 3)\), \((1, 3, 1)\), \((3, 1, 1)\), \((1, 2, 2)\), \((2, 1, 2)\), \((2, 2, 1)\).

This test input should help expose any bugs in the buggy code, especially those related to edge cases and boundary conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun
time_limit_ms: 2000
memory_limit_mb: 256
Description: Counting is Fun!

— satyam343

Given two integers $n$ and $x$, find the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Note that order matters (e.g. ($1, 1, 2$) and ($1, 2, 1$) are treated as different) and $a$, $b$, $c$ must be strictly greater than $0$.

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

Each test case contains two integers $n$ and $x$ ($1 \leq n,x \leq 10^6$).

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

Output: Output a single integer — the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Examples:
input:
4
7 4
10 5
7 1000
900000 400000
output:
4
10
7
1768016938

Note: In the first test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 2, 1$), and ($2, 1, 1$).

In the second test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 1, 3$), ($1, 2, 1$), ($1, 2, 2$), ($1, 3, 1$), ($2, 1, 1$), ($2, 1, 2$), ($2, 2, 1$), and ($3, 1, 1$).

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int mod = 1e9 + 7, N = 1e3 + 5;

main(){
    int tt; cin >> tt;
    while(tt--){
        int n, x;
        cin >> n >> x;
        int ans = 0;
        for(int a = 1; a <= 1e6; a++){
            for(int b = 1; b <= 1e6; b++){
                if(a * b > n || a + b > x) break;
                ans += max(0LL, min((n - (a * b)) / (a + b), x - a - b));
            }
        }
        cout << ans << '\n';
    }
}

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

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

Think step by step.

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

1. **Edge Cases for Small Values:**
   - Test with the smallest possible values for \( n \) and \( x \) to see if the code handles them correctly.
   - Example: \( n = 1 \) and \( x = 1 \).

2. **Edge Cases for Large Values:**
   - Test with the largest possible values for \( n \) and \( x \) to see if the code handles large inputs efficiently and correctly.
   - Example: \( n = 10^6 \) and \( x = 10^6 \).

3. **Critical Values:**
   - Test with values that are just below or just above the thresholds to ensure the code transitions smoothly.
   - Example: \( n = 10^6 - 1 \) and \( x = 10^6 - 1 \).

4. **Symmetric and Asymmetric Triplets:**
   - Test with values that create symmetric and asymmetric triplets to ensure the code counts all valid triplets.
   - Example: \( n = 10 \) and \( x = 5 \).

5. **Multiple Test Cases:**
   - Include multiple test cases to ensure the code handles different scenarios.
   - Example: A mix of small, large, and critical values.

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

```plaintext
5
1 1
7 4
10 5
999999 999999
1000000 1000000
```

This test input includes:
- A test case with the smallest possible values.
- Test cases from the provided examples.
- Test cases with large values close to the upper limits.
- Test cases with values that are just below the upper limits.

This should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun
time_limit_ms: 2000
memory_limit_mb: 256
Description: Counting is Fun!

— satyam343

Given two integers $n$ and $x$, find the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Note that order matters (e.g. ($1, 1, 2$) and ($1, 2, 1$) are treated as different) and $a$, $b$, $c$ must be strictly greater than $0$.

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

Each test case contains two integers $n$ and $x$ ($1 \leq n,x \leq 10^6$).

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

Output: Output a single integer — the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Examples:
input:
4
7 4
10 5
7 1000
900000 400000
output:
4
10
7
1768016938

Note: In the first test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 2, 1$), and ($2, 1, 1$).

In the second test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 1, 3$), ($1, 2, 1$), ($1, 2, 2$), ($1, 3, 1$), ($2, 1, 1$), ($2, 1, 2$), ($2, 2, 1$), and ($3, 1, 1$).

Buggy Code:
### ATG
# for _ in range(int(input())):
#     n,m = map(int,input().split())
#     a = list(map(int,input().split()))
#     n=int(input())
#     a=input()

def los(n):
    a = input()
    return list(a[:n])

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

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

def muls():
    return input().split()

def liststring(a,n):
    return list(map(int,a[:n]))

def bina(a):
    return bin(a)[2:]

# from math import log2
# from collections import defaultdict
# import sys
# input = sys.stdin.readline
# from collections import deque
#dequeue : append, appendleft, pop, popleft
# from queue import Queue
#Queue : put, get
# mo = 10**9+7

# loi : list of integer
# los : list of charachter
# muli : multiple integer input
# muls : multiple string input
# for _ in range(int(input())):

# import math
for _ in range(int(input())):

    n,x = muli()
    ans = 0
    if x>n:
        if n==1:
            print(0)
        else:
            print(n)
    else:

        for a in range(1,min(n,x)):
            b = 1
            while True:
                # print(a,b)
                if a*b<n and a+b<x:
                    ans += int(min(x-a-b,(n-a*b)/(a+b)))
                else:
                    break
                b += 1

        print(ans)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun
time_limit_ms: 2000
memory_limit_mb: 256
Description: Counting is Fun!

— satyam343

Given two integers $n$ and $x$, find the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Note that order matters (e.g. ($1, 1, 2$) and ($1, 2, 1$) are treated as different) and $a$, $b$, $c$ must be strictly greater than $0$.

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

Each test case contains two integers $n$ and $x$ ($1 \leq n,x \leq 10^6$).

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

Output: Output a single integer — the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Examples:
input:
4
7 4
10 5
7 1000
900000 400000
output:
4
10
7
1768016938

Note: In the first test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 2, 1$), and ($2, 1, 1$).

In the second test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 1, 3$), ($1, 2, 1$), ($1, 2, 2$), ($1, 3, 1$), ($2, 1, 1$), ($2, 1, 2$), ($2, 2, 1$), and ($3, 1, 1$).

Buggy Code:
from math import isqrt
import sys
input = sys.stdin.readline
MOD = 10**9 + 7

def read_array(factory):
    return [factory(num) for num in input().strip().split()]
def print_array(arr):
    print(" ".join(map(str, arr)))

def solve(n, x):
    cnt = 0
    # a = b = c
    for abc in range(3, x+1, 3):
        a = abc//3
        b = a
        c = a
        if a*b + a*c + b*c > n:
            break
        cnt += 1
    # a = b != c
    # aa + 2ac <= n  &&  2a+c <= x
    # c <= (n-aa)/2a  &&  c <= x-2a
    for a in range(1, isqrt(n)+2):
        max_c = max(0, min(x-2*a, (n-a*a)//(2*a)))
        if a <= max_c:
            cnt += 3 * (max_c - 1)
        else:
            cnt += 3 * max_c

    # a < b < c
    for a in range(1, isqrt(n)+2):
        # b <= x - a - b  &&  ab+ac+bc <= n
        # b <= (x - a)/2  &&  ab+ab+ab <= n
        for b in range(a+1, min((x-a)//2, n//(3*a))):
            # c <= x - a - b  &&  c <= (n - ab)/(a+b)
            max_c = max(0, min(x-a-b, (n-a*b)//(a+b)))
            if max_c < b:
                break
            cnt += 6 * max(0, max_c - b)
    return cnt

if __name__ == "__main__":
    # Read in input:
    t = int(input())

    # Call solution:
    for _ in range(t):
        n, x = read_array(int)
        ans = solve(n, x)
        print(ans)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun
time_limit_ms: 2000
memory_limit_mb: 256
Description: Counting is Fun!

— satyam343

Given two integers $n$ and $x$, find the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Note that order matters (e.g. ($1, 1, 2$) and ($1, 2, 1$) are treated as different) and $a$, $b$, $c$ must be strictly greater than $0$.

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

Each test case contains two integers $n$ and $x$ ($1 \leq n,x \leq 10^6$).

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

Output: Output a single integer — the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Examples:
input:
4
7 4
10 5
7 1000
900000 400000
output:
4
10
7
1768016938

Note: In the first test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 2, 1$), and ($2, 1, 1$).

In the second test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 1, 3$), ($1, 2, 1$), ($1, 2, 2$), ($1, 3, 1$), ($2, 1, 1$), ($2, 1, 2$), ($2, 2, 1$), and ($3, 1, 1$).

Buggy Code:
//Modulus properties
 //1 (a+b)%y = (a%y+b%y)%y also
// <if to find pairs st (a+b)%y==0 one is (a%y) and other is b%y==((y-(a%y))%y)>
 // < if (a+b)%y == (c+b)%y  then a%y = c%y (in  terms of remainder) >
// 2 (a-b)%y = (a%y-b%y+y)%y also
//   < if (a-b)%y ==0 then a%y = b%y  >
import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.StringTokenizer;
import java.util.LinkedList;
import java.util.Arrays;
import java.util.*;

public class one {

    static final int INF = Integer.MAX_VALUE;
    static final long MOD = 1000000007;

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

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

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

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

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

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

    static class Graph {
        int V;
        LinkedList<Integer>[] adjList;

        @SuppressWarnings("unchecked")
        Graph(int V) {
            this.V = V;
            adjList = new LinkedList[V];
            for (int i = 0; i < V; i++)
                adjList[i] = new LinkedList<>();
        }

        void addEdge(int u, int v) {
            adjList[u].add(v);
        }

        // Add more graph algorithms as needed
    }

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

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

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

        void union(int x, int y) {
            int xRoot = find(x);
            int yRoot = find(y);
            if (xRoot == yRoot)
                return;
            if (rank[xRoot] < rank[yRoot])
                parent[xRoot] = yRoot;
            else if (rank[xRoot] > rank[yRoot])
                parent[yRoot] = xRoot;
            else {
                parent[yRoot] = xRoot;
                rank[xRoot]++;
            }
        }
    }

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

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

    static long power(long x, long y, long p) {
        long res = 1;
        x = x % p;

        while (y > 0) {
            if (y % 2 == 1)
                res = (res * x) % p;
            y = y >> 1;
            x = (x * x) % p;
        }
        return res;
    }

    static boolean[] sieveOfEratosthenes(int n) {
        boolean[] prime = new boolean[n + 1];
        Arrays.fill(prime, true);
        prime[0] = prime[1] = false;
        for (int p = 2; p * p <= n; p++) {
            if (prime[p]) {
                for (int i = p * p; i <= n; i += p)
                    prime[i] = false;
            }
        }
        return prime;
    }

    public static void main(String[] args) {
        FastScanner sc = new FastScanner(System.in);
        int t = sc.nextInt();
        while (t > 0) {
            long n  = sc.nextLong();
            long k = sc.nextLong();
            long c = 0;
             for (long i = 1; i<=1000000; i++) {
                long s = 1;
                long e = 1000000;
                long ans  = -1;
                while(s<=e){
                    long mid = (s+e)/2;
                    long a = i;
                    long b = mid;
                    long pp = ((n-(a*b))/(a+b));
                    if((k-(a+b))>0&&pp>0){
                        ans  = mid;
                        s = mid+1;
                    }
                    else{
                        e = mid-1;
                    }
                }
                if(ans!=-1){
                    long a = i;
                    long b = ans;
                    for (int j = 1; j <= b; j++) {
                    long pp = ((n-(a*j))/(a+j));
                    long min =  Math.min(k-(a+j),pp);
                    c+= (min);
                    }
                    // System.out.println(a+" "+b+" "+min);

                }
             }
            System.out.println(c);

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 buggy code, we need to consider edge cases and scenarios that are likely to cause issues. The code seems to have a nested loop structure and uses binary search to find valid triplets. Let's consider the following scenarios:

1. **Small values of \( n \) and \( x \)**: These can help us verify the basic logic of the code.
2. **Large values of \( n \) and \( x \)**: These can test the efficiency and correctness of the code, especially the binary search and the loop conditions.
3. **Boundary values**: Values at the limits of the input constraints can help identify off-by-one errors or other boundary conditions.
4. **Values where \( n \) and \( x \) are equal**: This can test the code's handling of equal constraints.
5. **Values where \( n \) is much larger than \( x \)**: This can test how the code handles cases where the sum constraint is more restrictive than the product constraint.

Let's generate a small-scale test input that covers some of these scenarios:

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

### Explanation:
1. **Test Case 1: \( n = 1 \), \( x = 1 \)**
   - The only possible triplet is \((1, 1, 1)\), but since \( 1 + 1 + 1 > 1 \), there are no valid triplets.
   - Expected output: \( 0 \)

2. **Test Case 2: \( n = 2 \), \( x = 2 \)**
   - The possible triplets are \((1, 1, 1)\), but since \( 1 + 1 + 1 > 2 \), there are no valid triplets.
   - Expected output: \( 0 \)

3. **Test Case 3: \( n = 3 \), \( x = 3 \)**
   - The possible triplets are \((1, 1, 1)\), but since \( 1 + 1 + 1 = 3 \) and \( 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 = 3 \), there is one valid triplet.
   - Expected output: \( 1 \)

This test input is small and complete, and it should help expose any issues in the buggy code, especially in handling boundary conditions and small input values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun
time_limit_ms: 2000
memory_limit_mb: 256
Description: Counting is Fun!

— satyam343

Given two integers $n$ and $x$, find the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Note that order matters (e.g. ($1, 1, 2$) and ($1, 2, 1$) are treated as different) and $a$, $b$, $c$ must be strictly greater than $0$.

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

Each test case contains two integers $n$ and $x$ ($1 \leq n,x \leq 10^6$).

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

Output: Output a single integer — the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Examples:
input:
4
7 4
10 5
7 1000
900000 400000
output:
4
10
7
1768016938

Note: In the first test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 2, 1$), and ($2, 1, 1$).

In the second test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 1, 3$), ($1, 2, 1$), ($1, 2, 2$), ($1, 3, 1$), ($2, 1, 1$), ($2, 1, 2$), ($2, 2, 1$), and ($3, 1, 1$).

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

public class Main {
    static RealFastReader sc = new RealFastReader(System.in);
    static PrintWriter out = new PrintWriter(System.out);
    static long dp[][][];
    public static void main(String[] args) {
        //int n=sc.ni();
        //double d=sc.nd();

        int T = sc.ni();
        while (T-- > 0) solve();

//        solve();
        out.close();
    }

    public static void solve() {
        int n=sc.ni();
        int x=sc.ni();
        long ans=0;
        for(int a=1;a<=1000000;a++){
            for(int b=1;b<=Math.min(x-a,n/a);b++){
                int t=(n-a*b)/(a+b);
                t=Math.min(x-a-b,t);
                ans+=t;
            }
        }
        System.out.println(ans);

    }

    //方法.快读快写---------------------------------------------------------------------------------
    public static class RealFastReader {
        InputStream is;

        public RealFastReader(final InputStream is) {
            this.is = is;
        }

        private byte[] inbuf = new byte[8192];
        public int lenbuf = 0, ptrbuf = 0;

        public int readByte() {
            if (lenbuf == -1) {
                throw new InputMismatchException();
            }
            if (ptrbuf >= lenbuf) {
                ptrbuf = 0;
                try {
                    lenbuf = is.read(inbuf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (lenbuf <= 0) {
                    return -1;
                }
            }
            return inbuf[ptrbuf++];
        }

        private boolean isSpaceChar(int c) {
            return !(c >= 33 && c <= 126);
        }

        private int skip() {
            int b;
            while ((b = readByte()) != -1 && isSpaceChar(b)) ;
            return b;
        }

        public double nd() {
            return Double.parseDouble(ns());
        }

        public char nc() {
            return (char) skip();
        }

        public String ns() {
            int b = skip();
            StringBuilder sb = new StringBuilder();
            while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ')
                sb.appendCodePoint(b);
                b = readByte();
            }
            return sb.toString();
        }

        public char[] ns(int n) {
            char[] buf = new char[n];
            int b = skip(), p = 0;
            while (p < n && !(isSpaceChar(b))) {
                buf[p++] = (char) b;
                b = readByte();
            }
            return n == p ? buf : Arrays.copyOf(buf, p);
        }

        public int[] na(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        }

        public long[] nal(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = nl();
            }
            return a;
        }

        public char[][] nm(int n, int m) {
            char[][] map = new char[n][];
            for (int i = 0; i < n; i++) {
                map[i] = ns(m);
            }
            return map;
        }

        public int[][] nmi(int n, int m) {
            int[][] map = new int[n][];
            for (int i = 0; i < n; i++) {
                map[i] = na(m);
            }
            return map;
        }

        public int ni() {
            int num = 0;
            int b;
            boolean minus = false;
            while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
                ;
            if (b == '-') {
                minus = true;
                b = readByte();
            }

            while (true) {
                if (b >= '0' && b <= '9') {
                    num = num * 10 + (b - '0');
                } else {
                    return minus ? -num : num;
                }
                b = readByte();
            }
        }

        public long nl() {
            long num = 0;
            int b;
            boolean minus = false;
            while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
                ;
            if (b == '-') {
                minus = true;
                b = readByte();
            }

            while (true) {
                if (b >= '0' && b <= '9') {
                    num = num * 10 + (b - '0');
                } else {
                    return minus ? -num : num;
                }
                b = readByte();
            }
        }
    }
}

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

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

Think step by step.

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

1. **Small Test Cases**: Start with small values of `n` and `x` to ensure the test input is complete and valid.
2. **Edge Cases**: Consider edge cases where the values of `a`, `b`, and `c` are at the boundaries.
3. **Boundary Conditions**: Test with values that are just below, at, and just above the limits.
4. **Symmetry and Order**: Ensure that the test input includes cases where the order of `a`, `b`, and `c` matters.

Given the constraints \(1 \leq n, x \leq 10^6\), let's create a small-scale test input that includes edge cases and boundary conditions.

### Test Input
```plaintext
4
1 1
10 10
100 100
1000 1000
```

### Explanation
1. **Test Case 1: (1, 1)**
   - This is the smallest possible input. The only valid triplet is (1, 1, 1).
   - Expected output: 1

2. **Test Case 2: (10, 10)**
   - This is a small input that allows for more triplets.
   - Expected output: 10 (as given in the example)

3. **Test Case 3: (100, 100)**
   - This is a larger input that still keeps the values manageable.
   - Expected output: 7 (as given in the example)

4. **Test Case 4: (1000, 1000)**
   - This is a larger input that is still within the constraints.
   - Expected output: 7 (as given in the example)

### Challenging Aspects
- **Edge Case**: The first test case (1, 1) checks the smallest possible values.
- **Symmetry and Order**: The second and third test cases include values that allow for multiple valid triplets, testing the order of `a`, `b`, and `c`.
- **Boundary Conditions**: The fourth test case (1000, 1000) is large enough to test the limits of the algorithm without exceeding the constraints.

This test input should help expose any bugs in the buggy code by covering a range of scenarios from small to large inputs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun
time_limit_ms: 2000
memory_limit_mb: 256
Description: Counting is Fun!

— satyam343

Given two integers $n$ and $x$, find the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Note that order matters (e.g. ($1, 1, 2$) and ($1, 2, 1$) are treated as different) and $a$, $b$, $c$ must be strictly greater than $0$.

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

Each test case contains two integers $n$ and $x$ ($1 \leq n,x \leq 10^6$).

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

Output: Output a single integer — the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Examples:
input:
4
7 4
10 5
7 1000
900000 400000
output:
4
10
7
1768016938

Note: In the first test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 2, 1$), and ($2, 1, 1$).

In the second test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 1, 3$), ($1, 2, 1$), ($1, 2, 2$), ($1, 3, 1$), ($2, 1, 1$), ($2, 1, 2$), ($2, 2, 1$), and ($3, 1, 1$).

Buggy Code:
T = int(input())
ANS = []

for t in range(T):
    N, X = map(int, input().split())
    # a <= b <= c
    ans = 0
    for a in range(1, 10**6+1):
        for b in range(a, N//a+2):
            if a == b:
                coef = 1
            else:
                coef = 2
            if a*b >= N:
                break
            x = (N - a * b) // (a + b)
            y = X - a - b
            cntc = max(0, min(x, y))
            ans += cntc * coef
    ANS.append(ans)

for i in range(T):
    print(ANS[i])

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

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

Think step by step.

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

1. **Small Values of \( n \) and \( x \)**: These can help verify the correctness of the base cases.
2. **Large Values of \( n \) and \( x \)**: These can test the performance and correctness for larger inputs.
3. **Equal Values of \( n \) and \( x \)**: These can test the boundary conditions.
4. **Specific Values that Cause Breaks in Loops**: Values that might cause the loop to break early or skip important calculations.

Given the constraints, we will generate a small-scale test input with a few test cases that cover these scenarios.

```plaintext
4
1 1
2 2
1000 1000
1000000 1000000
```

### Explanation:
1. **Test Case 1: \( n = 1 \) and \( x = 1 \)**
   - There are no valid triplets \((a, b, c)\) because \(a, b, c\) must be positive integers and \(a + b + c \le 1\) is impossible.
   - Expected output: 0

2. **Test Case 2: \( n = 2 \) and \( x = 2 \)**
   - The only valid triplet is \((1, 1, 1)\).
   - Expected output: 1

3. **Test Case 3: \( n = 1000 \) and \( x = 1000 \)**
   - This tests a larger value where \(a + b + c \le 1000\) and \(ab + ac + bc \le 1000\).
   - Expected output: A large number (but less than the maximum possible for \(10^6\))

4. **Test Case 4: \( n = 1000000 \) and \( x = 1000000 \)**
   - This tests the maximum values within the constraints.
   - Expected output: A very large number

These test cases should help expose any issues in the buggy code, especially with the loop breaks and the calculations for larger values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun
time_limit_ms: 2000
memory_limit_mb: 256
Description: Counting is Fun!

— satyam343

Given two integers $n$ and $x$, find the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Note that order matters (e.g. ($1, 1, 2$) and ($1, 2, 1$) are treated as different) and $a$, $b$, $c$ must be strictly greater than $0$.

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

Each test case contains two integers $n$ and $x$ ($1 \leq n,x \leq 10^6$).

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

Output: Output a single integer — the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Examples:
input:
4
7 4
10 5
7 1000
900000 400000
output:
4
10
7
1768016938

Note: In the first test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 2, 1$), and ($2, 1, 1$).

In the second test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 1, 3$), ($1, 2, 1$), ($1, 2, 2$), ($1, 3, 1$), ($2, 1, 1$), ($2, 1, 2$), ($2, 2, 1$), and ($3, 1, 1$).

Buggy Code:
from bisect import bisect_left, bisect_right
from heapq import heappush, heappop, heapify
from math import *
from collections import defaultdict, Counter, deque
from functools import lru_cache
from sys import stdin, stdout
import random
import io, os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
RANDOM = random.randrange(2**62)
ninput = lambda: int(input())
sinput = lambda: input().decode().strip()
ainput = lambda: list(map(int, input().split()))
print = lambda *x: stdout.write(" ".join(map(str, x)) + '\n')
ceil = lambda a, b: a // b if (a >= 0) ^ (b > 0) else (abs(a) + abs(b) - 1) // abs(b)
wr = lambda x: x ^ RANDOM

def main():
    for _ in range(ninput()):
        n, x = ainput()
        ans = 0
        for a in range(1, 1000000):
            for sm in range(2, 1000000):
                if a * sm > n-1 or a + sm > x: break
                l, r = 1, sm // 2
                A = -1
                while l <= r:
                    mid = (l+r) // 2
                    if mid * (sm - mid) <= (n - a * sm): l = mid + 1; A = mid
                    else: r = mid - 1
                if A != -1: ans += 2 * A - int(A == sm - A)
        print(ans)

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 buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are some considerations:

1. **Small Values of \( n \) and \( x \)**: Test cases with small values can help identify issues with the loop limits and conditions.
2. **Large Values of \( n \) and \( x \)**: Test cases with large values can help identify performance issues and incorrect logic.
3. **Boundary Conditions**: Test cases where \( n \) and \( x \) are at the boundary of the allowed range.
4. **Special Cases**: Test cases where \( n \) and \( x \) have specific relationships, such as \( n = x \) or \( n \) being a multiple of \( x \).

Given the constraints and the nature of the problem, a small-scale test input that can expose the bug might look like this:

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

Let's break down why this test input is challenging:

1. **Test Case 1: \( n = 1 \), \( x = 1 \)**:
   - The only possible triplet is \((1, 1, 1)\), but it does not satisfy \( ab + ac + bc \le 1 \) and \( a + b + c \le 1 \). The expected output is \(0\).

2. **Test Case 2: \( n = 2 \), \( x = 2 \)**:
   - The only possible triplet is \((1, 1, 1)\), which satisfies both conditions. The expected output is \(1\).

3. **Test Case 3: \( n = 3 \), \( x = 3 \)**:
   - The possible triplets are \((1, 1, 1)\), \((1, 1, 2)\), \((1, 2, 1)\), and \((2, 1, 1)\). The expected output is \(4\).

These test cases are small but cover a range of conditions that can help identify issues with the loop limits and conditions in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun
time_limit_ms: 2000
memory_limit_mb: 256
Description: Counting is Fun!

— satyam343

Given two integers $n$ and $x$, find the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Note that order matters (e.g. ($1, 1, 2$) and ($1, 2, 1$) are treated as different) and $a$, $b$, $c$ must be strictly greater than $0$.

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

Each test case contains two integers $n$ and $x$ ($1 \leq n,x \leq 10^6$).

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

Output: Output a single integer — the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Examples:
input:
4
7 4
10 5
7 1000
900000 400000
output:
4
10
7
1768016938

Note: In the first test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 2, 1$), and ($2, 1, 1$).

In the second test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 1, 3$), ($1, 2, 1$), ($1, 2, 2$), ($1, 3, 1$), ($2, 1, 1$), ($2, 1, 2$), ($2, 2, 1$), and ($3, 1, 1$).

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
using namespace chrono;
#define ll long long
#define V vector<ll>
#define VV vector<V>
#define p push
#define po pop
#define pb push_back
#define ppb pop_back
#define lb lower_bound
#define ub upper_bound
#define CNT(x) __builtin_popcountll(x)
#define MAX(x) *max_element(all(x))
#define MIN(x) *min_element(all(x))
#define SUM(x) accumulate(all(x), 0LL)
#define rep(i, a, b) for (int i = (a); i < (b); i++)
#define brep(i, a, b) for (int i = (a); i > (b); i--)
#define ordered_set tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update>
// order_of_key(x) -> number of elements strictly smaller than x
// find_by_order(k) -> kth element
/* --------------------------------------------------------- UNIVERSAL CONSTANTS --------------------------------------------------- */
const double eps = 1e-9;
const int mod = 1e9 + 7;
const double pi = 3.14159265358979323846264338327950;
const int dx[16] = {1, 0, 0, -1, 1, 1, -1, -1, 2, 2, 1, 1, -1, -1, -2, -2};
const int dy[16] = {0, -1, 1, 0, -1, 1, -1, 1, -1, 1, -2, 2, -2, 2, -1, 1};
#define ff first
#define ld long double
#define ss second
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
void solve()
{
    ll n, x;
    cin >> n >> x;
    ll ans = 0;
    for (ll i = 1; i < min(n, x); i++)
    {
        for (ll j = 1; j <= min(n /(i),x-i); j++)
        {
            if (i * 1LL * j >= n)
            {
                break;
            }
            ll lo = 0, hi = x + 1;
            while (hi - lo > 1)
            {
                ll mid = (hi + lo) / 2;
                ll cnt1 = i + j + mid;
                ll cnt2 = i * 1LL * j + j * 1LL * mid + i * 1LL * mid;
                if (cnt1 > x or cnt2 > n)
                {
                    hi = mid;
                }
                else
                {
                    lo = mid;
                }
            }
            ans += lo;
        }
    }
    cout << ans << endl;
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    ll t;
    t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }

    return 0;
}

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

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

Think step by step.

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

1. **Edge Cases**: Test the boundaries of the input constraints.
2. **Specific Conditions**: Test scenarios where the constraints \( ab + ac + bc \le n \) and \( a + b + c \le x \) are tight.
3. **Symmetry and Order**: Ensure that the order of triplets is considered (e.g., (1, 2, 3) and (3, 2, 1) are different).

Given the constraints \( 1 \leq n, x \leq 10^6 \), we should focus on values that are likely to cause issues in the code. Here are a few scenarios to consider:

- **Small Values**: Test with small values of \( n \) and \( x \) to ensure the basic logic works.
- **Large Values**: Test with values close to the upper limit to check for performance and correctness.
- **Equal Values**: Test with \( n \) and \( x \) being equal to see how the code handles this special case.
- **Boundary Values**: Test with values that are just below or above the constraints to ensure the code handles them correctly.

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

```plaintext
5
1 1
2 2
10 10
1000000 1000000
1000000 1000001
```

Explanation:
- **Test Case 1**: \( n = 1 \), \( x = 1 \). This is the smallest possible input, and the only valid triplet is (1, 1, 1).
- **Test Case 2**: \( n = 2 \), \( x = 2 \). This tests a slightly larger input, and the valid triplets are (1, 1, 1) and (1, 1, 2).
- **Test Case 3**: \( n = 10 \), \( x = 10 \). This tests a larger input where both \( n \) and \( x \) are equal.
- **Test Case 4**: \( n = 1000000 \), \( x = 1000000 \). This tests the upper limit of the constraints.
- **Test Case 5**: \( n = 1000000 \), \( x = 1000001 \). This tests a scenario where \( n \) is at the upper limit and \( x \) is just above the upper limit.

This test input 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: Fun
time_limit_ms: 2000
memory_limit_mb: 256
Description: Counting is Fun!

— satyam343

Given two integers $n$ and $x$, find the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Note that order matters (e.g. ($1, 1, 2$) and ($1, 2, 1$) are treated as different) and $a$, $b$, $c$ must be strictly greater than $0$.

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

Each test case contains two integers $n$ and $x$ ($1 \leq n,x \leq 10^6$).

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

Output: Output a single integer — the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Examples:
input:
4
7 4
10 5
7 1000
900000 400000
output:
4
10
7
1768016938

Note: In the first test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 2, 1$), and ($2, 1, 1$).

In the second test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 1, 3$), ($1, 2, 1$), ($1, 2, 2$), ($1, 3, 1$), ($2, 1, 1$), ($2, 1, 2$), ($2, 2, 1$), and ($3, 1, 1$).

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

using namespace std;
#define AYAALI ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define fo(i, a, b) for(ll i = ll(a); i < ll(b); i++)
#define yes cout<<"YES\n";
#define no cout<<"NO\n";
#define mnEl(vec) *min_element(vec.begin(), vec.end())
#define mxEl(vec) *max_element(vec.begin(), vec.end()
#define ll long long
#define in(v) for (auto & x : v) cin >> x
#define out(v) for (auto & x : v) cout << x << " "
#define pb(x) push_back(x)
#define MP(x, y) make_pair(x,y)
#define fi first
#define se second
#define endl '\n'
#define all(v) v.begin(), v.end()
#define mem(arrr, xx) memset(arrr,xx,sizeof arrr)
const long long INF = LONG_LONG_MAX;
const double EPS = 1e-9;
const ll N = 2005;
const ll MOD = 1e9 + 7;
int dx8[] = {+0, +0, -1, +1, +1, +1, -1, -1};
int dy8[] = {-1, +1, +0, +0, +1, -1, +1, -1};
int dx4[] = {+0, +0, -1, +1};
int dy4[] = {-1, +1, +0, +0};
int chessx[] = {1, 1, -1, -1, 2, 2, -2, -2};
int chessy[] = {2, -2, 2, -2, 1, -1, 1, -1};

//---------------------------------------------
void files() {
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
}

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

void solve() {
    int n, x;
    cin >> n >> x;
    int count = 0;
    for (int a = 1; a <= x; ++a) {
        for (int b = 1; b <= x - a; ++b) {
            if(a*b > n || x - a - b<1)break;
            for (int c = 1; c <= x - a - b; ++c) {
                if(c*b > n || a*c>n)break;
                if (a + b + c <= x && a * b + a * c + b * c <= n) {
                    count++;
                }else{
                    break;
                }
            }

        }
    }
    cout << count << endl;
}

int main() {
    ll t = 1;
    AYAALI;
    cin >> t;
    //files();
    while (t--) {

        solve();
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
10 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun
time_limit_ms: 2000
memory_limit_mb: 256
Description: Counting is Fun!

— satyam343

Given two integers $n$ and $x$, find the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Note that order matters (e.g. ($1, 1, 2$) and ($1, 2, 1$) are treated as different) and $a$, $b$, $c$ must be strictly greater than $0$.

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

Each test case contains two integers $n$ and $x$ ($1 \leq n,x \leq 10^6$).

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

Output: Output a single integer — the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Examples:
input:
4
7 4
10 5
7 1000
900000 400000
output:
4
10
7
1768016938

Note: In the first test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 2, 1$), and ($2, 1, 1$).

In the second test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 1, 3$), ($1, 2, 1$), ($1, 2, 2$), ($1, 3, 1$), ($2, 1, 1$), ($2, 1, 2$), ($2, 2, 1$), and ($3, 1, 1$).

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

public class Main {
//	static Scanner sc = new Scanner(System.in);
	static BufferedReader sc = new BufferedReader(new InputStreamReader(System.in));
	static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
	static String[] ss;

	public static void main(String[] args) throws IOException {
//        pre();
		int T = Integer.parseInt(sc.readLine());
//		int T = 1;
		while (T-- > 0) {
			solve();
		}
		out.flush();
		out.close();
		sc.close();
	}

	private static void pre() {

	}

	//  static int[][] dir = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};//上右下左
//	static int mod = 1000000007;

	private static void solve() throws IOException {
		ss = sc.readLine().split(" ");
		int n = Integer.parseInt(ss[0]);
		int x = Integer.parseInt(ss[1]);
		n = Math.min(n,x*x/3);
		long cnt =0;
		for (int a = 1; a <= n ; a++) {
			for (int b = 1; b <= n  ; b++) {
				if(n-a*b<(a+b)){
					break;
				}
//				cnt+=Math.max(0,(((n-(a*b)))/(a+b)));
				cnt+=Math.min(Math.max(0,x-a-b),((n-((long) a *b)))/(a+b));
			}
		}
		out.println(cnt);
	}

}

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

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

Think step by step.

Response:
```plaintext
1
10 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun
time_limit_ms: 2000
memory_limit_mb: 256
Description: Counting is Fun!

— satyam343

Given two integers $n$ and $x$, find the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Note that order matters (e.g. ($1, 1, 2$) and ($1, 2, 1$) are treated as different) and $a$, $b$, $c$ must be strictly greater than $0$.

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

Each test case contains two integers $n$ and $x$ ($1 \leq n,x \leq 10^6$).

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

Output: Output a single integer — the number of triplets ($a,b,c$) of positive integers such that $ab + ac + bc \le n$ and $a + b + c \le x$.

Examples:
input:
4
7 4
10 5
7 1000
900000 400000
output:
4
10
7
1768016938

Note: In the first test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 2, 1$), and ($2, 1, 1$).

In the second test case, the triplets are ($1, 1, 1$), ($1, 1, 2$), ($1, 1, 3$), ($1, 2, 1$), ($1, 2, 2$), ($1, 3, 1$), ($2, 1, 1$), ($2, 1, 2$), ($2, 2, 1$), and ($3, 1, 1$).

Buggy Code:
import java.util.Scanner;

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

        while (t-- > 0) {
            int n = in.nextInt();
            int k = in.nextInt();
            long count = 0;
            if(n==900000 && k==400000) {
                System.out.println(1768016938);
            }
            else {
            for (int a = 1; a <= k; a++) {
                for (int b = 1; b <= k; b++) {
                    int maxCBySum = k - a - b;
                    if (maxCBySum <= 0) continue;

                    int maxCByProduct = (n - a * b) / (a + b);
                    if (maxCByProduct < 0) maxCByProduct = 0;
                    int maxValidC = Math.min(maxCBySum, maxCByProduct);
                    if (maxValidC >= 1) {
                        count += maxValidC;
                }
            }}

            System.out.println(count);

        }
    }
        in.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
1000000 1000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Polygonal Segments
time_limit_ms: 8000
memory_limit_mb: 512
Description: You are given an array $a$ of size $n$.

A segment $[l, r](1 \le l < r \le n)$ is called a polygonal segment only if the following conditions hold:

  * $(r-l+1) \geq 3$;    * Considering $a_l, a_{l+1}, \ldots, a_r$ as side lengths, these sides can form a polygon with $(r-l+1)$ sides.

Process $q$ queries of two types:

  * "1 l r": find the length of the longest segment among all polygonal segments $[l_0,r_0]$ satisfying $l \le l_0 \le r_0 \le r$. If there is no such polygonal segment, output $-1$ instead;    * "2 i x": assign $a_i := x$.

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

For each test case:

  * The first line of each testcase contains two integers $n$, $q$ ($4 \le n \le 2\cdot 10^5$, $1 \le q \le 10^5$);    * The second line of each testcase contains $n$ integers $a_1,a_2,\ldots, a_n$ ($1 \le a_i \le 10^{12}$);    * The following $q$ lines contain the description of queries. Each line is of one of two types:     * "1 l r" ($1 \le l < r \le n$, $r-l+1\ge 3$);      * "2 i x" ($1 \le i \le n$, $1 \le x \le 10^{12}$).

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

Output: For each query, if there is no suitable segment, output $-1$ in a new line. Otherwise, output the length of the longest segment satisfying the condition above in a new line.

Examples:
input:
2
5 6
3 1 2 2 8
1 1 3
1 1 4
1 1 5
2 1 5
1 1 4
1 1 5
4 10
500000000000 500000000000 1000000000000 500000000000
1 1 3
1 2 4
1 1 4
2 1 499999999999
2 3 999999999999
1 1 3
1 2 4
1 1 4
2 3 1000000000000
1 1 3
output:
-1
4
4
3
5
-1
-1
4
-1
3
4
-1

Note: In the first query of the first test case, there is no polygonal segment under the given condition. For example, considering segment $[1,3]$, you can not form a triangle with side lengths of $a_1=3$, $a_2=1$, and $a_3=2$.

In the second query of the first test case, the longest polygonal segment is $[1,4]$. You can form a quadrilateral with side lengths of $a_1=3$, $a_2=1$, $a_3=2$, and $a_4=2$.

![](CDN_BASE_URL/589caa1d808f5b5cc88c1eb5f51f2eaf) An example of a quadrilateral with side lengths of $3$, $1$, $2$, and $2$.

Buggy Code:
/**
 * https://codeforces.com/contest/1990/submission/273642822
 *
 * (c) 2024 Diego Sogari
 */
#include <bits/stdc++.h>

using namespace std;
using namespace placeholders;
using i64 = int64_t;

#ifdef ONLINE_JUDGE
#define debug
#else
#include "debug.h"
init(__FILE__);
#endif

void println(auto &&...args) { ((cout << args << ' '), ...) << endl; }

template <typename T> struct Num {
  T x;
  Num() { cin >> x; }
  Num(T a) : x(a) {}
  operator T &() { return x; }
  operator T() const { return x; }
};
using Int = Num<int>;
using I64 = Num<i64>;

struct IntNode {
  int left, right, mid; // [left, right)
  set<array<int, 2>> by_l;
  set<array<int, 2>, greater<array<int, 2>>> by_r;
  void insert(int l, int r) {
    by_l.insert({l, r});
    by_r.insert({r, l});
  }
  void erase(int l, int r) {
    by_l.erase({l, r});
    by_r.erase({r, l});
  }
  void visit(int x, auto &&f) {
    x < mid ? visit(x, f, true, by_l) : visit(x, f, false, by_r);
  }
  void visit(int x, auto &&f, bool isleft, auto &set) {
    for (auto it = set.begin(); it != set.end();) {
      auto [l, r] = *it++;
      if ((l > x) ^ isleft) {
        isleft ? f(*this, l, r) : f(*this, r, l);
      } else {
        break;
      }
    }
  }
};

template <typename T> struct IntTree {
  vector<IntNode> nodes;
  IntTree(int n) : nodes(2 * n + 1) { build(0, n - 1); }
  void build(int l, int r, int i = 0) { // [l, r]
    int mid = (l + r) / 2;
    nodes.at(i) = {l, r, mid};
    if (l != r) {
      build(l, mid, 2 * i + 1);
      build(mid + 1, r, 2 * i + 2);
    }
  }
  void insert(int l, int r, int i = 0) { // [l, r]
    if (l <= nodes[i].mid && r >= nodes[i].mid) {
      nodes[i].insert(l, r);
    } else {
      insert(l, r, 2 * i + 1 + (l >= nodes[i].mid));
    }
  }
  void visit(int x, auto &&f, int i = 0) { // all covering point x
    nodes[i].visit(x, f);
    if (nodes[i].left != nodes[i].right) {
      visit(x, f, 2 * i + 1 + (x >= nodes[i].mid));
    }
  }
};

constexpr int lssb(unsigned x) { return countr_zero(x); }
constexpr int mssb(unsigned x) { return 31 - countl_zero(x); }

template <typename T, T unit = T{}> struct SegTree {
  int n;
  vector<T> nodes;
  SegTree(int n) : n(n), nodes(2 * n, unit) {}
  SegTree(int n, bool stable) : SegTree(stable ? 1 << (1 + mssb(n - 1)) : n) {}
  const T &full() const { return nodes[1]; }    // O(1)
  T &operator[](int i) { return nodes[i + n]; } // O(1)
  void update(auto &&f) {                       // O(n)
    for (int i = n - 1; i > 0; i--) {
      nodes[i] = f(nodes[2 * i], nodes[2 * i + 1]);
    }
  }
  void update(int i, auto &&f) { // O(log n)
    for (i = (i + n) / 2; i > 0; i /= 2) {
      nodes[i] = f(nodes[2 * i], nodes[2 * i + 1]);
    }
  }
  T query(int l, int r, auto &&f) const { // O(log n)
    T ans = unit;
    for (l += n, r += n; l <= r; l /= 2, r /= 2) {
      if (l % 2) {
        ans = f(ans, nodes[l++]);
      }
      if (r % 2 == 0) {
        ans = f(ans, nodes[r--]);
      }
    }
    return ans;
  }
};

struct Seg {
  i64 sum, mx;
  int mpos;
  Seg join(const Seg &other) const {
    auto ans = mx < other.mx ? other : *this;
    ans.sum = sum + other.sum;
    return ans;
  }
  bool good() const { return 2 * mx < sum; }
};

auto joinseg = bind(&Seg::join, _1, _2);

struct Query {
  Int type, x;
  I64 y;
};

void solve(int t) {
  Int n, q;
  vector<I64> a(n);
  vector<Query> queries(q);
  SegTree<Seg> segments(n);
  for (int i = 0; i < n; i++) {
    segments[i] = {a[i], a[i], i};
  }
  segments.update(joinseg);
  IntTree<int> intervals(n);
  map<array<int, 2>, int> cache;
  auto prune = [&](auto &node, int l, int r) {
    node.erase(l, r);
    cache.erase({l, r});
  };
  auto query = [&](auto &self, int l, int r) -> int { // O(log n)
    if (r - l < 2) {
      return -1;
    }
    auto [it, ok] = cache.emplace(array<int, 2>{l, r}, 0);
    auto &ans = it->second;
    if (!ok) {
      return ans;
    }
    intervals.insert(l, r);
    auto seg = segments.query(l, r, joinseg);
    if (seg.good()) {
      return ans = r - l + 1;
    }
    auto ansl = self(self, l, seg.mpos - 1);
    auto ansr = self(self, seg.mpos + 1, r);
    return ans = max(ansl, ansr);
  };
  for (auto &[type, x, y] : queries) {
    if (type == 1) {
      println(query(query, x - 1, y - 1));
    } else {
      segments[x - 1] = {y, y, x - 1};
      segments.update(x - 1, joinseg);
      intervals.visit(x - 1, prune);
    }
  }
}

int main() {
  cin.tie(nullptr)->tie(nullptr)->sync_with_stdio(false);
  Int t;
  for (int i = 1; i <= t; i++) {
    solve(i);
  }
}

**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 maximize the failure probability. Here are the steps to create such a test case:

1. **Edge Case with Small Array:**
   - Test with a small array where the conditions for forming a polygon are tight. This can help catch issues with the logic for checking polygonal segments.

2. **Edge Case with Large Values:**
   - Test with large values in the array to ensure the code handles large numbers correctly.

3. **Edge Case with Multiple Queries:**
   - Include a mix of update and query operations to ensure the code can handle dynamic changes in the array.

4. **Edge Case with No Valid Segments:**
   - Create a scenario where no valid polygonal segments exist to check if the code correctly returns -1.

5. **Edge Case with Longest Segment:**
   - Create a scenario where the longest valid polygonal segment is not immediately obvious to ensure the code can find it correctly.

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

```plaintext
1
5 8
3 1 2 2 8
1 1 3
1 1 4
1 1 5
2 1 5
1 1 4
1 1 5
2 3 1
1 1 5
```

Explanation of the test case:
- The array is `3 1 2 2 8`.
- The first query `1 1 3` checks for the longest polygonal segment in the range [1, 3]. There is no valid polygonal segment in this range, so the expected output is `-1`.
- The second query `1 1 4` checks for the longest polygonal segment in the range [1, 4]. The valid segment is [1, 4] with length 4, so the expected output is `4`.
- The third query `1 1 5` checks for the longest polygonal segment in the range [1, 5]. The valid segment is [1, 5] with length 5, so the expected output is `5`.
- The fourth query `2 1 5` updates the value at index 1 to 5, making the array `5 1 2 2 8`.
- The fifth query `1 1 4` checks for the longest polygonal segment in the range [1, 4]. The valid segment is [1, 4] with length 4, so the expected output is `4`.
- The sixth query `1 1 5` checks for the longest polygonal segment in the range [1, 5]. The valid segment is [1, 5] with length 5, so the expected output is `5`.
- The seventh query `2 3 1` updates the value at index 3 to 1, making the array `5 1 1 2 8`.
- The eighth query `1 1 5` checks for the longest polygonal segment in the range [1, 5]. The valid segment is [2, 5] with length 4, so the expected output is `4`.

This test case is designed to expose potential issues in the logic for checking polygonal segments and handling updates and queries.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Polygonal Segments
time_limit_ms: 8000
memory_limit_mb: 512
Description: You are given an array $a$ of size $n$.

A segment $[l, r](1 \le l < r \le n)$ is called a polygonal segment only if the following conditions hold:

  * $(r-l+1) \geq 3$;    * Considering $a_l, a_{l+1}, \ldots, a_r$ as side lengths, these sides can form a polygon with $(r-l+1)$ sides.

Process $q$ queries of two types:

  * "1 l r": find the length of the longest segment among all polygonal segments $[l_0,r_0]$ satisfying $l \le l_0 \le r_0 \le r$. If there is no such polygonal segment, output $-1$ instead;    * "2 i x": assign $a_i := x$.

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

For each test case:

  * The first line of each testcase contains two integers $n$, $q$ ($4 \le n \le 2\cdot 10^5$, $1 \le q \le 10^5$);    * The second line of each testcase contains $n$ integers $a_1,a_2,\ldots, a_n$ ($1 \le a_i \le 10^{12}$);    * The following $q$ lines contain the description of queries. Each line is of one of two types:     * "1 l r" ($1 \le l < r \le n$, $r-l+1\ge 3$);      * "2 i x" ($1 \le i \le n$, $1 \le x \le 10^{12}$).

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

Output: For each query, if there is no suitable segment, output $-1$ in a new line. Otherwise, output the length of the longest segment satisfying the condition above in a new line.

Examples:
input:
2
5 6
3 1 2 2 8
1 1 3
1 1 4
1 1 5
2 1 5
1 1 4
1 1 5
4 10
500000000000 500000000000 1000000000000 500000000000
1 1 3
1 2 4
1 1 4
2 1 499999999999
2 3 999999999999
1 1 3
1 2 4
1 1 4
2 3 1000000000000
1 1 3
output:
-1
4
4
3
5
-1
-1
4
-1
3
4
-1

Note: In the first query of the first test case, there is no polygonal segment under the given condition. For example, considering segment $[1,3]$, you can not form a triangle with side lengths of $a_1=3$, $a_2=1$, and $a_3=2$.

In the second query of the first test case, the longest polygonal segment is $[1,4]$. You can form a quadrilateral with side lengths of $a_1=3$, $a_2=1$, $a_3=2$, and $a_4=2$.

![](CDN_BASE_URL/589caa1d808f5b5cc88c1eb5f51f2eaf) An example of a quadrilateral with side lengths of $3$, $1$, $2$, and $2$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
using ld = long double;
#define int long long
const int INF = 1e17;
const int maxN = 1e6+5;
const int MOD = 1e9+7;

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int b){
	return (unsigned long long)rng()%b;
}
unordered_map<int,int>dp;
unordered_map<int,int>Time;
set<pair<int,int>>pairs;
vector<int>t1;
vector<int>t2;
vector<int>t3;
vector<int>a;
int n;
int TM = 0;
void build(int v, int tl, int tr){
	if(tl==tr){
		t1[v] = a[tl];
		t2[v] = tl;
		t3[v] = 0;
		return;
	}
	int mid = (tl+tr)/2;
	build(2*v,tl,mid);
	build(2*v+1,mid+1,tr);
	t1[v] = t1[2*v]+t1[2*v+1];

	if(t2[2*v]==-1)t2[v] =  t2[2*v+1];
	else if(t2[2*v+1]==-1)t2[v] =  t2[2*v];
	else if(a[t2[2*v]]<a[t2[2*v+1]])t2[v] =  t2[2*v+1];
	else t2[v] =  t2[2*v];

	t3[v] = 0;
}
int get1(int v, int tl, int tr, int l, int r){
	if(r<tl || tr<l){
		return 0;
	}
	if(l<=tl && tr<=r){
		return t1[v];
	}
	int mid = (tl+tr)/2;
	return get1(2*v,tl,mid,l,r)+get1(2*v+1,mid+1,tr,l,r);
}
int get3(int v, int tl, int tr, int l, int r){
	if(r<tl || tr<l){
		return 0;
	}
	if(l<=tl && tr<=r){
		return t3[v];
	}
	int mid = (tl+tr)/2;
	return max(get3(2*v,tl,mid,l,r),get3(2*v+1,mid+1,tr,l,r));
}
int get2(int v, int tl, int tr, int l, int r){
	if(r<tl || tr<l){
		return -1;
	}
	if(l<=tl && tr<=r){
		return t2[v];
	}
	int mid = (tl+tr)/2;

	int ans = -1;

	int ans2 = get2(2*v+1,mid+1,tr,l,r);
	int ans1 = get2(2*v,tl,mid,l,r);
	if(ans2==-1)ans =  ans1;
	else if(ans1==-1)ans =  ans2;
	else if(a[ans1]<a[ans2])ans =  ans2;
	else ans =  ans1;
	return ans;
}
void update(int v, int tl, int tr, int pos, int x){
	if(pos<tl || tr<pos){
		return;
	}
	if(tl==tr){
		t1[v] = x;
		t2[v] = tl;
		t3[v] = x;
		return;
	}
	int mid = (tl+tr)/2;
	//return get1(2*v,tl,mid,l,r)+get1(2*v+1,mid+1,tr,l,r);
	update(2*v,tl,mid,pos,x);
	update(2*v+1,mid+1,tr,pos,x);
	t1[v] = t1[2*v]+t1[2*v+1];
	if(t2[2*v]==-1)t2[v] =  t2[2*v+1];
	else if(t2[2*v+1]==-1)t2[v] =  t2[2*v];
	else if(a[t2[2*v]]<a[t2[2*v+1]])t2[v] =  t2[2*v+1];
	else t2[v] =  t2[2*v];

	t3[v] = max(t3[2*v],t3[2*v+1]);
}

int solve(int l, int r){
	if(r-l+1<3)return -1;
	if(get3(1,0,n-1,l,r)<Time[1e9*l+r] && dp.count(1e9*l+r))return dp[1e9*l+r];
	int i = get2(1,0,n-1,l,r);
	if(a[i]<r-l)return r-l+1;
	int sum = get1(1,0,n-1,l,r);
	if(2*a[i]<sum)return r-l+1;
	dp[1e9*l+r]= max(solve(l,i-1),solve(i+1,r));
	Time[dp[1e9*l+r]] = TM;
	return dp[1e9*l+r];
}
signed main(){
	//ios_base::sync_with_stdio(0);
	//cin.tie(0);
	int t;
	cin>>t;
	while(t--){
		int q;
		cin>>n>>q;
		TM = 1;
		dp = unordered_map<int,int>();
		t1 = vector<int>(4*n+4);
		t2 = vector<int>(4*n+4);
		t3 = vector<int>(4*n+4);
		a = vector<int>(n);
		for(int i = 0;i<n;i++)cin>>a[i];
		build(1,0,n-1);
		while(q--){
			int tp;
			cin>>tp;
			if(tp==1){
				int l,r;
				cin>>l>>r;
				l--;r--;
				//cerr<<"ANS IS:";
				cout<<solve(l,r)<<endl;
			}
			else{
				int pos,x;
				cin>>pos>>x;
				pos--;
				a[pos] = x;
				update(1,0,n-1,pos,x);
			}
			TM++;
		}
	}
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Polygonal Segments
time_limit_ms: 8000
memory_limit_mb: 512
Description: You are given an array $a$ of size $n$.

A segment $[l, r](1 \le l < r \le n)$ is called a polygonal segment only if the following conditions hold:

  * $(r-l+1) \geq 3$;    * Considering $a_l, a_{l+1}, \ldots, a_r$ as side lengths, these sides can form a polygon with $(r-l+1)$ sides.

Process $q$ queries of two types:

  * "1 l r": find the length of the longest segment among all polygonal segments $[l_0,r_0]$ satisfying $l \le l_0 \le r_0 \le r$. If there is no such polygonal segment, output $-1$ instead;    * "2 i x": assign $a_i := x$.

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

For each test case:

  * The first line of each testcase contains two integers $n$, $q$ ($4 \le n \le 2\cdot 10^5$, $1 \le q \le 10^5$);    * The second line of each testcase contains $n$ integers $a_1,a_2,\ldots, a_n$ ($1 \le a_i \le 10^{12}$);    * The following $q$ lines contain the description of queries. Each line is of one of two types:     * "1 l r" ($1 \le l < r \le n$, $r-l+1\ge 3$);      * "2 i x" ($1 \le i \le n$, $1 \le x \le 10^{12}$).

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

Output: For each query, if there is no suitable segment, output $-1$ in a new line. Otherwise, output the length of the longest segment satisfying the condition above in a new line.

Examples:
input:
2
5 6
3 1 2 2 8
1 1 3
1 1 4
1 1 5
2 1 5
1 1 4
1 1 5
4 10
500000000000 500000000000 1000000000000 500000000000
1 1 3
1 2 4
1 1 4
2 1 499999999999
2 3 999999999999
1 1 3
1 2 4
1 1 4
2 3 1000000000000
1 1 3
output:
-1
4
4
3
5
-1
-1
4
-1
3
4
-1

Note: In the first query of the first test case, there is no polygonal segment under the given condition. For example, considering segment $[1,3]$, you can not form a triangle with side lengths of $a_1=3$, $a_2=1$, and $a_3=2$.

In the second query of the first test case, the longest polygonal segment is $[1,4]$. You can form a quadrilateral with side lengths of $a_1=3$, $a_2=1$, $a_3=2$, and $a_4=2$.

![](CDN_BASE_URL/589caa1d808f5b5cc88c1eb5f51f2eaf) An example of a quadrilateral with side lengths of $3$, $1$, $2$, and $2$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
using ld = long double;
#define int long long
const int INF = 1e17;
const int maxN = 1e6+5;
const int MOD = 1e9+7;

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int b){
	return (unsigned long long)rng()%b;
}

vector<int>t1;
vector<int>t2;
vector<int>a;
int n;
void build(int v, int tl, int tr){
	if(tl==tr){
		t1[v] = a[tl];
		t2[v] = tl;
		return;
	}
	int mid = (tl+tr)/2;
	build(2*v,tl,mid);
	build(2*v+1,mid+1,tr);
	t1[v] = t1[2*v]+t1[2*v+1];

	if(t2[2*v]==-1)t2[v] =  t2[2*v+1];
	else if(t2[2*v+1]==-1)t2[v] =  t2[2*v];
	else if(a[t2[2*v]]<a[t2[2*v+1]])t2[v] =  t2[2*v+1];
	else t2[v] =  t2[2*v];
}
int get1(int v, int tl, int tr, int l, int r){
	if(r<tl || tr<l){
		return 0;
	}
	if(l<=tl && tr<=r){
		return t1[v];
	}
	int mid = (tl+tr)/2;
	return get1(2*v,tl,mid,l,r)+get1(2*v+1,mid+1,tr,l,r);
}
int get2(int v, int tl, int tr, int l, int r){
	if(r<tl || tr<l){
		return -1;
	}
	if(l<=tl && tr<=r){
		return t2[v];
	}
	int mid = (tl+tr)/2;

	int ans = -1;

	int ans2 = get2(2*v+1,mid+1,tr,l,r);
	int ans1 = get2(2*v,tl,mid,l,r);
	if(ans2==-1)ans =  ans1;
	else if(ans1==-1)ans =  ans2;
	else if(a[ans1]<a[ans2])ans =  ans2;
	else ans =  ans1;
	return ans;
}
void update(int v, int tl, int tr, int pos, int x){
	if(pos<tl || tr<pos){
		return;
	}
	if(tl==tr){
		t1[v] = x;
		t2[v] = tl;
		return;
	}
	int mid = (tl+tr)/2;
	//return get1(2*v,tl,mid,l,r)+get1(2*v+1,mid+1,tr,l,r);
	update(2*v,tl,mid,pos,x);
	update(2*v+1,mid+1,tr,pos,x);
	t1[v] = t1[2*v]+t1[2*v+1];
	if(t2[2*v]==-1)t2[v] =  t2[2*v+1];
	else if(t2[2*v+1]==-1)t2[v] =  t2[2*v];
	else if(a[t2[2*v]]<a[t2[2*v+1]])t2[v] =  t2[2*v+1];
	else t2[v] =  t2[2*v];
}

int solve(int l, int r){
	if(r-l+1<3)return -1;
	int i = get2(1,0,n-1,l,r);
	if(a[i]==1)return -1;
	int sum = get1(1,0,n-1,l,r);
	if(2*a[i]<sum)return r-l+1;
	return max(solve(l,i-1),solve(i+1,r));
}
signed main(){
	//ios_base::sync_with_stdio(0);
	//cin.tie(0);
	int t;
	cin>>t;
	while(t--){
		int q;
		cin>>n>>q;
		t1 = vector<int>(4*n+4);
		t2 = vector<int>(4*n+4);
		a = vector<int>(n);
		for(int i = 0;i<n;i++)cin>>a[i];
		build(1,0,n-1);
		while(q--){
			int tp;
			cin>>tp;
			if(tp==1){
				int l,r;
				cin>>l>>r;
				l--;r--;
				//cerr<<"ANS IS:";
				cout<<solve(l,r)<<endl;
			}
			else{
				int pos,x;
				cin>>pos>>x;
				pos--;
				a[pos] = x;
				update(1,0,n-1,pos,x);
			}
		}
	}
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Polygonal Segments
time_limit_ms: 8000
memory_limit_mb: 512
Description: You are given an array $a$ of size $n$.

A segment $[l, r](1 \le l < r \le n)$ is called a polygonal segment only if the following conditions hold:

  * $(r-l+1) \geq 3$;    * Considering $a_l, a_{l+1}, \ldots, a_r$ as side lengths, these sides can form a polygon with $(r-l+1)$ sides.

Process $q$ queries of two types:

  * "1 l r": find the length of the longest segment among all polygonal segments $[l_0,r_0]$ satisfying $l \le l_0 \le r_0 \le r$. If there is no such polygonal segment, output $-1$ instead;    * "2 i x": assign $a_i := x$.

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

For each test case:

  * The first line of each testcase contains two integers $n$, $q$ ($4 \le n \le 2\cdot 10^5$, $1 \le q \le 10^5$);    * The second line of each testcase contains $n$ integers $a_1,a_2,\ldots, a_n$ ($1 \le a_i \le 10^{12}$);    * The following $q$ lines contain the description of queries. Each line is of one of two types:     * "1 l r" ($1 \le l < r \le n$, $r-l+1\ge 3$);      * "2 i x" ($1 \le i \le n$, $1 \le x \le 10^{12}$).

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

Output: For each query, if there is no suitable segment, output $-1$ in a new line. Otherwise, output the length of the longest segment satisfying the condition above in a new line.

Examples:
input:
2
5 6
3 1 2 2 8
1 1 3
1 1 4
1 1 5
2 1 5
1 1 4
1 1 5
4 10
500000000000 500000000000 1000000000000 500000000000
1 1 3
1 2 4
1 1 4
2 1 499999999999
2 3 999999999999
1 1 3
1 2 4
1 1 4
2 3 1000000000000
1 1 3
output:
-1
4
4
3
5
-1
-1
4
-1
3
4
-1

Note: In the first query of the first test case, there is no polygonal segment under the given condition. For example, considering segment $[1,3]$, you can not form a triangle with side lengths of $a_1=3$, $a_2=1$, and $a_3=2$.

In the second query of the first test case, the longest polygonal segment is $[1,4]$. You can form a quadrilateral with side lengths of $a_1=3$, $a_2=1$, $a_3=2$, and $a_4=2$.

![](CDN_BASE_URL/589caa1d808f5b5cc88c1eb5f51f2eaf) An example of a quadrilateral with side lengths of $3$, $1$, $2$, and $2$.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Codeforces1990F {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

    static class Ent {
        long sum;
        long max;
        int maxIndex;
    }

    static long[] readIntArray() throws IOException {
        String line = br.readLine();
        String[] linesplit = line.split(" ");
        long[] is = new long[linesplit.length];
        for (int i = 0; i < linesplit.length; i++) {
            is[i] = Long.parseLong(linesplit[i]);
        }
        return is;
    }

    static Ent[][] sums;

    public static void main(String[] args) throws NumberFormatException, IOException {
        int tests = Integer.parseInt(br.readLine());
        for (int t = 0; t < tests; t++) {
            long[] nqLine = readIntArray();
            int n = (int) nqLine[0];
            int q = (int) nqLine[1];

            long[] a = readIntArray();

            Ent[] ent = new Ent[n];
            for (int i = 0; i < n; i++) {
                Ent e = new Ent();
                e.sum = a[i];
                e.max = a[i];
                e.maxIndex = i;
                ent[i] = e;
            }

            sums = new Ent[20][];
            int sumIndex = 0;
            while (true) {
                sums[sumIndex] = ent;
                if (ent.length == 1) {
                    break;
                }
                sumIndex++;
                Ent[] nextent = new Ent[(ent.length + 1) / 2];
                for (int i = 0; i < ent.length; i++) {
                    if (nextent[i / 2] == null) {
                        Ent e = new Ent();
                        e.sum = ent[i].sum;
                        e.max = ent[i].max;
                        e.maxIndex = ent[i].maxIndex;
                        nextent[i / 2] = e;
                    } else {
                        nextent[i / 2].sum += ent[i].sum;
                        if (nextent[i / 2].max < ent[i].max) {
                            nextent[i / 2].max = ent[i].max;
                            nextent[i / 2].maxIndex += ent[i].maxIndex;
                        }
                    }
                }
                ent = nextent;
            }

            for (int i = 0; i < q; i++) {
                long[] qLine = readIntArray();
                if (qLine[0] == 1) {
                    int l = (int) qLine[1] - 1;
                    int r = (int) qLine[2] - 1;
                    int b = getBest(l, r);
                    System.out.println(b);
                } else {
                    int index = (int) qLine[1] - 1;
                    long x = qLine[2];
                    Ent other = sums[0][index];
                    other.sum = x;
                    other.max = x;
                    for (sumIndex = 1; sumIndex < sums.length; sumIndex++) {
                        if (sums[sumIndex] == null) {
                            break;
                        }
                        index /= 2;
                        Ent e = sums[sumIndex][index];
                        Ent o1 = sums[sumIndex - 1][index * 2];
                        Ent o2 = null;
                        if (index * 2 + 1 < sums[sumIndex - 1].length) {
                            o2 = sums[sumIndex - 1][index * 2 + 1];
                        }
                        e.sum = o1.sum;
                        e.max = o1.max;
                        e.maxIndex = o1.maxIndex;
                        if (o2 != null) {
                            e.sum += o2.sum;
                            if (e.max < o2.max) {
                                e.max = o2.max;
                                e.maxIndex += o2.maxIndex;
                            }
                        }
                    }
                }

            }
        }
    }

    public static int getBest(int l, int r) {
        if (r - l + 1 < 3) {
            return -1;
        }

        Ent e = getEnt(l, r);
        if (e.max * 2 < e.sum) {
            return r - l + 1;
        }

        int b1 = getBest(l, e.maxIndex - 1);
        int b2 = getBest(e.maxIndex + 1, r);
        return Math.max(b1, b2);
    }

    public static Ent getEnt(int l, int r) {
        //        System.out.print("SUM: " + l + "; " + r);
        Ent e = new Ent();
        while (l <= r) {
            int sumIndex = 0;
            int sumLength = 1;
            while (true) {
                if ((l & sumLength) == 0) {
                    if (sumLength * 2 <= r - l + 1) {
                        sumIndex++;
                        sumLength <<= 1;
                        continue;
                    }
                }
                break;
            }
            e.sum += sums[sumIndex][l >> sumIndex].sum;
            if (e.max < sums[sumIndex][l >> sumIndex].max) {
                e.max = sums[sumIndex][l >> sumIndex].max;
                e.maxIndex = sums[sumIndex][l >> sumIndex].maxIndex;
            }
            l += sumLength;
        }

        //        System.out.println(" = " + e.sum + "; " + e.max + "; " + e.maxIndex);

        return e;
    }
}

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

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

Think step by step.

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