**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Speedbreaker
time_limit_ms: 2000
memory_limit_mb: 256
Description: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)

There are $n$ cities in a row, numbered $1, 2, \ldots, n$ left to right.

  * At time $1$, you conquer exactly one city, called the starting city.    * At time $2, 3, \ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it.

You win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of cities.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the deadlines for conquering the cities.

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

Output: For each test case, output a single integer: the number of starting cities that allow you to win.

Examples:
input:
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
output:
3
0
1

Note: In the first test case, cities $2$, $3$, and $4$ are good starting cities.

In the second test case, there are no good starting cities.

In the third test case, the only good starting city is city $5$.

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

public class CodeForcesSolution {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    static StringTokenizer st;
    /*static int mod = (int) 1e9+7;*/
    static int MM = (int) 1e6, tree[], lazy[];
    public static void main(String[] args) throws IOException {
        int t = readInt();
        while (t-- > 0) {
            tree = new int[MM]; lazy = new int[MM];
            int n = readInt();
            int[] a = new int[n + 1];
            for (int i = 1; i <= n; i++) {
                a[i] = readInt();
                updateRange(1, 1, n, i, i, a[i] - i + 1);
            }
            int ans = 0;
            for (int i = 1; i <= n; i++) {
                TreeSet<int[]> ts = new TreeSet<>((f, s) -> f[0] == s[0] ? f[1] - s[1] : f[0] - s[0]);
                for (int j = 1; j <= n; j++) {
                    ts.add(new int[] {a[j], j});
                }
                int time = 1, l = i, r = i;
                boolean flg = true;
                while (!ts.isEmpty()) {
                    int[] cur = ts.first(); ts.remove(cur);
                    if (l <= cur[1] && cur[1] <= r) continue; // conquered
                    if (time + Math.min(Math.abs(cur[1] - l), Math.abs(cur[1] - r)) > cur[0]) {
                        flg = false; break;
                    }
                    time += Math.min(Math.abs(cur[1] - l), Math.abs(cur[1] - r));
                    l = Math.min(l, cur[1]);
                    r = Math.max(r, cur[1]);
                }
                if (flg) ans++;
            }
            System.out.println(ans);
        }
    }
    // l -> update left bound, r -> update right bound
    static void updateRange(int node, int start, int end, int l, int r, int val) {
        if(start > end || l > r) return;        // Out of range
        if(lazy[node] != 0) { // Propagate lazy flag (if exists one)
            // This node needs to be updated
            tree[node] += (end - start + 1) * lazy[node];    // Update it
            if(start != end) {
                lazy[node*2] += lazy[node];                  // Propagate lazy flag
                lazy[node*2+1] += lazy[node];                // Propagate lazy flag
            }
            lazy[node] = 0;                                  // Reset lazy flag
        }
        if(start >= l && end <= r) { // reached the segment where we want to update
            // Segment is fully within range
            tree[node] += (end - start + 1) * val;
            if(start != end) { // Propagate
                // Not leaf node
                lazy[node*2] += val;
                lazy[node*2+1] += val;
            }
            return;
        }
        int mid = (start + end) / 2;
        if (l > mid) updateRange(node*2 + 1, mid + 1, end, l, r, val);   // Updating right child
        else if (r<=mid) updateRange(node*2, start, mid, l, r, val);        // Updating left child
        else {
            updateRange(2*node, start, mid, l, mid, val);
            updateRange(2*node+1, mid+1, end, mid+1, r, val);
        }
        tree[node] = Math.min(tree[node*2], tree[node*2+1]);        // Updating root with min value
    }

    // l -> query left bound, r -> query right bound
    static int queryRange(int node, int start, int end, int l, int r) {
        if(start > end) return 0;       // Out of range
        if(lazy[node] != 0) {
            // This node needs to be updated
            tree[node] += (end - start + 1) * lazy[node];            // Update it
            if(start != end)
            {
                lazy[node*2] += lazy[node];         // Mark child as lazy
                lazy[node*2+1] += lazy[node];    // Mark child as lazy
            }
            lazy[node] = 0;                 // Reset it
        }
        if (start == l && end == r) return tree[node];
        int m = (start+end)/2;
        if (l > m) return queryRange(2*node+1, m+1, end, l, r);
        else if (r<=m) return queryRange(2*node, start, m, l, r);
        else return Math.min(queryRange(2*node, start, m, l, m), queryRange(2*node+1, m+1, end, m+1, r));
    }
    static String next() throws IOException {
        while (st == null || !st.hasMoreTokens())
            st = new StringTokenizer(br.readLine().trim());
        return st.nextToken();
    }
    static long readLong() throws IOException {
        return Long.parseLong(next());
    }
    static int readInt() throws IOException {
        return Integer.parseInt(next());
    }
    static double readDouble() throws IOException {
        return Double.parseDouble(next());
    }
    static char readCharacter() throws IOException {
        return next().charAt(0);
    }
    static String readLine() throws IOException {
        return br.readLine().trim();
    }
    static int readALotInt() throws IOException{
        int x = 0, c;
        while((c = br.read()) != ' ' && c != '\n')
            x = x * 10 + (c - '0');
        return 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:
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. Let's break down the problem and the code to identify potential issues:

1. **Understanding the Problem:**
   - We need to determine the number of starting cities that allow us to win the game.
   - Each city has a deadline by which it must be conquered.
   - We can only conquer adjacent cities to the ones already conquered.

2. **Identifying Potential Bugs:**
   - The code uses a segment tree for range updates and queries.
   - The logic for updating and querying the segment tree might have issues.
   - The logic for determining if a city can be conquered within the deadline might be flawed.

3. **Challenging Scenarios:**
   - **Edge Cases:**
     - Single city.
     - All cities have the same deadline.
     - Deadlines are in a strictly increasing or decreasing order.
   - **Complex Cases:**
     - Deadlines are such that it's impossible to conquer all cities.
     - Deadlines are such that only a specific subset of cities can be conquered.

4. **Generating the Test Input:**
   - Let's create a test case where the deadlines are such that it's impossible to conquer all cities, which should expose any issues in the logic.

Here is a small-scale test input that is complete and valid:

```plaintext
1
6
5 6 4 1 4 5
```

This test case is similar to one of the examples provided, where the expected output is `0` (no starting cities allow you to win). This should help expose any issues in the code's logic for determining the winning strategy.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Speedbreaker
time_limit_ms: 2000
memory_limit_mb: 256
Description: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)

There are $n$ cities in a row, numbered $1, 2, \ldots, n$ left to right.

  * At time $1$, you conquer exactly one city, called the starting city.    * At time $2, 3, \ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it.

You win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of cities.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the deadlines for conquering the cities.

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

Output: For each test case, output a single integer: the number of starting cities that allow you to win.

Examples:
input:
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
output:
3
0
1

Note: In the first test case, cities $2$, $3$, and $4$ are good starting cities.

In the second test case, there are no good starting cities.

In the third test case, the only good starting city is city $5$.

Buggy Code:
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <fstream>
#include <cmath>
#include <set>
#include <cassert>
#include <map>
using namespace std;

#define int long long

typedef pair <int, int> ii;

vector <int> suf, pref;

int Right(int left, int pos, int right) {
	return suf[right + 1] + right - (pos - left) - (right - pos);
}

int Left(int left, int pos, int right) {
	return pref[left - 1] - left - (right - pos) - (pos - left);
}

signed main() {

#ifdef _DEBUG
	fstream cin("input.txt");
	ofstream cout("output.txt");
#endif

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

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

		suf.clear(); pref.clear();

		int n, ans = 0, success = -1; cin >> n;
		vector <int> val(n), sufIdx(n), prefIdx(n);
		suf.resize(n); pref.resize(n);

		for (int i = 0; i < n; i++)
			cin >> val[i];

		if (n == 1) {
			cout << 1 << '\n';
			continue;
		}

		for (int i = n - 1; i >= 0; i--) {
			suf[i] = val[i] - i;
			sufIdx[i] = i;
		}
		for (int i = n - 2; i >= 0; i--) {
			if (suf[i] > suf[i + 1])
				sufIdx[i] = sufIdx[i + 1];
			suf[i] = min(suf[i], suf[i + 1]);
		}
		for (int i = 0; i < n; i++) {
			pref[i] = val[i] + i;
			prefIdx[i] = i;
			bool ok = suf[i] + i > 0;
			if (i) {
				if (pref[i] > pref[i - 1])
					prefIdx[i] = prefIdx[i - 1];
				pref[i] = min(pref[i], pref[i - 1]);
				ok = min(ok, pref[i] - i > 0);
			}
			if (ok && success == 1) ans++;
			else if (ok && success == -1) {
				int left = i, right = i;
				while (true) {
					if (left == 0) {
						if (Right(left, i, right) > 0) {
							success = 1;
							ans++;
						}
						else success = 0;
						break;
					}
					else if (right == n - 1) {
						if (Left(left, i, right) > 0) {
							success = 1;
							ans++;
						}
						else success = 0;
						break;
					}
					int l = Left(left, i, right), r = Right(left, i, right);
					if (min(l, r) <= 0) {
						success = 0;
						break;
					}
					else if (l > r) {
						right = sufIdx[right + 1];
					}
					else
						left = prefIdx[left - 1];

				}
			}
		}

		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:
```plaintext
1
5
3 2 1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Speedbreaker
time_limit_ms: 2000
memory_limit_mb: 256
Description: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)

There are $n$ cities in a row, numbered $1, 2, \ldots, n$ left to right.

  * At time $1$, you conquer exactly one city, called the starting city.    * At time $2, 3, \ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it.

You win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of cities.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the deadlines for conquering the cities.

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

Output: For each test case, output a single integer: the number of starting cities that allow you to win.

Examples:
input:
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
output:
3
0
1

Note: In the first test case, cities $2$, $3$, and $4$ are good starting cities.

In the second test case, there are no good starting cities.

In the third test case, the only good starting city is city $5$.

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

int main(){
    int t;
    cin >> t;
    while(t--) {
        int a;
        cin >> a;
        int arr[a], j;
        vector<int> vec,vec1,vec2;
        for(int i = 0; i < a; i++) {
            cin >> arr[i];
        }
        int minn = *min_element(arr, arr + a);
        for(int i = 0; i < a; i++) {
            if (arr[i] == minn) {
                vec.push_back(i);
                j = i;
            }
        }
       float minz = vec[0];
        float leftend = j, rightend = j,  x, y, z = 0, n = 0, k = a;
        float  t2 = 1e9, t1 = 1e9;
        while(k--) {
            t1 = 1e9;
            t2 = 1e9;
            for(int i = 0; i < leftend; i++) {
                if(i < leftend) {
                    t1 = min((arr[i] - z - 1) / (leftend-i), t1);
                    if(t1 < 1) {
                        cout << 0 << endl;
                        n=1;
                        goto b;
                    }
                    if((arr[i] - z - 1)/ (leftend - i) == t1) {
                        vec1.push_back(i);
                        x = i;
                    }
                }
            }
            for(int i = rightend+1; i < a; i++){
                   if(i > rightend) {
                    t2 = min((arr[i] - z - 1) / (i - rightend), t2);
                    if(t2 < 1) {
                        cout << 0 << endl;
                        n=1;
                        goto b;
                    }
                    if((arr[i] - z - 1) /(i - rightend) == t2) {
                            y = i;
                            // cout<<t2 <<" "<<y<<endl;
                    }
                }
            }
            if(t1 < t2) {
                z = (leftend - x) + z;
                leftend = x;
                // cout<<"l "<<leftend<<" "<<t2<<" "<<t1<<" "<<endl;
            } if(t2<t1) {
                z = (y - rightend) + z;
                rightend = y;
                //  cout<<"r "<<rightend<<" "<<t2<<" "<<t1<<" "<<endl;
            }
        }
        if(n==0){
            int ans = 0 , h;
            h=arr[j]-1;
            if(j==minz){
                if(j-h<0){
                    ans=ans+j;
                }
                else{
                   ans = h +ans;
                }
                if(j+h>a-1){
                    ans = (a-1)-j;
                }
                else{
                    ans = ans + h;
                }

                cout<<ans+1<<endl;}
            else{
                if(j-h<0){
                    ans = j + ans;
                }
                else{
                    ans = ans +h;
                }
                if(minz+h>a-1){
                    ans = a - 1 - minz;
                }
                else{
                    ans = ans + h;
                }
                ans = ans - (j - minz - 1);
                  cout<<ans<<endl;
            }
        }
        b:;
}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. The buggy code has several potential issues, including incorrect handling of minimum elements, incorrect boundary conditions, and potential off-by-one errors.

Here are the steps to create a challenging test input:

1. **Edge Case with Minimum Element**: The code assumes that the minimum element in the array is the starting point. We can create a test case where the minimum element is not a valid starting point.

2. **Boundary Conditions**: The code might have issues with boundary conditions, especially when the array is small or when the deadlines are very strict.

3. **Off-by-One Errors**: The code might have off-by-one errors when calculating the number of valid starting points.

Let's create a test case that includes these elements:

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

### Explanation:

1. **Test Case 1:**
   - Array: `[1, 2, 3, 4]`
   - Minimum element: `1` at index `0`
   - This is a simple case where the minimum element is the only valid starting point.

2. **Test Case 2:**
   - Array: `[2, 3, 4, 1]`
   - Minimum element: `1` at index `3`
   - This case is similar to the first but with the minimum element at the end of the array, which should still be the only valid starting point.

3. **Test Case 3:**
   - Array: `[3, 1, 4, 2, 5]`
   - Minimum element: `1` at index `1`
   - This case has the minimum element in the middle, and the valid starting points should be `1` and `2` (indices `1` and `2`).

These test cases should help expose any issues in the buggy code, particularly how it handles the minimum element and boundary conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Speedbreaker
time_limit_ms: 2000
memory_limit_mb: 256
Description: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)

There are $n$ cities in a row, numbered $1, 2, \ldots, n$ left to right.

  * At time $1$, you conquer exactly one city, called the starting city.    * At time $2, 3, \ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it.

You win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of cities.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the deadlines for conquering the cities.

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

Output: For each test case, output a single integer: the number of starting cities that allow you to win.

Examples:
input:
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
output:
3
0
1

Note: In the first test case, cities $2$, $3$, and $4$ are good starting cities.

In the second test case, there are no good starting cities.

In the third test case, the only good starting city is city $5$.

Buggy Code:
#include<bits/stdc++.h>
#include<bitset>
#include<nmmintrin.h>
using namespace std;
#define oOo {ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);}
#define ll long long
#define vc vector
#define str string
#define e "\n"
#define all(v) (v).begin() , (v).end()
#define File freopen("two2.in", "r", stdin);
#define ONES(x) _mm_popcnt_u64(x)
#define ONESs(x) __builtin_popcountll(x)
#define MEM(x , y) memset(x , y , sizeof(x))
#define MASE(data , n , m , df) vc<vc<data>>(n + 1, vc<data>(m + 1, df))
const int mod = 1e9 + 7;
const ll modd = 998244353;
const int oo = 0x3f3f3f3f;
bool ok = false, sat = false, ent = false;
int dx[8] = { 0, 0, 1, -1, -1, -1, 1, 1 };
int dy[8] = { -1, 1, 0, 0, -1, 1, -1, 1 };
ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); }
ll lcm(ll a, ll b) { return a * (b / gcd(a, b)); }
ll add(ll a, ll b, ll md) {
	return (((a % md) + (b % md)) % md);
}
ll mul(ll a, ll b, ll md) {
	return (((a % md) * (b % md)) % md);
}
ll fix_mod(ll a, ll b, ll md) {
	return ((a % md - b % md) + md) % md;
}
//---------------------------- [ We need a learn ] -------------------------</end>
const int N = 1e6 + 5, M = 21;
vc<int>v,v2;
//class segTree {
//private:
//	int n;
//	vc<int>tree , tree2;
//public:
//	segTree(int _n) {
//		n = _n;
//		while (ONES(n) != 1)
//		{
//			n++; v.push_back(0); v2.push_back(0);
//		}
//		tree=tree2 = vc<int>(n * 2);
//	}
//	int up_to() {
//		return n;
//	}
//	ll buildSegTree(ll idx, ll fm, ll to) {
//		if (fm == to) {
//			return  tree[idx] = v[idx - n];
//		}
//		ll mid = (fm + to) / 2;
//		return tree[idx] = min(buildSegTree(idx * 2, fm, mid) , buildSegTree(idx * 2 + 1, mid + 1, to));
//
//	}
//
//	ll quary(int idx, int fm, int to, int fm_q, int to_q) {
//		if (fm_q <= fm && to_q >= to) {
//			return tree[idx];
//		}
//		if (to < fm_q || to_q < fm) {
//			return INT_MAX;
//		}
//		ll mid = (fm + to) / 2;
//		return min(quary(idx * 2, fm, mid, fm_q, to_q) , quary(idx * 2 + 1, mid + 1, to, fm_q, to_q));
//
//	}
//	///
//	ll buildSegTree2(ll idx, ll fm, ll to) {
//		if (fm == to) {
//			return  tree2[idx] = v2[idx - n];
//		}
//		ll mid = (fm + to) / 2;
//		return tree2[idx] = min(buildSegTree2(idx * 2, fm, mid), buildSegTree2(idx * 2 + 1, mid + 1, to));
//
//	}
//
//	ll quary2(int idx, int fm, int to, int fm_q, int to_q) {
//		if (fm_q <= fm && to_q >= to) {
//			return tree2[idx];
//		}
//		if (to < fm_q || to_q < fm) {
//			return INT_MAX;
//		}
//		ll mid = (fm + to) / 2;
//		return min(quary2(idx * 2, fm, mid, fm_q, to_q), quary2(idx * 2 + 1, mid + 1, to, fm_q, to_q));
//
//	}
//	///
//	void update(ll idx, int val) {
//		tree[(n + idx - 1)] = val;
//		for (ll a = (n + idx - 1) / 2; a >= 1; a /= 2) {
//			tree[a] = tree[a * 2] + tree[a * 2 + 1];
//		}
//	}
//	/*void quary2(int idx, int fm, int to, int fm_q, int to_q , int g) {
//		if (geet)return;
//		if (fm == to && (fm_q <= fm && to_q >= to) && gcd(tree[idx], g) != g) {
//			geet = fm;
//			return ;
//		}
//		if (fm_q <= fm && to_q >= to && gcd(tree[idx] , g) == g) {
//			return ;
//		}
//		if (to < fm_q || to_q < fm) {
//			return ;
//		}
//		ll mid = (fm + to) / 2;
//		quary2(idx * 2, fm, mid, fm_q, to_q , g); quary2(idx * 2 + 1, mid + 1, to, fm_q, to_q , g);
//	}*/
//	/*void lazy(int idx, int fm, int to, int fm_upt, int to_upt, ll ad) {
//		if (fm == fm_upt && to == to_upt) {
//			tree[idx] += ad;
//			return;
//		}
//		int mid = (fm + to) / 2;
//
//		if (fm_upt <= mid) {
//			lazy(idx * 2, fm, mid, fm_upt, min(mid, to_upt), ad);
//		}
//
//		if (to_upt >= mid + 1) {
//			lazy(idx * 2 + 1, mid + 1, to, max(mid + 1, fm_upt), to_upt, ad);
//		}
//	}*/
//	/*void lazy2(int idx, int fm, int to, int fm_upt, int to_upt, ll ad) {
//		if (fm == fm_upt && to == to_upt) {
//			tree2[idx] = max(tree2[idx], ad);
//			return;
//		}
//		int mid = (fm + to) / 2;
//		/
//		if (fm_upt <= mid) {
//			lazy2(idx * 2, fm, mid, fm_upt, min(mid, to_upt), ad);
//		}
//
//		if (to_upt >= mid + 1) {
//			lazy2(idx * 2 + 1, mid + 1, to, max(mid + 1, fm_upt), to_upt, ad);
//		}
//	}*/
//};

int main() {
	oOo;
	int t; cin >> t;
	while (t--)
	{
		int n; cin >> n;
		v=v2 = vc<int>(n);
		for (int a = 0; a < n; a++) {
			cin >> v[a];
		}
		/*segTree sg(n) , sg2(n);
		sg.buildSegTree(1, 1, sg.up_to());
		sg2.buildSegTree(1, 1, sg2.up_to());*/
		int cnt = 0;
		priority_queue<pair<int, int>, vc<pair<int, int>>, greater<pair<int, int>>>pq, pq2;
		set<int>st;
		for (int a = n-1; a >=0; a--) {
			pq2.push({ v[a] + (n - a) , -(a + 1) });
		}
		pq.push({ v[0] + 1 ,  1 }); st.insert(1);
		for (int a = 1; a < n-1; a++) {
			st.insert(a + 1);
			while (pq2.size())
			{
				if (st.count(-pq2.top().second))pq2.pop();
				else break;
			}
			int mn = pq.top().first, mn2 = pq2.top().first, idx = pq.top().second, idx2 = -pq2.top().second;
			if (mn > a +1 && mn2 > n - a) {
				int len = idx2 - idx + 1;
				if (v[idx - 1] >= len || v[idx2 - 1] >= len) {
					cnt++;
				}
			}
			pq.push({ v[a] + a + 1 , a + 1 });
		}
		int ad = 0;
		ok = 1;
		for (int a = 0; a < n; a++) {
			if (v[a] < a + 1) {
				ok = 0;
			}
		}
		ad += ok;
		ok = 1;
		for (int a = n - 1; a >= 0; a--) {
			if (v[a] < n - a) {
				ok = 0;
			}
		}
		ad += ok;
		if (n == 1)ad--;
		cout << cnt+ad << 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:
To generate a challenging test input that exposes the bug in the provided buggy code, let's focus on edge cases and scenarios that maximize the failure probability. Here are some key points to consider:

1. **Single City Case**: The simplest case with only one city.
2. **All Deadlines are 1**: This will test the case where all cities have the minimum deadline.
3. **Deadlines are in Reverse Order**: This will test the case where the deadlines are in reverse order, which might expose issues with the logic for checking the feasibility of starting cities.
4. **Mixed Deadlines**: A mix of deadlines that are close to the city index and some that are far, to test the logic for both early and late deadlines.
5. **Edge Case with Middle Deadlines**: A case where the deadlines are centered around the middle of the city range, which can test the logic for handling middle starting cities.

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

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

Explanation:
1. **Single City Case**: The first test case has only one city with a deadline of 1. The expected output is 1, as there is only one city and it can be conquered at time 1.
2. **All Deadlines are 1**: The second test case has all deadlines set to 1. The expected output is 0, as it is impossible to conquer all cities within their deadlines.
3. **Deadlines in Reverse Order**: The third test case has deadlines in reverse order. The expected output is 0, as it is impossible to conquer all cities within their deadlines.
4. **Mixed Deadlines**: The fourth test case has a mix of deadlines. The expected output is 2, as cities 1 and 6 are the only starting cities that allow you to win.
5. **Edge Case with Middle Deadlines**: The fifth test case has deadlines centered around the middle. The expected output is 3, as cities 2, 3, and 4 are the only starting cities that allow you to win.

This test input should effectively expose any bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Speedbreaker
time_limit_ms: 2000
memory_limit_mb: 256
Description: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)

There are $n$ cities in a row, numbered $1, 2, \ldots, n$ left to right.

  * At time $1$, you conquer exactly one city, called the starting city.    * At time $2, 3, \ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it.

You win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of cities.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the deadlines for conquering the cities.

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

Output: For each test case, output a single integer: the number of starting cities that allow you to win.

Examples:
input:
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
output:
3
0
1

Note: In the first test case, cities $2$, $3$, and $4$ are good starting cities.

In the second test case, there are no good starting cities.

In the third test case, the only good starting city is city $5$.

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

    mnmtm = float('inf')
    count = 0

    for i in range(n - 1, -1, -1):
        mnmtm = min(mnmtm, a[i])
        if mnmtm >= i + 1:
            count += 1

    print(count-1)

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

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

Think step by step.

Response:
```plaintext
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Speedbreaker
time_limit_ms: 2000
memory_limit_mb: 256
Description: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)

There are $n$ cities in a row, numbered $1, 2, \ldots, n$ left to right.

  * At time $1$, you conquer exactly one city, called the starting city.    * At time $2, 3, \ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it.

You win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of cities.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the deadlines for conquering the cities.

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

Output: For each test case, output a single integer: the number of starting cities that allow you to win.

Examples:
input:
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
output:
3
0
1

Note: In the first test case, cities $2$, $3$, and $4$ are good starting cities.

In the second test case, there are no good starting cities.

In the third test case, the only good starting city is city $5$.

Buggy Code:
for _ in range(int(input())):
    n = int(input())
    array = [int(i) for i in input().split()]
    l_a = [float('inf')]
    r_a = [float('inf')]
    result = 0
    for i in range(n):
        l_a.append(min(l_a[-1]-1, array[i]))
    for i in range(n-1,-1,-1):
        r_a.append(min(r_a[-1]-1, array[i]))
    r_a.reverse()

    for i in range(n):
        count = 1
        if l_a[i] >0 and r_a[i]>0:
            l,r = i,i+1
            while(l_a[l]>count and r_a[r]>count):
                if l_a[l]<r_a[r]:
                    l-=1
                    count+=1
                else:
                    r+=1
                    count +=1
                if l==0 and r==n :
                    result+=1
                    break

    print(result)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Speedbreaker
time_limit_ms: 2000
memory_limit_mb: 256
Description: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)

There are $n$ cities in a row, numbered $1, 2, \ldots, n$ left to right.

  * At time $1$, you conquer exactly one city, called the starting city.    * At time $2, 3, \ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it.

You win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of cities.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the deadlines for conquering the cities.

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

Output: For each test case, output a single integer: the number of starting cities that allow you to win.

Examples:
input:
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
output:
3
0
1

Note: In the first test case, cities $2$, $3$, and $4$ are good starting cities.

In the second test case, there are no good starting cities.

In the third test case, the only good starting city is city $5$.

Buggy Code:
import java.io.*;

/**
 * @version 1.0
 * @Author 杨雨彤
 * @Date 2024/10/13 19:57
 * @注释
 */
public class Main {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
    static StreamTokenizer st = new StreamTokenizer(br);

    static int nextInt() throws IOException {
        st.nextToken();
        return (int) st.nval;
    }

    static long nextLong() throws IOException {
        st.nextToken();
        return (long) st.nval;
    }

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

    public static void main(String[] args) throws IOException{
        int T=nextInt();
        while(T-->0) {
            int n = nextInt();//城市数
            int[] arr = new int[n + 1];
            int ll = 1;
            int rr = n;
            for (int i = 1; i <= n; i++) {
                arr[i] = nextInt();
                ll = Math.max(ll, i - arr[i] + 1);
                rr = Math.min(rr, i + arr[i] - 1);
            }
            int ans=rr-ll+1;
            int l=1;
            int r=n;
            a:while(l<=r) {
                if (arr[l] >= r - l + 1 && arr[r] >= r - l + 1) {
                    r--;
                    l++;
                } else if (arr[l] >= r - l + 1) {
                    r--;
                } else if (arr[r] >= r - l + 1) {
                    l++;
                } else {
                    ans=0;
                    break a;
                }
            }
                pw.println(ans);
               pw.flush();
            }
        pw.close();
    }
}

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

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

Think step by step.

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

1. **Understand the Bug**: The buggy code calculates the range of possible starting cities and then tries to validate these starting cities using a while loop. The key issue is in the validation logic, which may not correctly handle all edge cases.

2. **Identify Edge Cases**:
   - **Single City**: A test case with only one city.
   - **All Cities Have the Same Deadline**: A test case where all cities have the same deadline.
   - **Deadlines in Reverse Order**: A test case where the deadlines are in reverse order.
   - **Interleaved Deadlines**: A test case with interleaved deadlines to test the range calculation.
   - **No Valid Starting Cities**: A test case where no starting city allows you to win.

3. **Generate the Test Input**:
   - **Single City**: This should be straightforward and should always return 1.
   - **All Cities Have the Same Deadline**: This should return the number of cities.
   - **Deadlines in Reverse Order**: This should test the logic for handling decreasing deadlines.
   - **Interleaved Deadlines**: This should test the logic for handling mixed deadlines.
   - **No Valid Starting Cities**: This should test the logic for handling invalid deadlines.

Here is a small-scale test input that covers these edge cases:

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

### Explanation:
1. **Single City**:
   - Input: `1 1`
   - Expected Output: `1`
   - Reason: Only one city, and it must be conquered at time 1.

2. **All Cities Have the Same Deadline**:
   - Input: `3 1 1 1`
   - Expected Output: `3`
   - Reason: All cities have the same deadline, so any city can be a starting city.

3. **Deadlines in Reverse Order**:
   - Input: `3 3 2 1`
   - Expected Output: `1`
   - Reason: The only valid starting city is city 3.

4. **Interleaved Deadlines**:
   - Input: `3 1 3 2`
   - Expected Output: `1`
   - Reason: The only valid starting city is city 2.

5. **No Valid Starting Cities**:
   - Input: `3 2 1 3`
   - Expected Output: `0`
   - Reason: No starting city allows you to win.

This test input should help expose the bug in the provided code by covering various edge cases and challenging the logic of the solution.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Speedbreaker
time_limit_ms: 2000
memory_limit_mb: 256
Description: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)

There are $n$ cities in a row, numbered $1, 2, \ldots, n$ left to right.

  * At time $1$, you conquer exactly one city, called the starting city.    * At time $2, 3, \ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it.

You win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of cities.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the deadlines for conquering the cities.

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

Output: For each test case, output a single integer: the number of starting cities that allow you to win.

Examples:
input:
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
output:
3
0
1

Note: In the first test case, cities $2$, $3$, and $4$ are good starting cities.

In the second test case, there are no good starting cities.

In the third test case, the only good starting city is city $5$.

Buggy Code:
import java.io.*;

/**
 * @version 1.0
 * @Author 杨雨彤
 * @Date 2024/10/13 19:57
 * @注释
 */
public class Main {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
    static StreamTokenizer st = new StreamTokenizer(br);

    static int nextInt() throws IOException {
        st.nextToken();
        return (int) st.nval;
    }

    static long nextLong() throws IOException {
        st.nextToken();
        return (long) st.nval;
    }

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

    public static void main(String[] args) throws IOException{
        int T=nextInt();
        while(T-->0){
            int n=nextInt();//城市数
            int []arr=new int[n+1];
            int l=1;
            int r=n;
            for(int i=1;i<=n;i++){
                arr[i]=nextInt();
                l=Math.max(l,i-arr[i]+1);
                r=Math.min(r,i+arr[i]-1);
            }

            if(arr[1]<n&&arr[n]<n){
                pw.println("0");
            }else{
                int ans=r-l+1;
                pw.println(ans);
            }
        }
        pw.flush();
        pw.close();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases or scenarios that can highlight the issues in the logic. The code provided has a few potential issues:

1. **Incorrect Calculation of `l` and `r`**: The code calculates `l` and `r` based on the formula `i - arr[i] + 1` and `i + arr[i] - 1`, which might not correctly capture the valid range of starting cities.
2. **Edge Cases**: The code might not handle edge cases properly, such as when the array `arr` has specific values that make it difficult to win from any starting city.

Let's create a test case that can expose these issues:

### Test Case 1: All Cities Have the Same Deadline
This scenario can expose issues with the logic if the code incorrectly handles uniform deadlines.

### Test Case 2: Deadlines That Prevent Winning from Any Starting City
This scenario can expose issues if the code incorrectly handles cases where no starting city allows a win.

### Test Case 3: Deadlines That Allow Winning from a Specific Starting City
This scenario can expose issues if the code incorrectly handles cases where only one specific starting city allows a win.

### Test Case 4: Deadlines That Allow Winning from Multiple Starting Cities
This scenario can expose issues if the code incorrectly handles cases where multiple starting cities allow a win.

### Test Case 5: Edge Case with Small `n`
This scenario can expose issues if the code incorrectly handles small values of `n`.

### Chosen Test Case
For this example, let's focus on a test case that prevents winning from any starting city, as this is a clear edge case that can expose the bug:

```plaintext
1
5
5 5 5 5 5
```

### Explanation
- **Input**: `1 5 5 5 5 5`
- **Expected Output**: `0`
- **Reason**: In this test case, all cities have the same deadline of 5. Since the maximum time to conquer all cities is 5, and you need to conquer each city in order, it is impossible to start from any city and win. This should output `0`.

This test case will help expose the issue in the code where it incorrectly calculates the range of valid starting cities.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Speedbreaker
time_limit_ms: 2000
memory_limit_mb: 256
Description: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)

There are $n$ cities in a row, numbered $1, 2, \ldots, n$ left to right.

  * At time $1$, you conquer exactly one city, called the starting city.    * At time $2, 3, \ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it.

You win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of cities.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the deadlines for conquering the cities.

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

Output: For each test case, output a single integer: the number of starting cities that allow you to win.

Examples:
input:
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
output:
3
0
1

Note: In the first test case, cities $2$, $3$, and $4$ are good starting cities.

In the second test case, there are no good starting cities.

In the third test case, the only good starting city is city $5$.

Buggy Code:
import java.util.*;

public class D_Speedbreaker {

    public static void main(String[] args) {
        Scanner sc =new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0){
		    int n = sc.nextInt();
			int[] a = new int[n];
            for(int i=0;i<n;i++) a[i] = sc.nextInt();
            Map<Integer,Integer> first = new HashMap<>();
            Map<Integer,Integer> last = new HashMap<>();
            for(int i=0;i<n;i++){
                first.putIfAbsent(a[i], i);
                last.put(a[i],i);
            }
            int interval = 0;
            int l = -1, r = -1;
            boolean f = true;
            boolean flag = true;
            for(int i=1;i<=n;i++){
                if(first.containsKey(i)){
                    if(f){
                        f = false;
                        interval = i;
                        l = first.get(i);
                        r = last.get(i);
                    }
                    l = Math.min(l,first.get(i));
                    r = Math.max(r,last.get(i));
                    if(r-l+1>i){
                        flag = false;
                        break;
                    }
                }
            }
            if(!flag) System.out.println(0);
            else System.out.println(interval);
		}
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to consider edge cases and scenarios that can maximize the failure probability. The code seems to have issues with handling intervals and the logic for determining valid starting cities. Here are some key points to consider:

1. **Single City Case**: This should always return 1 because there's only one city and it can always be conquered on time.
2. **All Cities Have the Same Deadline**: This should return 1 because the only valid starting city is the one that can be conquered on time.
3. **Cities with Increasing Deadlines**: This should return 1 because the only valid starting city is the first city.
4. **Cities with Decreasing Deadlines**: This should return 1 because the only valid starting city is the last city.
5. **Mixed Deadlines**: This should return a number based on the valid starting cities that can conquer all cities on time.

Let's create a small-scale test input that covers these scenarios:

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

### Explanation:
1. **Single City Case**:
   - Input: `1 1`
   - Expected Output: `1`

2. **All Cities Have the Same Deadline**:
   - Input: `2 2 1`
   - Expected Output: `1`

3. **Cities with Decreasing Deadlines**:
   - Input: `3 3 2 1`
   - Expected Output: `1`

4. **Cities with Increasing Deadlines**:
   - Input: `3 1 2 3`
   - Expected Output: `1`

This test input should help expose any issues in the code's logic for determining valid starting cities.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Speedbreaker
time_limit_ms: 2000
memory_limit_mb: 256
Description: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)

There are $n$ cities in a row, numbered $1, 2, \ldots, n$ left to right.

  * At time $1$, you conquer exactly one city, called the starting city.    * At time $2, 3, \ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it.

You win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of cities.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the deadlines for conquering the cities.

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

Output: For each test case, output a single integer: the number of starting cities that allow you to win.

Examples:
input:
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
output:
3
0
1

Note: In the first test case, cities $2$, $3$, and $4$ are good starting cities.

In the second test case, there are no good starting cities.

In the third test case, the only good starting city is city $5$.

Buggy Code:
import java.util.*;

public class D_Speedbreaker {

    public static void main(String[] args) {
        Scanner sc =new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0){
		    int n = sc.nextInt();
			int[] a = new int[n];
            for(int i=0;i<n;i++) a[i] = sc.nextInt();
            Map<Integer,Integer> first = new HashMap<>();
            Map<Integer,Integer> last = new HashMap<>();
            for(int i=0;i<n;i++){
                first.putIfAbsent(a[i], i);
                last.put(a[i],i);
            }
            int interval = 0;
            int l = -1, r = -1;
            boolean f = true;
            boolean flag = true;
            for(int i=1;i<=n;i++){
                if(first.containsKey(i)){
                    if(f){
                        f = false;
                        interval = last.get(i)-first.get(i)+1;
                        l = first.get(i);
                        r = last.get(i);
                    }
                    l = Math.min(l,first.get(i));
                    r = Math.max(r,last.get(i));
                    if(r-l+1>i){
                        flag = false;
                        break;
                    }
                }
            }
            if(!flag) System.out.println(0);
            else System.out.println(interval);
		}
    }
}

**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 3 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Speedbreaker
time_limit_ms: 2000
memory_limit_mb: 256
Description: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)

There are $n$ cities in a row, numbered $1, 2, \ldots, n$ left to right.

  * At time $1$, you conquer exactly one city, called the starting city.    * At time $2, 3, \ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it.

You win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of cities.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the deadlines for conquering the cities.

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

Output: For each test case, output a single integer: the number of starting cities that allow you to win.

Examples:
input:
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
output:
3
0
1

Note: In the first test case, cities $2$, $3$, and $4$ are good starting cities.

In the second test case, there are no good starting cities.

In the third test case, the only good starting city is city $5$.

Buggy Code:
class Solution():
    def __init__(self):
        pass

    def getSegem(self):
        self.segs = {}
        for i in range(len(self.xs)):
            if self.xs[i] in self.segs.keys():
                old_seg = self.segs[self.xs[i]]
                self.segs[self.xs[i]] = (old_seg[0], i + 1)
            else:
                self.segs[self.xs[i]] = (i + 1, i + 1)

    def getVal(self):
        val = (1, self.n)
        for key in self.keys:
            if key < self.segs[key][1] - self.segs[key][0] + 1:
                return 0
            good_seg = (self.segs[key][1] - key + 1, self.segs[key][0] + key - 1)
            # print(good_seg)
            val = (max(good_seg[0], val[0]), min(good_seg[1], val[1]))
            if val[0] > val[1]:
                return 0
            else:
                pass
        return val[1] - val[0] + 1

    def solve(self):
        self.n = int(input())
        self.xs = tuple(map(int, input().split()))

        self.getSegem()
        self.keys = self.segs.keys()
        self.keys = sorted(self.keys)

        print(self.getVal())
        # print(self.segs)

def main():
    case_size = int(input())
    solution = Solution()
    for _ in range(case_size):
        solution.solve()

"""
4
3
5 4 5
3
4 5 4
10
3 3 3 3 4 1 2 3 4 5
9
17 89 92 42 29 92 14 70 45
"""

main()
# test()

**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
3
1 2 3
3
3 2 1
5
1 2 3 4 5
5
5 4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Speedbreaker
time_limit_ms: 2000
memory_limit_mb: 256
Description: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)

There are $n$ cities in a row, numbered $1, 2, \ldots, n$ left to right.

  * At time $1$, you conquer exactly one city, called the starting city.    * At time $2, 3, \ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it.

You win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of cities.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the deadlines for conquering the cities.

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

Output: For each test case, output a single integer: the number of starting cities that allow you to win.

Examples:
input:
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
output:
3
0
1

Note: In the first test case, cities $2$, $3$, and $4$ are good starting cities.

In the second test case, there are no good starting cities.

In the third test case, the only good starting city is city $5$.

Buggy Code:
for _ in range(int(input())):
    n = int(input())
    a = list(map(int,input().split(' ')))
    started = []
    mn = n
    mx = 0
    # m_num = 2*10**5 +1
    # p_sum = [0 for _ in range (m_num)]
    # print('m:',m_num)
    for i in range(n):
        l = max(0,i-a[i]+1)
        r = min(n,i+a[i]-1)
        started.append([l,r])
        mx = max(mx,l)
        mn = min(mn,r)
        # p_sum[a[i]] +=1
    l = mx
    r = mn
    cost = 1
    can =1
    while l>=0 and r<n:
        # print('leftL',l,'right:',r,'cost:',cost)
        cost +=2
        if l >0:
            if cost - a[l-1] >1:
                can = 0
                # print('me:', cost)
                break
        if r<=n-2:
            if cost - a[r+1] >1:
                can = 0
                # print('me:', cost)
                break

        r+=1
        l-=1
    # print('can:',can)

    # print(started)
    # print(mx,mn)
    print(mn-mx+1 if can else can)

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

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

Think step by step.

Response:
```plaintext
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Speedbreaker
time_limit_ms: 2000
memory_limit_mb: 256
Description: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)

There are $n$ cities in a row, numbered $1, 2, \ldots, n$ left to right.

  * At time $1$, you conquer exactly one city, called the starting city.    * At time $2, 3, \ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it.

You win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of cities.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the deadlines for conquering the cities.

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

Output: For each test case, output a single integer: the number of starting cities that allow you to win.

Examples:
input:
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
output:
3
0
1

Note: In the first test case, cities $2$, $3$, and $4$ are good starting cities.

In the second test case, there are no good starting cities.

In the third test case, the only good starting city is city $5$.

Buggy Code:
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define ll long long
#define vl vector<ll>
#define vs vector<string>
#define mll map<ll,ll>
#define pll pair<ll,ll>
#define sl set<ll>
#define pql priority_queue<ll>
#define rep(i,n) for(i=0;i<n;i++)
#define repr(i,n) for(i=n-1;i>=0;i--)
#define REP(i,a,b) for(i=a;i<b;i++)
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(), a.rend()
#define vsort(a) sort(all(a))
#define rvsort(a) sort(rall(a))
#define F first
#define endl "\n"
#define S second
#define pi acos(-1)
#define inf 10000000000000000
#define mod 1000000007
#define MA LLONG_MAX
#define MI LLONG_MIN
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

long long power(long long x,ll y)
{
long long res = 1;
x = x % mod;
while (y > 0)
{
if (y & 1)
res = (res * x) % mod;
y = y >> 1;
x = (x * x) % mod;
}
return res;
}

long modInverse(long long n)
{
    return power(n, mod - 2);
}

long long ncr(long long n, ll r)
{
if (n < r)
return 0;
if (r == 0)
return 1;
long long fac[n + 1];
fac[0] = 1;
for (int i = 1; i <= n; i++)
fac[i] = (fac[i - 1] * i) % mod;

return (fac[n] * modInverse(fac[r]) % mod*modInverse(fac[n - r])% mod)% mod;
}

struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};

void solve()
{
    ll n,i,j,ans=-1,leftans=0,rightans=0,firstleft,firstright,leftextra,rightextra;
    cin>>n;
    vector<ll> a(n),first(n+1,-1),last(n+1,-1);
    rep(i,n)
    {
        cin>>a[i];
        if(first[a[i]]==-1)
        first[a[i]]=i;
        last[a[i]]=i;
    }
    ll curleft=n,curright=0;
    for(i=1;i<=n;i++)
    {
        if(last[i]!=-1)
        {
            curleft=min(curleft,first[i]);
            curright=max(curright,last[i]);
            if(curright-curleft+1>i)
            {
                cout<<0<<endl;
                return;
            }
            if(ans==-1)
            {
                ans=curright-curleft+1;
                firstleft=first[i];
                firstright=last[i];
                leftextra=i-ans;
                rightextra=i-ans;
            }

            if(curleft<firstleft)
            {
                leftans=max(leftans,min(leftextra+leftans,firstleft-curleft));

                // firstleft=firstleft-temp;
                // leftans+=temp;
            }
            if(curright>firstright)
            {
                rightans=max(rightans,min(rightextra+rightans,curright-firstright));
                // firstright=firstright+temp;
                // rightans+=temp;
            }
            ll temp=max(0ll,i-(curright-curleft+1));
            leftextra=max(0ll,min(leftextra-leftans,temp));
            rightextra=max(0ll,min(rightextra-rightans,temp));
            //cout<<ans<<" "<<leftans<<" "<<rightans<<" "<<leftextra<<" "<<rightextra<<endl;
        }
    }
    cout<<ans+leftans+rightans<<endl;

}

int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL); cout.tie(NULL);
int 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 code, we need to consider edge cases and scenarios that can highlight potential issues. The code seems to handle the logic of finding the number of starting cities that allow you to win, but there might be edge cases where it fails.

### Step-by-Step Analysis:
1. **Single City Case**: The simplest case where there is only one city.
2. **Consecutive Deadlines**: All cities have consecutive deadlines.
3. **Non-Consecutive Deadlines**: Deadlines are not consecutive, which can cause issues in the logic.
4. **High Deadlines**: Deadlines are much higher than the city index.
5. **Low Deadlines**: Deadlines are much lower than the city index.
6. **Mixed Deadlines**: A combination of high and low deadlines.

### Chosen Test Input:
Let's create a test input that includes a mix of these scenarios to ensure the code is robust.

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

### Explanation:
1. **Single City Case**:
   - `1 1`: Only one city with a deadline of 1. The only starting city is 1, so the answer is 1.
2. **Consecutive Deadlines**:
   - `3 1 2 3`: Cities with deadlines 1, 2, and 3. All cities can be conquered in order, so the answer is 1 (starting city 1).
3. **Non-Consecutive Deadlines**:
   - `3 3 1 2`: Cities with deadlines 3, 1, and 2. The only starting city that allows you to win is city 2, so the answer is 1.
4. **Low Deadlines**:
   - `3 1 3 2`: Cities with deadlines 1, 3, and 2. The only starting city that allows you to win is city 1, so the answer is 1.
5. **High Deadlines**:
   - `5 5 4 3 2 1`: Cities with deadlines 5, 4, 3, 2, and 1. All cities can be conquered in reverse order, so the answer is 1 (starting city 5).

This test input covers a variety of scenarios and should help expose any bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Speedbreaker
time_limit_ms: 2000
memory_limit_mb: 256
Description: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)

There are $n$ cities in a row, numbered $1, 2, \ldots, n$ left to right.

  * At time $1$, you conquer exactly one city, called the starting city.    * At time $2, 3, \ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it.

You win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of cities.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the deadlines for conquering the cities.

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

Output: For each test case, output a single integer: the number of starting cities that allow you to win.

Examples:
input:
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
output:
3
0
1

Note: In the first test case, cities $2$, $3$, and $4$ are good starting cities.

In the second test case, there are no good starting cities.

In the third test case, the only good starting city is city $5$.

Buggy Code:
#include<bits/stdc++.h>
#define FOR(i,a,b) for(int i=(a); i<=(b); ++i)
#define ROF(i,a,b) for(int i=(a); i>=(b); --i)
#define sz(x) (int)(x).size()
#define all(x) (x).begin(),(x).end()
#define make_unique(x) sort(all(x)), (x).erase(unique(all(x)), (x).end())
#define x first
#define y second
#define pb push_back
#define eb emplace_back

using namespace std;

typedef pair<int,int> pii;
typedef long long i64;
typedef __int128 i128;
typedef tuple<int,int,int> iii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<i64> vl;
typedef long double ld;

int readInt(){ int a; scanf("%d",&a); return a;}
i64 readLong(){ i64 a; scanf("%lld",&a); return a;}
void readString(char *s){ scanf(" %s",s); }

const int mod = 1000000007;
int add(int a, int b){ return ((a+=b)>=mod)?a-mod:a; }
void adding(int &a, int b){ a = add(a,b); }
int mul(int a, int b){ return a*1ll*b%mod; }
int modpw(int a, int b){
    if(a == 0) return 0;
    if(b == 0) return 1;
    int ans = 1;
    for(i64 i=1,res=a; i<=b; i<<=1, res=mul(res,res)){
        if(b&i) ans = mul(ans,res);
    }
    return ans;
}

const int N = 200005;
int MiL[N*4],MiR[N*4];
int num[N];
void init(int l, int r, int now) {
    if(l == r) {
        MiL[now] = MiR[now] = num[l];
        return;
    }
    int mid = (l+r)/2;
    init(l,mid,now*2); init(mid+1,r,now*2+1);
    MiL[now] = min(MiL[now*2], MiL[now*2+1]-(mid-l+1));
    MiR[now] = min(MiR[now*2]-(r-mid), MiR[now*2+1]);
}
int queryL(int ll, int rr, int l, int r, int now) {
    if(l > r) return (1<<30);
    if(l > rr || r < ll) return (1<<30);
    if(ll <= l && r <= rr) {
        //printf("L (%d, %d) = %d\n",l,r,MiR[now]);
        return MiL[now];
    }
    int mid = (l+r)/2;
    if(rr <= mid) return queryL(ll, rr, l, mid, now*2);
    if(mid < ll) return queryL(ll, rr, mid+1, r, now*2+1);
    return min(queryL(ll,rr,l,mid,now*2), queryL(ll,rr,mid+1,r,now*2+1)-(mid-ll+1));
}
int queryR(int ll, int rr, int l, int r, int now) {
    if(l > r) return (1<<30);
    if(l > rr || r < ll) return (1<<30);
    if(ll <= l && r <= rr){
        //printf("R (%d, %d) = %d\n",l,r,MiR[now]);
        return MiR[now];
    }
    int mid = (l+r)/2;
    if(rr <= mid) return queryR(ll, rr, l, mid, now*2);
    if(mid < ll) return queryR(ll, rr, mid+1, r, now*2+1);
    return min(queryR(ll,rr,l,mid,now*2)-(rr-mid), queryR(ll,rr,mid+1,r,now*2+1));
}

void solve(){
    int n = readInt();
    FOR(i,1,n) num[i] = readInt();
    init(1,n,1);
    int l = 1, r = n;
    int ans = 0;
    while(l <= r) {
        //printf("%d %d\n",l,r);
        if(num[l] >= r-l+1) {
            int x = queryL(l+1, r, 1, n, 1);
            if(x > 1) {
                //printf("[%d]\n",x);
                //puts("F");
                ++ans;
            }
            ++l;
        } else if(num[r] >= r-l+1) {
            int x = queryR(l, r-1, 1, n, 1);
            if(x > 1) {
                //printf("[%d\n",x);
                //puts("F");
                ++ans;
            }
            --r;
        } else {
            break;
        }
    }
    printf("%d\n",ans);

    // clean up
    FOR(i,1,4*n) MiL[i] = MiR[i] = 0;
}

int main(){
    int t = 1;
    t = readInt(); // comment this out is q = 1
    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 focus on edge cases and scenarios that maximize the failure probability. The code is designed to determine the number of starting cities that allow you to win given the deadlines for conquering each city.

### Key Points to Consider:
1. **Deadlines and Conquest Order**: The code must correctly handle the deadlines and the order in which cities can be conquered.
2. **Edge Cases**: Consider edge cases such as:
   - All cities have the same deadline.
   - Deadlines are in strictly increasing or decreasing order.
   - Deadlines are very close to the city indices.
   - Deadlines are very far from the city indices.
3. **Boundary Conditions**: Test the boundaries of the input constraints, such as the smallest and largest possible values for `n` and `a_i`.

### Challenging Test Input:
Let's create a test case where the deadlines are such that the bug in the code is likely to be exposed. We will use a small-scale test input to ensure it is manageable.

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

### Explanation:
1. **First Test Case**:
   - Deadlines: `[1, 2, 3, 4, 5]`
   - This is a straightforward case where the deadlines are in increasing order. The correct answer should be 1 (starting from city 1).

2. **Second Test Case**:
   - Deadlines: `[5, 4, 3, 2, 1]`
   - This is a case where the deadlines are in decreasing order. The correct answer should be 1 (starting from city 5).

3. **Third Test Case**:
   - Deadlines: `[3, 3, 3, 3, 3]`
   - This is a case where all deadlines are the same. The correct answer should be 3 (starting from cities 3, 4, or 5).

This test input covers a range of scenarios that should help expose any bugs in the code, especially those related to handling different deadline patterns and boundary conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Speedbreaker
time_limit_ms: 2000
memory_limit_mb: 256
Description: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)

There are $n$ cities in a row, numbered $1, 2, \ldots, n$ left to right.

  * At time $1$, you conquer exactly one city, called the starting city.    * At time $2, 3, \ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it.

You win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of cities.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the deadlines for conquering the cities.

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

Output: For each test case, output a single integer: the number of starting cities that allow you to win.

Examples:
input:
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
output:
3
0
1

Note: In the first test case, cities $2$, $3$, and $4$ are good starting cities.

In the second test case, there are no good starting cities.

In the third test case, the only good starting city is city $5$.

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

public class B_Speedbreaker {
    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        int t = sc.nextInt();

        while(t-- > 0) {
            int n = sc.nextInt();
            int[] a = new int[n];

            for(int i = 0; i < n; i++)
                a[i] = sc.nextInt();

            int count = 0;
            for(int i = 0; i < n; i++) {
                HashSet<Integer> set = new HashSet<>();
                set.add(i);
                if(helper(set, a, 2)) {
                    count++;
                }
                set.remove(i);
            }

            System.out.println(count);
        }

    }
    private static boolean helper(HashSet<Integer> set, int[] a, int time) {
        if(set.size() == a.length)  return true;

        boolean ans = false;
        for(int i = 0; i < a.length; i++) {
            if(!set.contains(i)) {
                if(a[i] < time) return false;

                if(set.contains(i + 1) || set.contains(i - 1)) {
                    set.add(i);
                    ans |= helper(set, a, time + 1);
                    set.remove(i);
                }
            }
        }

        return ans;
    }
    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;
        public FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt() {
            return Integer.parseInt(next());
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on scenarios that are likely to cause the code to fail. The primary issues in the buggy code are:

1. **Inefficient Use of HashSet**: The use of `HashSet` to track conquered cities is inefficient and can lead to performance issues, especially for large inputs.
2. **Incorrect Recursion**: The recursive function `helper` does not correctly handle the adjacency constraints and can lead to incorrect results.

Given these issues, we should create a test case that:

1. **Tests the Edge Cases**: Cities with tight deadlines.
2. **Tests the Adjacency Constraints**: Cities that must be conquered in a specific order due to adjacency.
3. **Tests the Performance**: A small-scale test case that can still reveal the inefficiency.

Here is a small-scale test input that meets these criteria:

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

### Explanation:
- **Test Case**: There is one test case with 5 cities.
- **Deadlines**: The deadlines for the cities are `[5, 4, 3, 2, 1]`.
- **Expected Output**: The expected output is `1`.

### Why This Test Case is Challenging:
- **Tight Deadlines**: The deadlines are decreasing, which means the cities must be conquered in a specific order to meet the deadlines. This tests the code's ability to handle tight deadlines.
- **Specific Order**: The only valid starting city is city 5, because it must be conquered at time 1 to meet its deadline of 1. From there, the cities must be conquered in reverse order (5, 4, 3, 2, 1) to meet all the deadlines.
- **Performance**: While this is a small-scale test, it is sufficient to expose the inefficiency in the use of `HashSet` and the incorrect handling of adjacency constraints in the recursive function.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Speedbreaker
time_limit_ms: 2000
memory_limit_mb: 256
Description: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)

There are $n$ cities in a row, numbered $1, 2, \ldots, n$ left to right.

  * At time $1$, you conquer exactly one city, called the starting city.    * At time $2, 3, \ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it.

You win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of cities.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the deadlines for conquering the cities.

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

Output: For each test case, output a single integer: the number of starting cities that allow you to win.

Examples:
input:
3
6
6 3 3 3 5 5
6
5 6 4 1 4 5
9
8 6 4 2 1 3 5 7 9
output:
3
0
1

Note: In the first test case, cities $2$, $3$, and $4$ are good starting cities.

In the second test case, there are no good starting cities.

In the third test case, the only good starting city is city $5$.

Buggy Code:
import java.util.*;
import java.io.*;
public class SpeedBreaker {

    static HashSet<Integer> set;
    public static void counter(int arr[],int s,int e,int t){
        if( s > e || t > e - s + 1) return;
        if(t >= 1 && s == e){
            set.add(s);
        }

        counter(arr, s + 1, e, Math.min(t, arr[s]) - 1);
        counter(arr, s, e - 1, Math.min(t, arr[e]) - 1);
    }
    public static void main(String[] args) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw =  new PrintWriter(System.out);
        int tt = Integer.parseInt(br.readLine());
        while(tt-- > 0){
            int n = Integer.parseInt(br.readLine());
            int arr[] = new int[n];
            StringTokenizer st = new StringTokenizer(br.readLine());
            for(int i = 0; i < n;i++){
                arr[i] = Integer.parseInt(st.nextToken());
            }
            set = new HashSet<>();

            counter(arr, 0, n - 1, n);
            pw.println(set.size());

        }
        pw.flush();

    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios where the code might fail. The code uses a recursive approach to explore all possible starting cities and their corresponding conquering strategies. The main issues to consider are:

1. **Edge Cases with Small Inputs**: These can help us verify the basic logic.
2. **Inputs with Specific Patterns**: These can help us identify issues with the recursive logic and the use of the `HashSet`.
3. **Inputs with Large Constraints**: These can help us identify performance issues, although the test input should be small-scale.

Given the constraints and the nature of the problem, let's create a test input that includes a mix of these scenarios.

### Step-by-Step Reasoning:

1. **Small Input with a Single City**:
   - This should be a straightforward case where the starting city is the only city.
   - Expected output: 1

2. **Small Input with Two Cities**:
   - This should test the basic logic of conquering adjacent cities.
   - Expected output: 2

3. **Input with a Linear Deadline Pattern**:
   - This should test the logic where the deadlines are in a linear sequence.
   - Expected output: 1

4. **Input with a Non-Linear Deadline Pattern**:
   - This should test the logic where the deadlines are not in a linear sequence.
   - Expected output: 0

5. **Input with a Single Good Starting City**:
   - This should test the logic where only one city is a good starting city.
   - Expected output: 1

### Challenging Test Input:

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

### Explanation:

1. **Test Case 1**:
   - **Input**: `1 1`
   - **Output**: `1`
   - **Reason**: Only one city, and the deadline is 1, so the only starting city is valid.

2. **Test Case 2**:
   - **Input**: `2 1 2`
   - **Output**: `2`
   - **Reason**: Both cities can be valid starting cities because the deadlines are 1 and 2, respectively.

3. **Test Case 3**:
   - **Input**: `3 1 2 3`
   - **Output**: `1`
   - **Reason**: Only city 1 can be a valid starting city because the deadlines are in a linear sequence.

4. **Test Case 4**:
   - **Input**: `4 1 3 2 4`
   - **Output**: `0`
   - **Reason**: No city can be a valid starting city because the deadlines are not in a linear sequence.

5. **Test Case 5**:
   - **Input**: `5 3 1 2 4 5`
   - **Output**: `1`
   - **Reason**: Only city 2 can be a valid starting city because it allows conquering the cities in the required order.

This test input should help expose any bugs in the provided code by covering a variety of edge cases and scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Stickogon
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:

- Each side of a polygon is formed by exactly one stick.
- No stick is used in more than $$$1$$$ polygon.

Note: Sticks cannot be broken.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the number of sticks available.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 100$$$) — the stick lengths.

Output: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.

Examples:
input:
4
1
1
2
1 1
6
2 2 3 3 3 3
9
4 2 2 2 2 4 2 4 4
output:
0
0
1
2

Note: In the first test case, we only have one stick, hence we can't form any polygon.

In the second test case, the two sticks aren't enough to form a polygon either.

In the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.

In the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.

Buggy Code:
#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
int n,m,ans,sum;
int a[1000];
int main()
{
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>m;
		if(m<3){
			for(int k=0;k<m;k++){
			cin>>a[k];
			}
			cout<<0<<endl;
			continue;
		}
		if(m>2){
			sum=0;
			ans=0;
			for(int k=0;k<m;k++){
			cin>>a[k];
			}
			sort(a,a+m);
			for(int s=1;s<m;s++){
				if(a[s]==a[s-1]){
					ans+=1;
				//	printf("ans=%d\n",ans);
				}
				if(a[s]!=a[s-1]){
					if(ans>2){
						sum+=1;
					//	printf("sum=%d\n",sum);
						ans=0;
					}
					else ans=0;
				}
				if(s==m-1&&ans>1) sum+=1;
			}
		}
		cout<<sum<<endl;
	}
	return 0;
 }

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Stickogon
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:

- Each side of a polygon is formed by exactly one stick.
- No stick is used in more than $$$1$$$ polygon.

Note: Sticks cannot be broken.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the number of sticks available.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 100$$$) — the stick lengths.

Output: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.

Examples:
input:
4
1
1
2
1 1
6
2 2 3 3 3 3
9
4 2 2 2 2 4 2 4 4
output:
0
0
1
2

Note: In the first test case, we only have one stick, hence we can't form any polygon.

In the second test case, the two sticks aren't enough to form a polygon either.

In the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.

In the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.

Buggy Code:
#include<iostream>
#include<algorithm>
using namespace std;

int main()
{
    short int t;
    cin >> t;

    while(t--)
    {
       short int n, i, j, k, cot=0;
       cin >> n;
       short int stick[n];

       for(i=0; i<n; i++)
       {
           cin >> stick[i];
       }

       if(n>2)
       {
           sort(stick, stick+n);

           for(i=0; i<n-1; )
           {
               k=1;
               for(j=i+1; j<n; j++)
               {
                   if(stick[i]==stick[j])
                   {
                       k++;
                   }
                   else
                   {
                       break;
                   }
               }
               if(k>=3)
               {
                   cot++;
               }
               i=j;
           }

           cout << cot << '\n';

       }
       else
       {
           cout << "0" << '\n';
       }

    }

    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, let's consider the following points:

1. **Sorting and Counting**: The code sorts the sticks and then counts the number of consecutive equal sticks. It increments the count (`cot`) if the count of equal sticks is 3 or more.

2. **Bug Analysis**: The bug in the code is that it only checks for the presence of at least 3 consecutive equal sticks but does not consider the possibility of forming multiple polygons with the same stick length. For example, if there are 6 sticks of the same length, it should be possible to form two polygons (e.g., two triangles).

3. **Test Case**: We need a test case where the number of sticks of the same length is a multiple of the number of sides of a polygon (3, 4, 5, etc.) and the total number of sticks is more than 3. This will ensure that the bug is exposed.

4. **Example**: Let's consider a test case with 9 sticks of the same length. This should allow forming three triangles (3 sides each).

Here is the test input:

```plaintext
1
9
2 2 2 2 2 2 2 2 2
```

This test case will expose the bug because the code will only count the presence of at least 3 consecutive equal sticks once, instead of counting the possibility of forming multiple polygons with the same stick length. The expected output should be 3 (three triangles), but the buggy code will output 1.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Stickogon
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:

- Each side of a polygon is formed by exactly one stick.
- No stick is used in more than $$$1$$$ polygon.

Note: Sticks cannot be broken.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the number of sticks available.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 100$$$) — the stick lengths.

Output: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.

Examples:
input:
4
1
1
2
1 1
6
2 2 3 3 3 3
9
4 2 2 2 2 4 2 4 4
output:
0
0
1
2

Note: In the first test case, we only have one stick, hence we can't form any polygon.

In the second test case, the two sticks aren't enough to form a polygon either.

In the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.

In the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.

Buggy Code:
#include <iostream>
using namespace std;
int main()
{
int n;
cin>>n;
int e[n];
for(int i=0;i<n;i++)
{
    int m;
    cin>>m;
    int a[m];
    for(int j=0;j<m;j++)
    {
        cin>>a[j];
    }
    int y[101]={0};
    for(int k=0;k<m;k++)
    {
        y[a[k]]++;
    }
    int u;
    int s=0;
    for(int z=0;z<m;z++)
    {
        u=y[z]/4;
        s+=u;
    }
    e[i]=s;
}
for(int q=0;q<n;q++)
{
    cout<<e[q]<<endl;
}
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on the logic that counts the number of regular polygons. The buggy code only considers polygons with 4 sides (squares), which is incorrect. We need to test cases where the code should consider other types of polygons (e.g., triangles, pentagons, etc.).

Let's create a test input that includes sticks that can form various types of polygons, but the code will only count squares, leading to an incorrect result.

### Test Input:
```plaintext
3
4
1 1 1 1
5
2 2 2 2 2
6
3 3 3 3 3 3
```

### Explanation:
1. **Test Case 1:**
   - Input: `4 1 1 1 1`
   - Expected Output: `1`
   - Actual Output (Buggy Code): `1`
   - This case correctly forms one square, but we need to ensure the code handles other polygons as well.

2. **Test Case 2:**
   - Input: `5 2 2 2 2 2`
   - Expected Output: `1`
   - Actual Output (Buggy Code): `1`
   - This case correctly forms one pentagon, but the buggy code will not count it as a valid polygon.

3. **Test Case 3:**
   - Input: `6 3 3 3 3 3 3`
   - Expected Output: `2`
   - Actual Output (Buggy Code): `1`
   - This case can form two triangles, but the buggy code will only count one square, leading to an incorrect result.

By using these test cases, we can expose the bug in the code where it only considers squares and ignores other types of polygons.