**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Firefly's Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: Firefly is given an array $$$a$$$ of length $$$n$$$. Let $$$c_i$$$ denote the $$$i$$$'th cyclic shift$$$^{\text{∗}}$$$ of $$$a$$$. She creates a new array $$$b$$$ such that $$$b = c_1 + c_2 + \dots + c_n$$$ where $$$+$$$ represents concatenation$$$^{\text{†}}$$$.

Then, she asks you $$$q$$$ queries. For each query, output the sum of all elements in the subarray of $$$b$$$ that starts from the $$$l$$$-th element and ends at the $$$r$$$-th element, inclusive of both ends.

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n, q \leq 2 \cdot 10^5$$$) — the length of the array and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq 10^6$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n^2$$$) — the left and right bounds of the query.

Note that the large inputs may require the use of 64-bit integers.

It is guaranteed that the sum of $$$n$$$ does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ does not exceed $$$2 \cdot 10^5$$$.

Output: For each query, output the answer on a new line.

Examples:
input:
5
3 3
1 2 3
1 9
3 5
8 8
5 5
4 8 3 2 4
1 14
3 7
7 10
2 11
1 25
1 1
6
1 1
5 7
3 1 6 10 4
3 21
6 17
2 2
1 5
1 14
9 15
12 13
5 3
4 9 10 10 1
20 25
3 11
20 22
output:
18
8
1
55
20
13
41
105
6
96
62
1
24
71
31
14
44
65
15

Note: For the first test case, $$$b = [1, 2, 3, 2, 3, 1, 3, 1, 2]$$$.

Buggy Code:
#include <iostream>
using namespace std;
long long a[400010];

int main() {
	int t, n, q, l, r;
t=1;
	cin >> t;
	while (t--) {
		cin >> n >> q;
		long long sum = 0;
		for (int i = 1; i <= n; i++) {
			cin >> a[i];
			sum += a[i];
			a[n + i] = a[i];
		}
		for (int i = 1; i <= 2 * n; i++) {
			a[i] += a[i - 1];
		}
		long long ans = 0;
		while (q--) {
			ans = 0;
			cin >> l >> r;
			int num = (r - 1) / n - (l - 1) / n - 1;
			ans += sum * num;
			int f = (l - 1) / n + 1;
			int m = l - (f - 1) * n ;
			f = (f + m - 1) % n;
			m = n - m + 1;
			if (f == 0)f = n;
			ans += a[f + m - 1] - a[f - 1];

			f = (r - 1) / n + 1;
			m = r - (f - 1) * n;
			ans += a[f + m - 1] - a[f - 1];

			cout << ans << 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
1
5 5
1 2 3 4 5
1 25
1 5
6 10
11 15
16 20
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Firefly's Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: Firefly is given an array $$$a$$$ of length $$$n$$$. Let $$$c_i$$$ denote the $$$i$$$'th cyclic shift$$$^{\text{∗}}$$$ of $$$a$$$. She creates a new array $$$b$$$ such that $$$b = c_1 + c_2 + \dots + c_n$$$ where $$$+$$$ represents concatenation$$$^{\text{†}}$$$.

Then, she asks you $$$q$$$ queries. For each query, output the sum of all elements in the subarray of $$$b$$$ that starts from the $$$l$$$-th element and ends at the $$$r$$$-th element, inclusive of both ends.

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n, q \leq 2 \cdot 10^5$$$) — the length of the array and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq 10^6$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n^2$$$) — the left and right bounds of the query.

Note that the large inputs may require the use of 64-bit integers.

It is guaranteed that the sum of $$$n$$$ does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ does not exceed $$$2 \cdot 10^5$$$.

Output: For each query, output the answer on a new line.

Examples:
input:
5
3 3
1 2 3
1 9
3 5
8 8
5 5
4 8 3 2 4
1 14
3 7
7 10
2 11
1 25
1 1
6
1 1
5 7
3 1 6 10 4
3 21
6 17
2 2
1 5
1 14
9 15
12 13
5 3
4 9 10 10 1
20 25
3 11
20 22
output:
18
8
1
55
20
13
41
105
6
96
62
1
24
71
31
14
44
65
15

Note: For the first test case, $$$b = [1, 2, 3, 2, 3, 1, 3, 1, 2]$$$.

Buggy Code:
import java.io.*;

public class Main {
		static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		static StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
		public static void main(String[] args) throws IOException {
				in.nextToken();int t=(int)in.nval;
				while(t-->0){
						in.nextToken();int n=(int)in.nval;
						in.nextToken();int q=(int)in.nval;
						long[] sum=new long[n+1];
						long[] arr=new long[n+1];
						for(int i=1;i<=n;i++){
								in.nextToken();int a=(int)in.nval;
								sum[i]=sum[i-1]+a;
								arr[i-1]=a;
						}
						while(q-->0){
								in.nextToken();int l=(int)in.nval-1;
								in.nextToken();int r=(int)in.nval;
								long sumR=sum[n]*(r/n);
								int i=r%n;
								while(--i>=0){
										sumR+=arr[(i+(r/n))%n];
								}
								long sumL=sum[n]*(l/n);
								i=l%n;
								while(--i>=0){
										sumL+=arr[(i+(l/n))%n];
								}
								out.println(sumR-sumL);
						}
				}
				out.close();
		}
}

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

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

Think step by step.

Response:
```plaintext
1
5 3
1 2 3 4 5
1 25
2 24
13 17
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Firefly's Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: Firefly is given an array $$$a$$$ of length $$$n$$$. Let $$$c_i$$$ denote the $$$i$$$'th cyclic shift$$$^{\text{∗}}$$$ of $$$a$$$. She creates a new array $$$b$$$ such that $$$b = c_1 + c_2 + \dots + c_n$$$ where $$$+$$$ represents concatenation$$$^{\text{†}}$$$.

Then, she asks you $$$q$$$ queries. For each query, output the sum of all elements in the subarray of $$$b$$$ that starts from the $$$l$$$-th element and ends at the $$$r$$$-th element, inclusive of both ends.

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n, q \leq 2 \cdot 10^5$$$) — the length of the array and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq 10^6$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n^2$$$) — the left and right bounds of the query.

Note that the large inputs may require the use of 64-bit integers.

It is guaranteed that the sum of $$$n$$$ does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ does not exceed $$$2 \cdot 10^5$$$.

Output: For each query, output the answer on a new line.

Examples:
input:
5
3 3
1 2 3
1 9
3 5
8 8
5 5
4 8 3 2 4
1 14
3 7
7 10
2 11
1 25
1 1
6
1 1
5 7
3 1 6 10 4
3 21
6 17
2 2
1 5
1 14
9 15
12 13
5 3
4 9 10 10 1
20 25
3 11
20 22
output:
18
8
1
55
20
13
41
105
6
96
62
1
24
71
31
14
44
65
15

Note: For the first test case, $$$b = [1, 2, 3, 2, 3, 1, 3, 1, 2]$$$.

Buggy Code:
import java.util.Scanner;

public class Main {
    public static long compute(int start, int steps, long[] prefix) {
        if (steps == prefix.length) {
            return prefix[prefix.length - 1];
        }
        int end = (start + steps - 1 + prefix.length) % prefix.length;
        long prev = (start == 0) ? 0 : prefix[start - 1];
        if (end >= start) {
            return prefix[end] - prev;
        } else {
            return prefix[prefix.length - 1] - prev + prefix[end];
        }
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        for (int i = 0; i < t; i++) {
            int n = scanner.nextInt();
            int q = scanner.nextInt();
            int[] a = new int[n];
            for (int j = 0; j < n; j++) {
                a[j] = scanner.nextInt();
            }

            // prefix sum
            long[] prefix = new long[n];
            prefix[0] = a[0];
            for (int j = 1; j < n; j++) {
                prefix[j] = prefix[j - 1] + a[j];
            }

            for (int j = 0; j < q; j++) {
                int l = scanner.nextInt() - 1;
                int r = scanner.nextInt() - 1;

                long res = 0;

                int l_cycle = l / n;
                int r_cycle = r / n;

                // full arrays in between
                res += Math.max(0, (r_cycle - l_cycle - 1)) * prefix[n - 1];

                // if they are in the same cycle:
                if (l_cycle == r_cycle) {
                    int steps = r - l + 1;
                    int start = (l + l_cycle) % n;
                    res += compute(start, steps, prefix);
                } else {
                    int l_start = (l % n + l_cycle) % n;
                    int l_steps = n - l % n;
                    res += compute(l_start, l_steps, prefix);

                    int r_end = (r + r_cycle) % n;
                    int r_steps = r % n + 1;
                    int r_start = (r_end - r_steps + n + 1) % n;
                    res += compute(r_start, r_steps, prefix);

                }
                System.out.println(res);
            }
        }
        scanner.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
2
5 5
1 2 3 4 5
1 25
1 5
6 10
11 15
21 25
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Firefly's Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: Firefly is given an array $$$a$$$ of length $$$n$$$. Let $$$c_i$$$ denote the $$$i$$$'th cyclic shift$$$^{\text{∗}}$$$ of $$$a$$$. She creates a new array $$$b$$$ such that $$$b = c_1 + c_2 + \dots + c_n$$$ where $$$+$$$ represents concatenation$$$^{\text{†}}$$$.

Then, she asks you $$$q$$$ queries. For each query, output the sum of all elements in the subarray of $$$b$$$ that starts from the $$$l$$$-th element and ends at the $$$r$$$-th element, inclusive of both ends.

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n, q \leq 2 \cdot 10^5$$$) — the length of the array and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq 10^6$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n^2$$$) — the left and right bounds of the query.

Note that the large inputs may require the use of 64-bit integers.

It is guaranteed that the sum of $$$n$$$ does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ does not exceed $$$2 \cdot 10^5$$$.

Output: For each query, output the answer on a new line.

Examples:
input:
5
3 3
1 2 3
1 9
3 5
8 8
5 5
4 8 3 2 4
1 14
3 7
7 10
2 11
1 25
1 1
6
1 1
5 7
3 1 6 10 4
3 21
6 17
2 2
1 5
1 14
9 15
12 13
5 3
4 9 10 10 1
20 25
3 11
20 22
output:
18
8
1
55
20
13
41
105
6
96
62
1
24
71
31
14
44
65
15

Note: For the first test case, $$$b = [1, 2, 3, 2, 3, 1, 3, 1, 2]$$$.

Buggy Code:
#include <iostream>

using namespace std;
int long long t,n,p1,p2,st,dr,v[200100],p,r,q,s;
int main()
{
    cin>>t;
    while (t)
    {
        t--;
        cin>>n>>q;
        for (int i=1; i<=n; i++)
        cin>>v[i];

        while (q)
        {
            r=0;
            q--;
            cin>>p1>>p2;
            st=p1%n;
            if (st==0) st=n;
            dr=n;
            p=(p1-1)/n+1;
            p=p+st-1;
            p=p%n;
            if (p==0) p=n;
            for ( int i=st; i<=dr && p1<=p2; i++)
            {
                r=r+v[p];
                p=p%n+1;
                p1++;
            }

            dr=p2%n;
            if (dr==0) dr=n;
            st=1;
            p=(p2-1)/n+1;
            for ( int i=st; i<=dr && p2>=p1; i++)
            {
                r=r+v[p];
                p=p%n+1;
                p2--;
            }

            s=0;
            for (int i=1; i<=n; i++) s=s+v[i];
            r=r+s*((p2-p1+1)/n);

        cout<<r<<'\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:
```plaintext
1
5 3
1 2 3 4 5
1 25
13 19
21 25
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Firefly's Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: Firefly is given an array $$$a$$$ of length $$$n$$$. Let $$$c_i$$$ denote the $$$i$$$'th cyclic shift$$$^{\text{∗}}$$$ of $$$a$$$. She creates a new array $$$b$$$ such that $$$b = c_1 + c_2 + \dots + c_n$$$ where $$$+$$$ represents concatenation$$$^{\text{†}}$$$.

Then, she asks you $$$q$$$ queries. For each query, output the sum of all elements in the subarray of $$$b$$$ that starts from the $$$l$$$-th element and ends at the $$$r$$$-th element, inclusive of both ends.

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n, q \leq 2 \cdot 10^5$$$) — the length of the array and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq 10^6$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n^2$$$) — the left and right bounds of the query.

Note that the large inputs may require the use of 64-bit integers.

It is guaranteed that the sum of $$$n$$$ does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ does not exceed $$$2 \cdot 10^5$$$.

Output: For each query, output the answer on a new line.

Examples:
input:
5
3 3
1 2 3
1 9
3 5
8 8
5 5
4 8 3 2 4
1 14
3 7
7 10
2 11
1 25
1 1
6
1 1
5 7
3 1 6 10 4
3 21
6 17
2 2
1 5
1 14
9 15
12 13
5 3
4 9 10 10 1
20 25
3 11
20 22
output:
18
8
1
55
20
13
41
105
6
96
62
1
24
71
31
14
44
65
15

Note: For the first test case, $$$b = [1, 2, 3, 2, 3, 1, 3, 1, 2]$$$.

Buggy Code:
#include <bits/stdc++.h>
#define ll long long
using namespace std;
int main()
{
    ll t;
    cin >> t;
    while(t--)
    {
        ll n,q;
        cin >> n >> q;
        vector<ll> a(n + n),pfsum(n + n);
        for (ll i = 0;i < n + n;i++)
        {
            if (i < n)
            {
                cin >> a[i];
                a[i + n] = a[i];
            }
            if (i == 0) pfsum[i] = a[i];
            else pfsum[i] = pfsum[i - 1] + a[i];
        }
        //for (auto& el : pfsum) cout << el << ' ';
        //cout << endl;
        while(q--)
        {
            ll l,r;
            cin >> l >> r;
            l--;
            ll right,left;
            if (r < n) right = pfsum[r - 1];
            else
            {
                if (r % n == 0) right = pfsum[n - 1] * r / n;
                else right = r / n * pfsum[n - 1] + pfsum[r / n + r % n - 1] - pfsum[r / n - 1];
            }
            if (l < n) left = pfsum[l - 1];
            else
            {
                if (l % n == 0) left = pfsum[n - 1] * l / n;
                else left = l / n * pfsum[n - 1] + pfsum[l / n + l % n - 1] - pfsum[l / n - 1];
            }
            if (l == 0) left = 0;
            if (q == 200000 - 2728) cout << left << endl;
            cout << right - left << "\n";
        }
    }
    return 0;
}

// k1 3 5 6 8 13 k20
//  2 2 1 2 5 7
//  k2 4 15 17 19 k25
//  k2 4 5 7

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

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

Think step by step.

Response:
```plaintext
1
5 5
1 2 3 4 5
1 25
6 10
15 20
21 25
1 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Firefly's Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: Firefly is given an array $$$a$$$ of length $$$n$$$. Let $$$c_i$$$ denote the $$$i$$$'th cyclic shift$$$^{\text{∗}}$$$ of $$$a$$$. She creates a new array $$$b$$$ such that $$$b = c_1 + c_2 + \dots + c_n$$$ where $$$+$$$ represents concatenation$$$^{\text{†}}$$$.

Then, she asks you $$$q$$$ queries. For each query, output the sum of all elements in the subarray of $$$b$$$ that starts from the $$$l$$$-th element and ends at the $$$r$$$-th element, inclusive of both ends.

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n, q \leq 2 \cdot 10^5$$$) — the length of the array and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq 10^6$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n^2$$$) — the left and right bounds of the query.

Note that the large inputs may require the use of 64-bit integers.

It is guaranteed that the sum of $$$n$$$ does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ does not exceed $$$2 \cdot 10^5$$$.

Output: For each query, output the answer on a new line.

Examples:
input:
5
3 3
1 2 3
1 9
3 5
8 8
5 5
4 8 3 2 4
1 14
3 7
7 10
2 11
1 25
1 1
6
1 1
5 7
3 1 6 10 4
3 21
6 17
2 2
1 5
1 14
9 15
12 13
5 3
4 9 10 10 1
20 25
3 11
20 22
output:
18
8
1
55
20
13
41
105
6
96
62
1
24
71
31
14
44
65
15

Note: For the first test case, $$$b = [1, 2, 3, 2, 3, 1, 3, 1, 2]$$$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define R return
#define fi first
#define se second
#define yes cout<<"YES\n";
#define no cout<<"NO\n";
const int N=4e5+7;
ll T=1;
ll a[N];
ll n,q,sum=0;
ll get(ll x)
{
    ll Amount_n = x/n;
    ll rem = x%n;
    if(!rem)
        return Amount_n * a[n-1];
    return Amount_n * a[n-1] + a[rem - 1 + Amount_n] - (Amount_n>0)*a[Amount_n - 1];
}
void solve()
{
    sum=0;
    cin >> n >> q;
    for(int i=0;i<n;i++){
        cin >> a[i];
        a[i+n]=a[i];
        sum+=a[i];
   }
    for(int i=1;i<=2*n;i++)
        a[i]+=a[i-1];
    while(q--){
        int l,r;
        cin >> l >> r;
        cout << get(r) - get(l-1) << endl;
    }

}
int main()
{
    ios::sync_with_stdio(NULL);
    cin.tie(0);
    cout.tie(0);
    cin>>T;
    while(T--)
        solve();
    R 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5 5
1 2 3 4 5
1 25
5 15
10 10
15 20
20 25
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Firefly's Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: Firefly is given an array $$$a$$$ of length $$$n$$$. Let $$$c_i$$$ denote the $$$i$$$'th cyclic shift$$$^{\text{∗}}$$$ of $$$a$$$. She creates a new array $$$b$$$ such that $$$b = c_1 + c_2 + \dots + c_n$$$ where $$$+$$$ represents concatenation$$$^{\text{†}}$$$.

Then, she asks you $$$q$$$ queries. For each query, output the sum of all elements in the subarray of $$$b$$$ that starts from the $$$l$$$-th element and ends at the $$$r$$$-th element, inclusive of both ends.

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n, q \leq 2 \cdot 10^5$$$) — the length of the array and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq 10^6$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n^2$$$) — the left and right bounds of the query.

Note that the large inputs may require the use of 64-bit integers.

It is guaranteed that the sum of $$$n$$$ does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ does not exceed $$$2 \cdot 10^5$$$.

Output: For each query, output the answer on a new line.

Examples:
input:
5
3 3
1 2 3
1 9
3 5
8 8
5 5
4 8 3 2 4
1 14
3 7
7 10
2 11
1 25
1 1
6
1 1
5 7
3 1 6 10 4
3 21
6 17
2 2
1 5
1 14
9 15
12 13
5 3
4 9 10 10 1
20 25
3 11
20 22
output:
18
8
1
55
20
13
41
105
6
96
62
1
24
71
31
14
44
65
15

Note: For the first test case, $$$b = [1, 2, 3, 2, 3, 1, 3, 1, 2]$$$.

Buggy Code:
t=int(input())
for _ in range(t):
    n,m=list(map(int,input().split()))
    a=list(map(int,input().split()))
    length = n
    cnt=sum(a)
    def sum_a(q):
        ans = q // length * cnt
        i, j = 0, (q - 1) % length + 1
        k = (q - 1) // length
        if j % length == 0:
            return ans
        i, j = i + k, (j + k) % length
        if j >= i:
            return ans + sum(a[i:j])
        else:
            return ans + sum(a[0:j]) + sum(a[i:])
    for _ in range(m):
        q1,q2=list(map(int,input().split()))

        print(sum_a(q2)-sum_a(q1-1))

**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
1
5 5
1 2 3 4 5
1 25
13 19
20 24
1 12
11 15
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Firefly's Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: Firefly is given an array $$$a$$$ of length $$$n$$$. Let $$$c_i$$$ denote the $$$i$$$'th cyclic shift$$$^{\text{∗}}$$$ of $$$a$$$. She creates a new array $$$b$$$ such that $$$b = c_1 + c_2 + \dots + c_n$$$ where $$$+$$$ represents concatenation$$$^{\text{†}}$$$.

Then, she asks you $$$q$$$ queries. For each query, output the sum of all elements in the subarray of $$$b$$$ that starts from the $$$l$$$-th element and ends at the $$$r$$$-th element, inclusive of both ends.

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n, q \leq 2 \cdot 10^5$$$) — the length of the array and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq 10^6$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n^2$$$) — the left and right bounds of the query.

Note that the large inputs may require the use of 64-bit integers.

It is guaranteed that the sum of $$$n$$$ does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ does not exceed $$$2 \cdot 10^5$$$.

Output: For each query, output the answer on a new line.

Examples:
input:
5
3 3
1 2 3
1 9
3 5
8 8
5 5
4 8 3 2 4
1 14
3 7
7 10
2 11
1 25
1 1
6
1 1
5 7
3 1 6 10 4
3 21
6 17
2 2
1 5
1 14
9 15
12 13
5 3
4 9 10 10 1
20 25
3 11
20 22
output:
18
8
1
55
20
13
41
105
6
96
62
1
24
71
31
14
44
65
15

Note: For the first test case, $$$b = [1, 2, 3, 2, 3, 1, 3, 1, 2]$$$.

Buggy Code:
t = int(input())

for _ in range(t):
    n, q = map(int, input().split())

    a = [int(x) for x in input().split()]
    total_a = sum(a)

    for _ in range(q):
        l, r = map(int, input().split())
        left_cycle = (l-1) // n
        right_cycle = (r-1) // n
        left_idx = (l - 1) % n
        right_idx = (r - 1)%n + 1

        if left_cycle == right_cycle:
            b = a[left_cycle:] + a[:left_cycle]
            print(sum(b[left_idx:right_idx]))
        else:
            tmp = total_a * (right_cycle - left_cycle - 1)
            b1 = a[left_cycle:] + a[:left_cycle]
            b2 = a[right_cycle:] + a[:right_cycle]
            print(sum(b1[left_idx:]) + sum(b2[:right_idx]) + tmp)

**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
5 5
1 2 3 4 5
1 25
6 10
16 20
21 25
1 5
2 2
1 2
1 4
3 4
3 3 3
1 9
3 5
8 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Firefly's Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: Firefly is given an array $$$a$$$ of length $$$n$$$. Let $$$c_i$$$ denote the $$$i$$$'th cyclic shift$$$^{\text{∗}}$$$ of $$$a$$$. She creates a new array $$$b$$$ such that $$$b = c_1 + c_2 + \dots + c_n$$$ where $$$+$$$ represents concatenation$$$^{\text{†}}$$$.

Then, she asks you $$$q$$$ queries. For each query, output the sum of all elements in the subarray of $$$b$$$ that starts from the $$$l$$$-th element and ends at the $$$r$$$-th element, inclusive of both ends.

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n, q \leq 2 \cdot 10^5$$$) — the length of the array and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq 10^6$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n^2$$$) — the left and right bounds of the query.

Note that the large inputs may require the use of 64-bit integers.

It is guaranteed that the sum of $$$n$$$ does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ does not exceed $$$2 \cdot 10^5$$$.

Output: For each query, output the answer on a new line.

Examples:
input:
5
3 3
1 2 3
1 9
3 5
8 8
5 5
4 8 3 2 4
1 14
3 7
7 10
2 11
1 25
1 1
6
1 1
5 7
3 1 6 10 4
3 21
6 17
2 2
1 5
1 14
9 15
12 13
5 3
4 9 10 10 1
20 25
3 11
20 22
output:
18
8
1
55
20
13
41
105
6
96
62
1
24
71
31
14
44
65
15

Note: For the first test case, $$$b = [1, 2, 3, 2, 3, 1, 3, 1, 2]$$$.

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

public class Solution {

    static final long mod = (long) 1e9 + 7;
    // down, right, left, top, TR, BR, TL, BT
    static int dx[] = {0, 1, -1, 0, 1, 1, -1, -1};
    static int dy[] = {-1, 0, 0, 1, -1, 1, -1, 1};
    static FastReader in;
    static FastWriter out;

    public static void main(String[] args) {
        try {
            in = new FastReader();
            out = new FastWriter();

            long ttt = in.nextLong();
            while (ttt-- != 0l) {

                int n = in.nextInt(), q = in.nextInt();
                long arr[] = new long[n];
                for (int i = 0; i < n; i++) {
                    arr[i] = in.nextLong();
                }

                long sum[] = new long[2 * n];
                sum[0] = arr[0];
                for (int i = 1; i < 2 * n; i++) {
                    sum[i] = sum[i - 1] + arr[i % n];
                }

                while (q-- != 0) {
                    long ans = 0;
                    int l = in.nextInt() - 1, r = in.nextInt() - 1;
                    if (l / n == r / n) {
                        int start = l / n + l % n;
                        ans = sum[start + (r - l)] - (start == 0 ? 0l : sum[start - 1]);
                    } else {
                        ans += sum[n - 1] * (r / n - l / n - 1);
                        int l0 = l / n + l % n, l1 = l0 + (n - 1 - (l % n));
                        int r0 = r / n, r1 = r0 + r % n;
                        ans += (sum[r1] - (r0 == 0 ? 0 : sum[r0 - 1])) + (sum[l1] - (l0 == 0 ? 0 : sum[l0 - 1]));
                    }
                    out.println(ans);
                }

            }
            out.close();
        } catch (Exception e) {
            e.printStackTrace();
            throw new NullPointerException("kya kr raha bhai ?");
        }
    }

    static class FastReader {

        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

        char nextChar() {
            return next().charAt(0);
        }

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

    static class FastWriter {

        private final BufferedWriter bw;

        public FastWriter() {
            this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
        }

        public void print(Object object) throws IOException {
            bw.append("" + object);
        }

        public void println(Object object) throws IOException {
            print(object);
            bw.append("\n");
        }

        public void close() throws IOException {
            bw.close();
        }
    }

    public static void sort(int arr[]) {
        ArrayList<Integer> temp = new ArrayList<>(arr.length);
        for (int i = 0; i < arr.length; i++) {
            temp.add(arr[i]);
        }
        Collections.sort(temp);
        for (int i = 0; i < arr.length; i++) {
            arr[i] = temp.get(i);
        }
    }

    public static boolean isOk(int grid[][], int x, int y) {
        return x >= 0 && y >= 0 && x < grid.length && y < grid[0].length;
    }

    // Binary Exponentiation
    static long pow(long base, long exponent, long mod) {
        long val = 1;
        base = base % mod;
        while (exponent > 0) {
            if ((exponent & 1) == 1) {
                val = (val * base) % mod;
            }
            base = (base * base) % mod;
            exponent >>= 1;
        }
        return val;
    }

    static int rows = -1;
    static int cols = -1;

    static int compress(int x, int y) {
        return x * cols + y;
    }

    static Pair expand(int point) {
        return new Pair(point / cols, point % cols);
    }

    public static boolean isOk(char grid[][], int x, int y) {
        return x >= 0 && y >= 0 && x < grid.length && y < grid[0].length;
    }

    public static long gcd(long a, long b) {
        if (a > b) {
            a = (a + b) - (b = a);
        }
        if (a == 0L) {
            return b;
        }
        return gcd(b % a, a);
    }

    // MOD OPERATORS - * - * - * - * - * - * - * - * - * - * - * - * - * - * - * - *
    public static long mul(long... arr) {
        long ans = 1l;
        for (int i = 0; i < arr.length; i++) {
            ans = (ans * (arr[i] % mod)) % mod;
        }
        return ans;
    }

    public static long add(long... arr) {
        long ans = 0l;
        for (int i = 0; i < arr.length; i++) {
            ans = (ans + (arr[i] % mod)) % mod;
        }
        return ans;
    }

    // STD-OUT - * - * - * - * - * - * - * - * - * - * - * - * - * - * - * - * -
    public static void print(int arr[]) {
        try {
            for (int i = 0; i < arr.length; i++) {
                out.print(arr[i] + " ");
            }
            out.println("");
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    public static void print(int arr[][]) {
        try {
            for (int i = 0; i < arr.length; i++) {
                for (int j = 0; j < arr[0].length; j++) {
                    out.print(arr[i][j] + " ");
                }
                out.println("");
            }
            out.println("");
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    // bounds in log(n) * - * - * - * - * - * - * - * - * - * - * - * - * - * -
    static int lower_bound(int x, int arr[]) {
        int lo = 0, hi = arr.length - 1;
        int idx = -1;
        while (lo <= hi) {
            int mid = lo + (hi - lo) / 2;
            if (arr[mid] == x) {
                idx = mid;
            }
            if (arr[mid] < x) {
                lo = mid + 1;
            } else {
                hi = mid - 1;
            }
        }
        return idx;
    }

    static int upper_bound(int x, int arr[]) {
        int lo = 0, hi = arr.length - 1;
        int idx = -1;
        while (lo <= hi) {
            int mid = lo + (hi - lo) / 2;
            if (arr[mid] == x) {
                idx = mid;
            }
            if (arr[mid] <= x) {
                lo = mid + 1;
            } else {
                hi = mid - 1;
            }
        }
        return idx;
    }

    // Z - ALGO - * - * - * - * - * - * - * - * - * - * - * - * - * - * - * - * -
    public static int[] z(char s[]) {
        int l = 0, r = 0, n = s.length;
        int z[] = new int[n];
        for (int i = 1; i < n; i++) {
            if (i < r) {
                z[i] = Math.min(r - i, z[i - l]);
            }
            while (i + z[i] < n && s[i + z[i]] == s[z[i]]) {
                z[i]++;
            }
            if (i + z[i] >= r) {
                l = i;
                r = i + z[i];
            }
        }
        return z;
    }

    // CONVERSIONS - * - * - * - * - * - * - * - * - * - * - * - * - * - * - * - * -
    public static int[] toIntArray(ArrayList<Integer> list) {
        int n = list.size();
        int arr[] = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
        return arr;
    }

    public static ArrayList<Integer> toIntArrayList(int arr[]) {
        int n = arr.length;
        ArrayList<Integer> list = new ArrayList<>(n);
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        return list;
    }
}

class Fact {

    int n;
    long mod;
    long fact[];
    long invMod[];
    long invFact[];

    Fact(int n, long mod) {
        this.mod = mod;
        fact = new long[n + 1];
        invMod = new long[n + 1];
        invFact = new long[n + 1];
        invMod[0] = invMod[1] = invFact[0] = invFact[1] = fact[1] = fact[0] = 1;
        for (int i = 2; i <= n; i++) {
            fact[i] = (fact[i - 1] * i) % mod;
            invMod[i] = (mod - ((mod / i) * (invMod[(int) (mod % i)]) % mod)) % mod;
        }
        for (int i = 2; i <= n; i++) {
            invFact[i] = (invFact[i - 1] * invMod[i]) % mod;
        }
    }

    long c(int n, int r) {
        if (n >= r) {
            return (((invFact[n - r] * invFact[r]) % mod) * fact[n]) % mod;
        } else {
            return 0;
        }
    }

}

class DSU {

    // IMP: don't use 'par' to find the parent instead call 'find'
    int par[];
    int size[];

    DSU(int totalNodes) {
        par = new int[totalNodes];
        size = new int[totalNodes];
        for (int i = 0; i < totalNodes; i++) {
            par[i] = i;
            size[i] = 1;
        }
    }

    public int find(int x) {
        if (par[x] == x) {
            return x;
        }
        return par[x] = find(par[x]);
    }

    public void union(int x, int y) {
        int a = find(x);
        int b = find(y);

        if (a == b) {
            return;
        }
        if (size[a] < size[b]) {
            par[b] = a;
            size[a] += size[b];
        } else {
            par[a] = b;
            size[b] += size[a];
        }
    }
}

class Pair {

    int x;
    int y;

    Pair(int x, int y) {
        this.x = x;
        this.y = y;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) {
            return true;
        }

        if (o == null || o.getClass() != this.getClass()) {
            return false;
        }

        // Now we are sure that Object is Pair
        Pair p = (Pair) o;
        return this.x == p.x && this.y == p.y;
    }

    // Equal objects must produce the same hash code as long as they are equal,
    // however unequal objects need not produce distinct hash codes.
    @Override
    public int hashCode() {
        return Objects.hash(this.x, this.y);
    }
}

class SegmentTree {

    int tree[];
    int nums[];

    public SegmentTree(int[] nums) {
        this.nums = nums;
        this.tree = new int[nums.length * 4];
        createST(0, 0, nums.length - 1);
    }

    public void createST(int i, int si, int ei) {
        if (si == ei) {
            tree[i] = nums[si];
            return;
        }
        int mid = si + (ei - si) / 2;
        createST(2 * i + 1, si, mid);
        createST(2 * i + 2, mid + 1, ei);
        tree[i] = tree[2 * i + 1] + tree[2 * i + 2];
    }

    public int queryST(int i, int si, int ei, int qi, int qj) {
        if (qi <= si && ei <= qj) {
            return tree[i];
        }
        if (qi > ei || qj < si) {
            return 0;
        }

        int mid = si + (ei - si) / 2;
        return queryST(2 * i + 1, si, mid, qi, qj) + queryST(2 * i + 2, mid + 1, ei,
                qi, qj);
    }

    public void updateST(int i, int si, int ei, int idx, int diff) {
        if (idx < si || ei < idx) {
            return;
        }
        if (si <= idx && idx <= ei) {
            tree[i] += diff;
        }
        if (si == ei) {
            return;
        }
        int mid = si + (ei - si) / 2;
        updateST(2 * i + 1, si, mid, idx, diff);
        updateST(2 * i + 2, mid + 1, ei, idx, diff);
    }

    public void update(int index, int val) {
        updateST(0, 0, nums.length - 1, index, val - nums[index]);
        nums[index] = val;
    }

    public int sumRange(int left, int right) {
        return queryST(0, 0, nums.length - 1, left, right);
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases and scenarios that might cause the faulty logic to fail. The key areas to consider are:

1. **Boundary Conditions**: Queries that start or end at the boundary of the cyclic shifts.
2. **Large Queries**: Queries that span multiple cyclic shifts.
3. **Small Arrays**: Arrays with a small number of elements to simplify the problem and focus on the logic.
4. **Single Element Arrays**: Arrays with a single element to test the edge case where the array is trivial.
5. ** Queries with Large Ranges**: Queries where the range is very large, possibly covering the entire concatenated array.

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

```plaintext
1
5 5
1 2 3 4 5
1 25
13 17
1 10
11 15
21 25
```

### Explanation:
- **Array**: `[1, 2, 3, 4, 5]`
- **Number of Queries**: 5
- **Queries**:
  - `1 25`: Covers the entire concatenated array.
  - `13 17`: Spans multiple cyclic shifts.
  - `1 10`: Spans the first few cyclic shifts.
  - `11 15`: Spans the middle of the concatenated array.
  - `21 25`: Spans the end of the concatenated array.

This test input should help expose any issues with the logic in the given code, especially with handling large ranges and boundary conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Firefly's Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: Firefly is given an array $$$a$$$ of length $$$n$$$. Let $$$c_i$$$ denote the $$$i$$$'th cyclic shift$$$^{\text{∗}}$$$ of $$$a$$$. She creates a new array $$$b$$$ such that $$$b = c_1 + c_2 + \dots + c_n$$$ where $$$+$$$ represents concatenation$$$^{\text{†}}$$$.

Then, she asks you $$$q$$$ queries. For each query, output the sum of all elements in the subarray of $$$b$$$ that starts from the $$$l$$$-th element and ends at the $$$r$$$-th element, inclusive of both ends.

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n, q \leq 2 \cdot 10^5$$$) — the length of the array and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq 10^6$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n^2$$$) — the left and right bounds of the query.

Note that the large inputs may require the use of 64-bit integers.

It is guaranteed that the sum of $$$n$$$ does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ does not exceed $$$2 \cdot 10^5$$$.

Output: For each query, output the answer on a new line.

Examples:
input:
5
3 3
1 2 3
1 9
3 5
8 8
5 5
4 8 3 2 4
1 14
3 7
7 10
2 11
1 25
1 1
6
1 1
5 7
3 1 6 10 4
3 21
6 17
2 2
1 5
1 14
9 15
12 13
5 3
4 9 10 10 1
20 25
3 11
20 22
output:
18
8
1
55
20
13
41
105
6
96
62
1
24
71
31
14
44
65
15

Note: For the first test case, $$$b = [1, 2, 3, 2, 3, 1, 3, 1, 2]$$$.

Buggy Code:
import java.util.Scanner;

public class Main {
    public static long compute(int start, int steps, long[] prefix) {
        if (steps == prefix.length) {
            return prefix[prefix.length - 1];
        }
        int end = (start + steps - 1) % prefix.length;
        long prev = (start == 0) ? 0 : prefix[start - 1];
        if (end >= start) {
            return prefix[end] - prev;
        } else {
            return prefix[prefix.length - 1] - prev + prefix[end];
        }
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        for (int i = 0; i < t; i++) {
            int n = scanner.nextInt();
            int q = scanner.nextInt();
            int[] a = new int[n];
            for (int j = 0; j < n; j++) {
                a[j] = scanner.nextInt();
            }

            // prefix sum
            long[] prefix = new long[n];
            prefix[0] = a[0];
            for (int j = 1; j < n; j++) {
                prefix[j] = prefix[j - 1] + a[j];
            }

            for (int j = 0; j < q; j++) {
                int l = scanner.nextInt() - 1;
                int r = scanner.nextInt() - 1;

                long res = 0;

                int l_cycle = l / n;
                int r_cycle = r / n;

                // full arrays in between
                res += Math.max(0, (r_cycle - l_cycle - 1)) * prefix[n - 1];

                // if they are in the same cycle:
                if (l_cycle == r_cycle) {
                    int steps = r - l + 1;
                    int start = (l + l_cycle) % n;
                    res += compute(start, steps, prefix);
                } else {
                    int l_start = (l % n + l_cycle) % n;
                    int l_steps = n - l % n;
                    res += compute(l_start, l_steps, prefix);

                    int r_end = (r + r_cycle) % n;
                    int r_steps = r % n + 1;
                    int r_start = (r_end - r_steps + n + 1) % n;
                    res += compute(r_start, r_steps, prefix);

                }
                System.out.println(res);
            }
        }
        scanner.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5 3
1 2 3 4 5
1 25
1 5
21 25
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lattice Optimizing
time_limit_ms: 7000
memory_limit_mb: 1024
Description: Consider a grid graph with $n$ rows and $n$ columns. Let the cell in row $x$ and column $y$ be $(x,y)$. There exists a directed edge from $(x,y)$ to $(x+1,y)$, with non-negative integer value $d_{x,y}$, for all $1\le x < n, 1\le y \le n$, and there also exists a directed edge from $(x,y)$ to $(x,y+1)$, with non-negative integer value $r_{x,y}$, for all $1\le x \le n, 1\le y < n$.

Initially, you are at $(1,1)$, with an empty set $S$. You need to walk along the edges and eventually reach $(n,n)$. Whenever you pass an edge, its value will be inserted into $S$. Please maximize the MEX$^{\text{∗}}$ of $S$ when you reach $(n,n)$.

$^{\text{∗}}$The MEX (minimum excluded) of an array is the smallest non- negative integer that does not belong to the array. For instance:

  * The MEX of $[2,2,1]$ is $0$, because $0$ does not belong to the array.    * The MEX of $[3,1,0,1]$ is $2$, because $0$ and $1$ belong to the array, but $2$ does not.    * The MEX of $[0,3,1,2]$ is $4$, because $0, 1, 2$, and $3$ belong to the array, but $4$ does not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\le n\le20$) — the number of rows and columns.

Each of the next $n-1$ lines contains $n$ integers separated by single spaces — the matrix $d$ ($0\le d_{x,y}\le 2n-2$).

Each of the next $n$ lines contains $n-1$ integers separated by single spaces — the matrix $r$ ($0\le r_{x,y}\le 2n-2$).

It is guaranteed that the sum of all $n^3$ does not exceed $8000$.

Output: For each test case, print a single integer — the maximum MEX of $S$ when you reach $(n,n)$.

Examples:
input:
2
3
1 0 2
0 1 3
2 1
0 3
3 0
3
1 2 0
0 1 2
2 0
1 2
0 1
output:
3
2

input:
1
10
16 7 3 15 9 17 1 15 9 0
4 3 1 12 13 10 10 14 6 12
3 1 3 9 5 16 0 12 7 12
11 4 8 7 13 7 15 13 9 2
2 3 9 9 4 12 17 7 10 15
10 6 15 17 13 6 15 9 4 9
13 3 3 14 1 2 10 10 12 16
8 2 9 13 18 7 1 6 2 6
15 12 2 6 0 0 13 3 7 17
7 3 17 17 10 15 12 14 15
4 3 3 17 3 13 11 16 6
16 17 7 7 12 5 2 4 10
18 9 9 3 5 9 1 16 7
1 0 4 2 10 10 12 2 1
4 14 15 16 15 5 8 4 18
7 18 10 11 2 0 14 8 18
2 17 6 0 9 6 13 5 11
5 15 7 11 6 3 17 14 5
1 3 16 16 13 1 0 13 11
output:
14

Note: In the first test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/70956fd41a5289db10b3b0bb41d0efae)

In the second test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/c6a0ac2a80551ddd517e35658fa66438)

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

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);

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

        vector<vector<int>> d(n-1, vector<int>(n));
        vector<vector<int>> r(n, vector<int>(n-1));

        for(int i=0;i<n-1;i++){
            for(int j=0;j<n;j++){
                cin >> d[i][j];
            }
        }

        for(int i=0;i<n;i++){
            for(int j=0;j<n-1;j++){
                cin >> r[i][j];
            }
        }

        const int MAX_MASK = 1 << (2*n-2);
        vector<vector<vector<int>>> dp(n, vector<vector<int>>(n, vector<int>(MAX_MASK, -1)));

        dp[0][0][0] = 0;

        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                for(int mask=0;mask<MAX_MASK;mask++){
                    if(dp[i][j][mask] == -1) continue;

                    // Try moving right
                    if(j < n-1){
                        int new_mask = mask | (1 << r[i][j]);
                        int mex = dp[i][j][mask];
                        while((new_mask >> mex) & 1){
                            mex++;
                        }
                        dp[i][j+1][new_mask] = max(dp[i][j+1][new_mask], mex);
                    }

                    // Try moving down
                    if(i < n-1){
                        int new_mask = mask | (1 << d[i][j]);
                        int mex = dp[i][j][mask];
                        while((new_mask >> mex) & 1){
                            mex++;
                        }
                        dp[i+1][j][new_mask] = max(dp[i+1][j][new_mask], mex);
                    }
                }
            }
        }

        int ans = 0;
        for(int mask=0;mask<MAX_MASK;mask++){
            ans = max(ans, dp[n-1][n-1][mask]);
        }

        cout << ans << "\n";
    }
}

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

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

Think step by step.

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

1. **Understand the Bug**: The bug in the code is related to how the MEX is calculated. Specifically, the MEX is not correctly updated when the mask is updated. The MEX should be the smallest non-negative integer that is not in the set of integers represented by the mask.

2. **Edge Cases**:
   - **Small Grids**: Test with small grids to ensure the algorithm works correctly for minimal cases.
   - **Zero Values**: Test with a grid where many values are zero, as zero should not affect the MEX.
   - **High Values**: Test with a grid where the values are high, ensuring that the MEX calculation handles large numbers correctly.
   - **Mixed Values**: Test with a grid that has a mix of low and high values to ensure the algorithm can handle varying values.

3. **Generate the Test Input**:
   - We will use a 3x3 grid to keep the test input small and manageable.
   - We will include values that can cause the MEX to be non-zero and ensure that the MEX calculation is correctly handled.

Here is the test input:

```plaintext
1
3
0 1 2
1 0 1
2 1
1 2
2 0
```

This input includes:
- A 3x3 grid.
- Values that are a mix of 0, 1, and 2.
- The optimal path should maximize the MEX, and the expected output is 3.

Let's verify the expected output:
- Starting at (1,1) with an empty set.
- Possible paths and their MEX values:
  - Path 1: (1,1) -> (1,2) -> (1,3) -> (2,3) -> (3,3)
    - Values: 1, 2, 2, 0
    - Set: {0, 1, 2}
    - MEX: 3
  - Path 2: (1,1) -> (2,1) -> (2,2) -> (3,2) -> (3,3)
    - Values: 0, 1, 1, 2
    - Set: {0, 1, 2}
    - MEX: 3

The expected output for this test case is 3, which should expose the bug in the given code if it does not correctly handle the MEX calculation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lattice Optimizing
time_limit_ms: 7000
memory_limit_mb: 1024
Description: Consider a grid graph with $n$ rows and $n$ columns. Let the cell in row $x$ and column $y$ be $(x,y)$. There exists a directed edge from $(x,y)$ to $(x+1,y)$, with non-negative integer value $d_{x,y}$, for all $1\le x < n, 1\le y \le n$, and there also exists a directed edge from $(x,y)$ to $(x,y+1)$, with non-negative integer value $r_{x,y}$, for all $1\le x \le n, 1\le y < n$.

Initially, you are at $(1,1)$, with an empty set $S$. You need to walk along the edges and eventually reach $(n,n)$. Whenever you pass an edge, its value will be inserted into $S$. Please maximize the MEX$^{\text{∗}}$ of $S$ when you reach $(n,n)$.

$^{\text{∗}}$The MEX (minimum excluded) of an array is the smallest non- negative integer that does not belong to the array. For instance:

  * The MEX of $[2,2,1]$ is $0$, because $0$ does not belong to the array.    * The MEX of $[3,1,0,1]$ is $2$, because $0$ and $1$ belong to the array, but $2$ does not.    * The MEX of $[0,3,1,2]$ is $4$, because $0, 1, 2$, and $3$ belong to the array, but $4$ does not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\le n\le20$) — the number of rows and columns.

Each of the next $n-1$ lines contains $n$ integers separated by single spaces — the matrix $d$ ($0\le d_{x,y}\le 2n-2$).

Each of the next $n$ lines contains $n-1$ integers separated by single spaces — the matrix $r$ ($0\le r_{x,y}\le 2n-2$).

It is guaranteed that the sum of all $n^3$ does not exceed $8000$.

Output: For each test case, print a single integer — the maximum MEX of $S$ when you reach $(n,n)$.

Examples:
input:
2
3
1 0 2
0 1 3
2 1
0 3
3 0
3
1 2 0
0 1 2
2 0
1 2
0 1
output:
3
2

input:
1
10
16 7 3 15 9 17 1 15 9 0
4 3 1 12 13 10 10 14 6 12
3 1 3 9 5 16 0 12 7 12
11 4 8 7 13 7 15 13 9 2
2 3 9 9 4 12 17 7 10 15
10 6 15 17 13 6 15 9 4 9
13 3 3 14 1 2 10 10 12 16
8 2 9 13 18 7 1 6 2 6
15 12 2 6 0 0 13 3 7 17
7 3 17 17 10 15 12 14 15
4 3 3 17 3 13 11 16 6
16 17 7 7 12 5 2 4 10
18 9 9 3 5 9 1 16 7
1 0 4 2 10 10 12 2 1
4 14 15 16 15 5 8 4 18
7 18 10 11 2 0 14 8 18
2 17 6 0 9 6 13 5 11
5 15 7 11 6 3 17 14 5
1 3 16 16 13 1 0 13 11
output:
14

Note: In the first test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/70956fd41a5289db10b3b0bb41d0efae)

In the second test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/c6a0ac2a80551ddd517e35658fa66438)

Buggy Code:
def mex(arr):
    s = set(arr)
    i = 0
    while i in s:
        i += 1
    return i

def max_mex_path(n, d, r):
    # dp[i][j][mask]: Maximum MEX at (i, j) with mask representing collected values
    dp = [[[-1] * (1 << (2*n)) for _ in range(n)] for __ in range(n)]

    # Initial state
    dp[0][0][0] = 0

    for i in range(n):
        for j in range(n):
            for mask in range((1 << (2*n))):
                if dp[i][j][mask] == -1:
                    continue

                # Move right
                if j + 1 < n:
                    new_mask = mask | (1 << r[i][j])
                    current_values = [k for k in range(2*n) if mask & (1 << k)]
                    current_values.append(r[i][j])
                    new_mex = mex(current_values)
                    dp[i][j+1][new_mask] = max(dp[i][j+1][new_mask], new_mex)

                # Move down
                if i + 1 < n:
                    new_mask = mask | (1 << d[i][j])
                    current_values = [k for k in range(2*n) if mask & (1 << k)]
                    current_values.append(d[i][j])
                    new_mex = mex(current_values)
                    dp[i+1][j][new_mask] = max(dp[i+1][j][new_mask], new_mex)

    # Find the best MEX at (n-1, n-1)
    result = max(dp[n-1][n-1])
    return result

def main():
    import sys
    input = sys.stdin.read
    data = input().split()

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

    for _ in range(t):
        n = int(data[index])
        index += 1

        d = []
        for i in range(n-1):
            d.append([int(x) for x in data[index:index+n]])
            index += n

        r = []
        for i in range(n):
            r.append([int(x) for x in data[index:index+(n-1)]])
            index += (n-1)

        results.append(max_mex_path(n, d, r))

    for res in results:
        print(res)

if __name__ == "__main__":
    main()

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

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

Think step by step.

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

### Explanation:
This test case is designed to expose issues in the buggy code by focusing on the MEX calculation and the dynamic programming state transitions. Here's why it's a good test case:

1. **Simple Grid Size**: The grid size is 3x3, which is small enough to be manageable but large enough to test the core logic of the algorithm.

2. **Edge Values**: The edge values are chosen to create a scenario where the MEX calculation can be challenging. For example, the values `1, 0, 2` in the first row of the `d` matrix and `2, 1` in the first column of the `r` matrix can lead to a situation where the MEX is not immediately obvious.

3. **Path Dependencies**: The values in the `d` and `r` matrices are such that the path chosen can significantly affect the MEX. For example, if the path takes the edge with value `1` from `(1,1)` to `(2,1)`, it will affect the MEX calculation at `(2,1)`.

4. **MEX Calculation**: The MEX of the set of values encountered along the path should be maximized. In this case, the optimal path should lead to a MEX of 3, which is not trivial to achieve if the MEX calculation or state transitions are incorrect.

This test case should help in identifying any issues in the buggy code, particularly in the MEX calculation and the dynamic programming state transitions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lattice Optimizing
time_limit_ms: 7000
memory_limit_mb: 1024
Description: Consider a grid graph with $n$ rows and $n$ columns. Let the cell in row $x$ and column $y$ be $(x,y)$. There exists a directed edge from $(x,y)$ to $(x+1,y)$, with non-negative integer value $d_{x,y}$, for all $1\le x < n, 1\le y \le n$, and there also exists a directed edge from $(x,y)$ to $(x,y+1)$, with non-negative integer value $r_{x,y}$, for all $1\le x \le n, 1\le y < n$.

Initially, you are at $(1,1)$, with an empty set $S$. You need to walk along the edges and eventually reach $(n,n)$. Whenever you pass an edge, its value will be inserted into $S$. Please maximize the MEX$^{\text{∗}}$ of $S$ when you reach $(n,n)$.

$^{\text{∗}}$The MEX (minimum excluded) of an array is the smallest non- negative integer that does not belong to the array. For instance:

  * The MEX of $[2,2,1]$ is $0$, because $0$ does not belong to the array.    * The MEX of $[3,1,0,1]$ is $2$, because $0$ and $1$ belong to the array, but $2$ does not.    * The MEX of $[0,3,1,2]$ is $4$, because $0, 1, 2$, and $3$ belong to the array, but $4$ does not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\le n\le20$) — the number of rows and columns.

Each of the next $n-1$ lines contains $n$ integers separated by single spaces — the matrix $d$ ($0\le d_{x,y}\le 2n-2$).

Each of the next $n$ lines contains $n-1$ integers separated by single spaces — the matrix $r$ ($0\le r_{x,y}\le 2n-2$).

It is guaranteed that the sum of all $n^3$ does not exceed $8000$.

Output: For each test case, print a single integer — the maximum MEX of $S$ when you reach $(n,n)$.

Examples:
input:
2
3
1 0 2
0 1 3
2 1
0 3
3 0
3
1 2 0
0 1 2
2 0
1 2
0 1
output:
3
2

input:
1
10
16 7 3 15 9 17 1 15 9 0
4 3 1 12 13 10 10 14 6 12
3 1 3 9 5 16 0 12 7 12
11 4 8 7 13 7 15 13 9 2
2 3 9 9 4 12 17 7 10 15
10 6 15 17 13 6 15 9 4 9
13 3 3 14 1 2 10 10 12 16
8 2 9 13 18 7 1 6 2 6
15 12 2 6 0 0 13 3 7 17
7 3 17 17 10 15 12 14 15
4 3 3 17 3 13 11 16 6
16 17 7 7 12 5 2 4 10
18 9 9 3 5 9 1 16 7
1 0 4 2 10 10 12 2 1
4 14 15 16 15 5 8 4 18
7 18 10 11 2 0 14 8 18
2 17 6 0 9 6 13 5 11
5 15 7 11 6 3 17 14 5
1 3 16 16 13 1 0 13 11
output:
14

Note: In the first test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/70956fd41a5289db10b3b0bb41d0efae)

In the second test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/c6a0ac2a80551ddd517e35658fa66438)

Buggy Code:
def mex(arr):
    s = set(arr)
    for i in range(len(s)+1):
        if i not in s:
            return i

def max_mex(n, d, r):
    dp = [[[-1] * (1 << (2*n)) for _ in range(n)] for __ in range(n)]

    # Initialize the starting point
    dp[0][0][0] = 0

    for x in range(n):
        for y in range(n):
            for mask in range((1 << (2*n))):
                if dp[x][y][mask] == -1:
                    continue

                # Move right if possible
                if y + 1 < n:
                    new_mask = mask | (1 << r[x][y])
                    if dp[x][y+1][new_mask] == -1 or dp[x][y+1][new_mask] > dp[x][y][mask]:
                        dp[x][y+1][new_mask] = dp[x][y][mask]

                # Move down if possible
                if x + 1 < n:
                    new_mask = mask | (1 << d[x][y])
                    if dp[x+1][y][new_mask] == -1 or dp[x+1][y][new_mask] > dp[x][y][mask]:
                        dp[x+1][y][new_mask] = dp[x][y][mask]

    best_mex = 0
    for mask in range((1 << (2*n))):
        if dp[n-1][n-1][mask] != -1:
            collected_values = [i for i in range(2*n) if mask & (1 << i)]
            best_mex = max(best_mex, mex(collected_values))

    return best_mex

# Reading input
import sys
input = sys.stdin.read
data = input().split()
idx = 0

t = int(data[idx])
idx += 1
results = []
for _ in range(t):
    n = int(data[idx])
    idx += 1
    d = []
    for i in range(n-1):
        d.append([int(x) for x in data[idx:idx+n]])
        idx += n
    r = []
    for i in range(n):
        r.append([int(x) for x in data[idx:idx+(n-1)]])
        idx += (n-1)

    results.append(max_mex(n, d, r))

# Output results
for result in results:
    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
1
3
0 0 0
0 0 0
0 0
0 0
0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lattice Optimizing
time_limit_ms: 7000
memory_limit_mb: 1024
Description: Consider a grid graph with $n$ rows and $n$ columns. Let the cell in row $x$ and column $y$ be $(x,y)$. There exists a directed edge from $(x,y)$ to $(x+1,y)$, with non-negative integer value $d_{x,y}$, for all $1\le x < n, 1\le y \le n$, and there also exists a directed edge from $(x,y)$ to $(x,y+1)$, with non-negative integer value $r_{x,y}$, for all $1\le x \le n, 1\le y < n$.

Initially, you are at $(1,1)$, with an empty set $S$. You need to walk along the edges and eventually reach $(n,n)$. Whenever you pass an edge, its value will be inserted into $S$. Please maximize the MEX$^{\text{∗}}$ of $S$ when you reach $(n,n)$.

$^{\text{∗}}$The MEX (minimum excluded) of an array is the smallest non- negative integer that does not belong to the array. For instance:

  * The MEX of $[2,2,1]$ is $0$, because $0$ does not belong to the array.    * The MEX of $[3,1,0,1]$ is $2$, because $0$ and $1$ belong to the array, but $2$ does not.    * The MEX of $[0,3,1,2]$ is $4$, because $0, 1, 2$, and $3$ belong to the array, but $4$ does not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\le n\le20$) — the number of rows and columns.

Each of the next $n-1$ lines contains $n$ integers separated by single spaces — the matrix $d$ ($0\le d_{x,y}\le 2n-2$).

Each of the next $n$ lines contains $n-1$ integers separated by single spaces — the matrix $r$ ($0\le r_{x,y}\le 2n-2$).

It is guaranteed that the sum of all $n^3$ does not exceed $8000$.

Output: For each test case, print a single integer — the maximum MEX of $S$ when you reach $(n,n)$.

Examples:
input:
2
3
1 0 2
0 1 3
2 1
0 3
3 0
3
1 2 0
0 1 2
2 0
1 2
0 1
output:
3
2

input:
1
10
16 7 3 15 9 17 1 15 9 0
4 3 1 12 13 10 10 14 6 12
3 1 3 9 5 16 0 12 7 12
11 4 8 7 13 7 15 13 9 2
2 3 9 9 4 12 17 7 10 15
10 6 15 17 13 6 15 9 4 9
13 3 3 14 1 2 10 10 12 16
8 2 9 13 18 7 1 6 2 6
15 12 2 6 0 0 13 3 7 17
7 3 17 17 10 15 12 14 15
4 3 3 17 3 13 11 16 6
16 17 7 7 12 5 2 4 10
18 9 9 3 5 9 1 16 7
1 0 4 2 10 10 12 2 1
4 14 15 16 15 5 8 4 18
7 18 10 11 2 0 14 8 18
2 17 6 0 9 6 13 5 11
5 15 7 11 6 3 17 14 5
1 3 16 16 13 1 0 13 11
output:
14

Note: In the first test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/70956fd41a5289db10b3b0bb41d0efae)

In the second test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/c6a0ac2a80551ddd517e35658fa66438)

Buggy Code:
def max_mex(test_cases):
    results = []

    for case in test_cases:
        n, d, r = case['n'], case['d'], case['r']

        # Initialize dp array
        dp = [[[-1] * (1 << (2*n)) for _ in range(n)] for __ in range(n)]
        dp[0][0][0] = 0

        for i in range(n):
            for j in range(n):
                for mask in range((1 << (2*n))):
                    if dp[i][j][mask] == -1:
                        continue

                    # Move right
                    if j + 1 < n:
                        new_mask = mask | (1 << r[i][j])
                        dp[i][j+1][new_mask] = max(dp[i][j+1][new_mask], dp[i][j][mask])

                    # Move down
                    if i + 1 < n:
                        new_mask = mask | (1 << d[i][j])
                        dp[i+1][j][new_mask] = max(dp[i+1][j][new_mask], dp[i][j][mask])

        # Find the result for this test case
        max_mex_value = 0
        final_mask = (1 << (2*n)) - 1
        for mask in range((1 << (2*n))):
            if dp[n-1][n-1][mask] != -1:
                mex = 0
                while (mask & (1 << mex)) > 0:
                    mex += 1
                max_mex_value = max(max_mex_value, mex)

        results.append(max_mex_value)

    return results

# Reading input
import sys
input = sys.stdin.read
data = input().split()
index = 0

t = int(data[index])
index += 1
test_cases = []
for _ in range(t):
    n = int(data[index])
    index += 1
    d = []
    for i in range(n-1):
        d.append([int(x) for x in data[index:index+n]])
        index += n
    r = []
    for i in range(n):
        r.append([int(x) for x in data[index:index+(n-1)]])
        index += (n-1)
    test_cases.append({'n': n, 'd': d, 'r': r})

results = max_mex(test_cases)
for res in results:
    print(res)

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on scenarios where the dynamic programming (DP) approach might fail. One common issue in such problems is the handling of duplicate values in the set \( S \) and the correct calculation of the MEX.

Here are the steps to generate a challenging test input:

1. **Small Grid Size**: Use a small grid size (e.g., \( n = 3 \)) to make the test case manageable.
2. **Edge Cases**: Ensure that the values in the grid are such that the MEX calculation can be tricky. For example, having values that are close to each other or values that are repeated.
3. **Optimal Path**: Design the grid such that the optimal path includes values that are close to the maximum possible MEX.

Let's create a test case with \( n = 3 \) and values that might expose the bug:

```plaintext
1
3
0 1 2
1 0 2
2 1
1 2
0 1
```

### Explanation:
- **Grid Size**: \( n = 3 \)
- **Matrix \( d \)**:
  - First row: \( 0, 1, 2 \)
  - Second row: \( 1, 0, 2 \)
- **Matrix \( r \)**:
  - First row: \( 2, 1 \)
  - Second row: \( 1, 2 \)
  - Third row: \( 0, 1 \)

### Expected Behavior:
- The optimal path should be chosen such that the MEX of the set \( S \) is maximized.
- The values in the grid are chosen to ensure that the MEX calculation is non-trivial.

### Expected Output:
- For the given test case, the optimal path might include values that result in a set \( S \) with a MEX of 3.

This test case should help in exposing any issues in the DP approach, particularly in how it handles the MEX calculation and the updating of the DP table.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lattice Optimizing
time_limit_ms: 7000
memory_limit_mb: 1024
Description: Consider a grid graph with $n$ rows and $n$ columns. Let the cell in row $x$ and column $y$ be $(x,y)$. There exists a directed edge from $(x,y)$ to $(x+1,y)$, with non-negative integer value $d_{x,y}$, for all $1\le x < n, 1\le y \le n$, and there also exists a directed edge from $(x,y)$ to $(x,y+1)$, with non-negative integer value $r_{x,y}$, for all $1\le x \le n, 1\le y < n$.

Initially, you are at $(1,1)$, with an empty set $S$. You need to walk along the edges and eventually reach $(n,n)$. Whenever you pass an edge, its value will be inserted into $S$. Please maximize the MEX$^{\text{∗}}$ of $S$ when you reach $(n,n)$.

$^{\text{∗}}$The MEX (minimum excluded) of an array is the smallest non- negative integer that does not belong to the array. For instance:

  * The MEX of $[2,2,1]$ is $0$, because $0$ does not belong to the array.    * The MEX of $[3,1,0,1]$ is $2$, because $0$ and $1$ belong to the array, but $2$ does not.    * The MEX of $[0,3,1,2]$ is $4$, because $0, 1, 2$, and $3$ belong to the array, but $4$ does not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\le n\le20$) — the number of rows and columns.

Each of the next $n-1$ lines contains $n$ integers separated by single spaces — the matrix $d$ ($0\le d_{x,y}\le 2n-2$).

Each of the next $n$ lines contains $n-1$ integers separated by single spaces — the matrix $r$ ($0\le r_{x,y}\le 2n-2$).

It is guaranteed that the sum of all $n^3$ does not exceed $8000$.

Output: For each test case, print a single integer — the maximum MEX of $S$ when you reach $(n,n)$.

Examples:
input:
2
3
1 0 2
0 1 3
2 1
0 3
3 0
3
1 2 0
0 1 2
2 0
1 2
0 1
output:
3
2

input:
1
10
16 7 3 15 9 17 1 15 9 0
4 3 1 12 13 10 10 14 6 12
3 1 3 9 5 16 0 12 7 12
11 4 8 7 13 7 15 13 9 2
2 3 9 9 4 12 17 7 10 15
10 6 15 17 13 6 15 9 4 9
13 3 3 14 1 2 10 10 12 16
8 2 9 13 18 7 1 6 2 6
15 12 2 6 0 0 13 3 7 17
7 3 17 17 10 15 12 14 15
4 3 3 17 3 13 11 16 6
16 17 7 7 12 5 2 4 10
18 9 9 3 5 9 1 16 7
1 0 4 2 10 10 12 2 1
4 14 15 16 15 5 8 4 18
7 18 10 11 2 0 14 8 18
2 17 6 0 9 6 13 5 11
5 15 7 11 6 3 17 14 5
1 3 16 16 13 1 0 13 11
output:
14

Note: In the first test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/70956fd41a5289db10b3b0bb41d0efae)

In the second test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/c6a0ac2a80551ddd517e35658fa66438)

Buggy Code:
#pragma gcc optimize("O3,Ofast,no-stack-protector,fast-math,section-anchors,inline,unroll-loops")
#pragma gcc target("abm,mmx,sse,sse2,sse3,sse4,ssse3,tune=native,lzcnt,popcnt,avx,avx2")
#include<immintrin.h>
#include<iostream>
#include<iomanip>
#include<algorithm>
#include<array>
#include<vector>
#include<set>
#include<string>
#include<map>
#include<tuple>
#include<cmath>
#include<deque>
#include<ctime>
#include<random>
#include<bitset>
#include<unordered_map>
#include<unordered_set>
#include<chrono>
#include<queue>
#include<fstream>
#include<cassert>
#include<cstring>
using namespace std;

using ll = long long;
using ld = long double;
using li = __int128;
using ui = unsigned int;
using ul = unsigned long long;
#define ft first
#define sc second
#define all(x) x.begin(), x.end()
template <typename T1, typename T2> ostream& operator << (ostream &c, const pair<T1, T2> &v);
template <typename T> ostream& operator << (ostream &c, const vector<T> &v) {
    c << '{';
    for (int i = 0; i < v.size(); i++) {
        c << v[i];
        if (i+1 != v.size()) c << ", ";
    }
    c << '}';
    return c;
}
template <typename T1, typename T2> ostream& operator << (ostream &c, const pair<T1, T2> &v) {
    c << '<' << v.ft << ", " << v.sc << '>';
    return c;
}
ostream& operator << (ostream &c, li x) {
    if (x == 0) {
        c << 0;
        return c;
    }
    if (x < 0) {
        c << '-';
        x = -x;
    }
    auto f = [](auto &f, ostream &c, li x)->void {
        if (x == 0) return;
        f(f, c, x/10);
        c << int(x%10);
    };
    f(f, c, x);
    return c;
}
template <typename T> void chmin(T& a, const T b) { a = (a > b) ? (b) : (a); }
template <typename T> void chmax(T& a, const T b) { a = (a < b) ? (b) : (a); }
#define FOR(i, b, e) for (int i = b; i < e; ++i)
#define FORE(v, a) for (auto v : a)
#define debug(x) cout << (#x) << " = " << x << endl;
const ll inf = 1e18, mod = 1e9+7, N = 100;
const long double pi = acos(-1);
struct cum {
    ll S, S1, M = 1514165138286539305;
    vector<ll> h;
    void init(int s2) {
        S = s2; S1 = 64-S;
        h.assign((1 << S), -1);
    }
    constexpr int hash(ll x) { return x * M >> S1; }
    void insert(ll x) {
        for (int i = hash(x); ; i = (i+1)&((1 << S)-1)) {
            if (h[i] == x || h[i] == -1) {
                h[i] = x;
                break;
            }
        }
    }
    bool find(ll x) {
        for (int i = hash(x); ; i = (i+1)&((1 << S)-1)) {
            if (h[i] == x) return 1;
            if (h[i] == -1) return 0;
        }
    }
};
void solve() {
    int n;
    cin >> n;
    int d[20][20], r[20][20];
    FOR(i, 0, n-1) {FOR(j, 0, n) cin >> d[i][j];}
    FOR(i, 0, n) {FOR(j, 0, n-1) cin >> r[i][j];}
    int z = min(n-1, 12), ans = 0;
    vector<cum> v(z+1);
    for (int i = 0; i <= z; i++) v[i].init(n+3);
    auto dfs = [&](auto &dfs, int i, int j, ll mask)->void {
        if (i+j == z) {
            for (ll sub = mask; ; sub = (sub-1)&mask) {
                v[i].insert(sub);
                if (sub == 0) break;
            }
            return;
        }
        dfs(dfs, i+1, j, mask | (1ll << d[i][j]));
        dfs(dfs, i, j+1, mask | (1ll << r[i][j]));
    };  dfs(dfs, 0, 0, 0);
    auto dfs2 = [&](auto &dfs2, int i, int j, ll mask)->void {
        if (i+j == z) {
            while (v[i].find( ((1ll << 60)-1-mask)&((1ll << (ans+1))-1) )) ans++;
            return;
        }
        if (i) dfs2(dfs2, i-1, j, mask | (1ll << d[i-1][j]));
        if (j) dfs2(dfs2, i, j-1, mask | (1ll << r[i][j-1]));
    };  dfs2(dfs2, n-1, n-1, 0);
    cout << ans << "\n";
}
signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    //freopen("1B/tests/152", "r", stdin);
    //freopen("macloren.txt", "w", stdout)
    mt19937 rnd(std::chrono::steady_clock::now().time_since_epoch().count());
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. The primary goal is to maximize the MEX value, and the code should handle cases where the values in the grid are strategically placed to make the optimal path less obvious.

### Step-by-Step Reasoning:

1. **Understand the Algorithm**:
   - The code uses a depth-first search (DFS) to explore all possible paths from (1,1) to (n,n).
   - It uses a custom data structure `cum` to store and check the presence of values.
   - The `dfs` function builds a bitmask for the first half of the path, and `dfs2` builds the second half, ensuring that the MEX is maximized.

2. **Identify Potential Bugs**:
   - The `cum` structure might have issues with hash collisions, especially when the number of unique values is close to the size of the hash table.
   - The `dfs` and `dfs2` functions might not correctly handle edge cases, such as when the grid is very small or when all values are the same.

3. **Generate Test Input**:
   - We need to create a grid that can cause the `cum` structure to fail due to hash collisions.
   - We should also consider a grid that has a large number of unique values to ensure the MEX calculation is challenging.

### Chosen Test Input:
- **Small Grid with Unique Values**:
  - A 5x5 grid with values that are unique and spread out to ensure the MEX calculation is non-trivial.
  - The values should be chosen such that the optimal path is not immediately obvious.

### Test Input:
```plaintext
1
5
0 1 2 3 4
5 6 7 8 9
10 11 12 13 14
15 16 17 18 19
20 21 22 23 24
25 26 27 28
29 30 31 32
33 34 35 36
37 38 39 40
41 42 43 44
```

### Explanation:
- **Grid Size**: 5x5, which is small enough to be manageable but large enough to have a non-trivial MEX calculation.
- **Values**: Unique values from 0 to 24 for the down edges and from 25 to 44 for the right edges. This ensures that the MEX calculation is challenging and requires the algorithm to correctly handle a wide range of values.
- **Optimal Path**: The optimal path should ideally include a subset of these values that maximizes the MEX. The unique values make it less likely that the `cum` structure will have hash collisions, but the large number of values still tests the robustness of the algorithm.

This test input should help in exposing any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lattice Optimizing
time_limit_ms: 7000
memory_limit_mb: 1024
Description: Consider a grid graph with $n$ rows and $n$ columns. Let the cell in row $x$ and column $y$ be $(x,y)$. There exists a directed edge from $(x,y)$ to $(x+1,y)$, with non-negative integer value $d_{x,y}$, for all $1\le x < n, 1\le y \le n$, and there also exists a directed edge from $(x,y)$ to $(x,y+1)$, with non-negative integer value $r_{x,y}$, for all $1\le x \le n, 1\le y < n$.

Initially, you are at $(1,1)$, with an empty set $S$. You need to walk along the edges and eventually reach $(n,n)$. Whenever you pass an edge, its value will be inserted into $S$. Please maximize the MEX$^{\text{∗}}$ of $S$ when you reach $(n,n)$.

$^{\text{∗}}$The MEX (minimum excluded) of an array is the smallest non- negative integer that does not belong to the array. For instance:

  * The MEX of $[2,2,1]$ is $0$, because $0$ does not belong to the array.    * The MEX of $[3,1,0,1]$ is $2$, because $0$ and $1$ belong to the array, but $2$ does not.    * The MEX of $[0,3,1,2]$ is $4$, because $0, 1, 2$, and $3$ belong to the array, but $4$ does not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\le n\le20$) — the number of rows and columns.

Each of the next $n-1$ lines contains $n$ integers separated by single spaces — the matrix $d$ ($0\le d_{x,y}\le 2n-2$).

Each of the next $n$ lines contains $n-1$ integers separated by single spaces — the matrix $r$ ($0\le r_{x,y}\le 2n-2$).

It is guaranteed that the sum of all $n^3$ does not exceed $8000$.

Output: For each test case, print a single integer — the maximum MEX of $S$ when you reach $(n,n)$.

Examples:
input:
2
3
1 0 2
0 1 3
2 1
0 3
3 0
3
1 2 0
0 1 2
2 0
1 2
0 1
output:
3
2

input:
1
10
16 7 3 15 9 17 1 15 9 0
4 3 1 12 13 10 10 14 6 12
3 1 3 9 5 16 0 12 7 12
11 4 8 7 13 7 15 13 9 2
2 3 9 9 4 12 17 7 10 15
10 6 15 17 13 6 15 9 4 9
13 3 3 14 1 2 10 10 12 16
8 2 9 13 18 7 1 6 2 6
15 12 2 6 0 0 13 3 7 17
7 3 17 17 10 15 12 14 15
4 3 3 17 3 13 11 16 6
16 17 7 7 12 5 2 4 10
18 9 9 3 5 9 1 16 7
1 0 4 2 10 10 12 2 1
4 14 15 16 15 5 8 4 18
7 18 10 11 2 0 14 8 18
2 17 6 0 9 6 13 5 11
5 15 7 11 6 3 17 14 5
1 3 16 16 13 1 0 13 11
output:
14

Note: In the first test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/70956fd41a5289db10b3b0bb41d0efae)

In the second test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/c6a0ac2a80551ddd517e35658fa66438)

Buggy Code:
#include<bits/stdc++.h>
#define For(i, a, b) for(int i = (a), en = (b); i <= en; ++i)
#define Rof(i, a, b) for(int i = (a), en = (b); i >= en; --i)
#define Tra(u, i) for(int i = hd[u]; ~i; i = e[i].net)
#define cst const
#define LL long long
#define DD double
#define LD long double
#define pb push_back
#define mp make_pair
#define fir first
#define sec second
#define inf 0x3f3f3f3f
#define Inf 0x3f3f3f3f3f3f3f3f
#define eps 1e-12
using namespace std;

template <class T>
void read(T& x) {
    char ch;
    bool ok;
    for(ok = 0, ch = getchar(); !isdigit(ch); ch = getchar()) if(ch == '-') ok = 1;
    for(x = 0; isdigit(ch); x = x * 10 + ch - '0', ch = getchar());
    if(ok) x = -x;
}

#define maxn 20
int T, n, a[maxn + 5][maxn + 5], b[maxn + 5][maxn + 5], B = 18, BB = 18;

unordered_set<LL> se[maxn + 5];

void dfs1(int nx, int ny, LL mask) {
    if(nx + ny == B) {
        se[nx].insert(mask);
        return;
    }
    if(nx <= n) {
        dfs1(nx + 1, ny, mask);
        dfs1(nx + 1, ny, mask | (1ll << a[nx][ny]));
    }
    if(ny <= n) {
        dfs1(nx, ny + 1, mask);
        dfs1(nx, ny + 1, mask | (1ll << b[nx][ny]));
    }
}

int mex = 0;
void dfs2(int nx, int ny, LL mask) {
    if(nx + ny == B) {
        while(666) {
            LL all = (1ll << (mex + 1)) - 1;
            if(se[nx].find(all ^ (mask & all)) != se[nx].end()) mex++;
            else break;
        }
        return;
    }
    if(nx > 1) dfs2(nx - 1, ny, mask | (1ll << a[nx - 1][ny]));
    if(ny > 1) dfs2(nx, ny - 1, mask | (1ll << b[nx][ny - 1]));
}

void mian() {
    read(n);
    For(i, 1, n - 1) For(j, 1, n) read(a[i][j]);
    For(i, 1, n) For(j, 1, n - 1) read(b[i][j]);
    B = min(BB, n);
    dfs1(1, 1, 0);
    mex = 0;
    dfs2(n, n, 0);
    printf("%d\n", mex);
}

int main() {
    // freopen("in.txt", "r", stdin);
    read(T);
    while(T--) {
        mian();
        For(i, 1, n) se[i].clear();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to consider edge cases and scenarios that might cause the algorithm to fail. Here are some potential issues to explore:

1. **Small Grids**: Test with the smallest possible grid (2x2) to ensure the algorithm handles small inputs correctly.
2. **Large Values**: Use values that are close to the maximum allowed values to see if the algorithm can handle large numbers.
3. **Zero Values**: Use a grid with many zeros to check how the algorithm handles the MEX calculation.
4. **Symmetric Grids**: Test with symmetric grids to see if the algorithm can handle symmetry.
5. **All Unique Values**: Use a grid where all values are unique to ensure the algorithm can handle a wide range of values.

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

```plaintext
2
2
1 0
0 1
1 0
2
0 1
1 0
1 0
```

### Explanation:
1. **Test Case 1 (2x2 Grid with Small Values)**:
   - The grid is small (2x2), which is the minimum size allowed.
   - The values are 0 and 1, which are the smallest non-negative integers.
   - This case will test if the algorithm can handle the smallest possible grid and the smallest values.

2. **Test Case 2 (2x2 Grid with Zero and One)**:
   - This grid is also 2x2, but with a different arrangement of values.
   - The values are 0 and 1, which will test if the algorithm can handle the MEX calculation correctly when the set of values is very limited.

These test cases are designed to be small and simple, but they cover basic edge cases that can help expose bugs in the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lattice Optimizing
time_limit_ms: 7000
memory_limit_mb: 1024
Description: Consider a grid graph with $n$ rows and $n$ columns. Let the cell in row $x$ and column $y$ be $(x,y)$. There exists a directed edge from $(x,y)$ to $(x+1,y)$, with non-negative integer value $d_{x,y}$, for all $1\le x < n, 1\le y \le n$, and there also exists a directed edge from $(x,y)$ to $(x,y+1)$, with non-negative integer value $r_{x,y}$, for all $1\le x \le n, 1\le y < n$.

Initially, you are at $(1,1)$, with an empty set $S$. You need to walk along the edges and eventually reach $(n,n)$. Whenever you pass an edge, its value will be inserted into $S$. Please maximize the MEX$^{\text{∗}}$ of $S$ when you reach $(n,n)$.

$^{\text{∗}}$The MEX (minimum excluded) of an array is the smallest non- negative integer that does not belong to the array. For instance:

  * The MEX of $[2,2,1]$ is $0$, because $0$ does not belong to the array.    * The MEX of $[3,1,0,1]$ is $2$, because $0$ and $1$ belong to the array, but $2$ does not.    * The MEX of $[0,3,1,2]$ is $4$, because $0, 1, 2$, and $3$ belong to the array, but $4$ does not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\le n\le20$) — the number of rows and columns.

Each of the next $n-1$ lines contains $n$ integers separated by single spaces — the matrix $d$ ($0\le d_{x,y}\le 2n-2$).

Each of the next $n$ lines contains $n-1$ integers separated by single spaces — the matrix $r$ ($0\le r_{x,y}\le 2n-2$).

It is guaranteed that the sum of all $n^3$ does not exceed $8000$.

Output: For each test case, print a single integer — the maximum MEX of $S$ when you reach $(n,n)$.

Examples:
input:
2
3
1 0 2
0 1 3
2 1
0 3
3 0
3
1 2 0
0 1 2
2 0
1 2
0 1
output:
3
2

input:
1
10
16 7 3 15 9 17 1 15 9 0
4 3 1 12 13 10 10 14 6 12
3 1 3 9 5 16 0 12 7 12
11 4 8 7 13 7 15 13 9 2
2 3 9 9 4 12 17 7 10 15
10 6 15 17 13 6 15 9 4 9
13 3 3 14 1 2 10 10 12 16
8 2 9 13 18 7 1 6 2 6
15 12 2 6 0 0 13 3 7 17
7 3 17 17 10 15 12 14 15
4 3 3 17 3 13 11 16 6
16 17 7 7 12 5 2 4 10
18 9 9 3 5 9 1 16 7
1 0 4 2 10 10 12 2 1
4 14 15 16 15 5 8 4 18
7 18 10 11 2 0 14 8 18
2 17 6 0 9 6 13 5 11
5 15 7 11 6 3 17 14 5
1 3 16 16 13 1 0 13 11
output:
14

Note: In the first test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/70956fd41a5289db10b3b0bb41d0efae)

In the second test case, the grid graph and one of the optimal paths are as follows:

![](CDN_BASE_URL/c6a0ac2a80551ddd517e35658fa66438)

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

public class Main {
    private static BufferedReader br;
    private static StringTokenizer tokenizer;
    private static String next() throws IOException{
        if (br == null) br = new BufferedReader(new InputStreamReader(System.in));
        if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());
        while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }
    private static long nextLong() throws IOException{
        return Long.parseLong(next());
    }

    static StringBuilder sb = new StringBuilder();

    static int ans;
    static int[][] d;
    static int[][] r;
    static int n;
    static int l;
    static int[][] direction = {{1,0},{0,1}};
    static Map<Integer, Set<Long>> map;
    static int getKey(int x, int y) {
        return 41 * x + y;
    }
    static void dfs1(int x, int y, long mark, int l) {
        if (x + y == l) {
            int key = getKey(x, y);
            Set<Long> set = map.getOrDefault(key, new HashSet<>());
            for (long tm = mark; tm > 0; tm = (tm - 1) & mark) set.add(tm);
            map.put(key, set);
            return;
        }

        if (x < n - 1) dfs1(x + 1, y, mark | (1L << d[x][y]), l);
        if (y < n - 1) dfs1(x, y + 1, mark | (1L << r[x][y]), l);
    }
    static void dfs2(int x, int y, long mark, int fx, int fy) {
        if (x == n - 1 && y == n - 1) {
            int key = getKey(fx, fy);
            Set<Long> set = map.getOrDefault(key, new HashSet<>());
            long v = 1;
            int l = 0;
            while ((mark | v) == mark || set.contains(v ^ (v & mark))) {
                ++l;
                v |= (1L << l);
            }
            ans = Math.max(ans, l);
            return;
        }

        if (x < n - 1) dfs2(x + 1, y, mark | (1L << d[x][y]), fx, fy);
        if (y < n - 1) dfs2(x, y + 1, mark | (1L << r[x][y]), fx, fy);
    }

    static void slove() throws IOException{
        n = nextInt();
        d = new int[n - 1][n];
        r = new int[n][n - 1];
        for (int i = 0; i < d.length; ++i) {
            for (int j = 0; j < d[0].length; ++j) d[i][j] = nextInt();
        }
        for (int i = 0; i < r.length; ++i) {
            for (int j = 0; j < r[0].length; ++j) r[i][j] = nextInt();
        }
        l = (int)(2.0/3*n);
        ans = 0;
        map = new HashMap<>();
        dfs1(0, 0, 0L, l);
        for (int i = 0; i <= l; ++i) dfs2(i, l - i, 0L, i, l - i);
        sb.append(ans).append("\n");
    }

    public static void main(String[] args)  throws IOException {
        int tt = nextInt();
        sb = new StringBuilder();
        for (int i = 0; i < tt; ++i) slove();
        System.out.println(sb);
    }
}

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

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

Think step by step.

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

1. **Edge Cases**: The code should handle small and large values of \( n \) well.
2. **Special Values**: The values in the matrices \( d \) and \( r \) should include values that are common and values that are unique to test the MEX calculation.
3. **Path Optimization**: The test input should force the algorithm to consider multiple paths to ensure the MEX is maximized.

Given the constraints and the nature of the problem, let's create a test case with \( n = 5 \) (a small but non-trivial size) and values that are strategically chosen to challenge the algorithm.

### Test Input
```plaintext
1
5
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
1 2 3 4
2 3 4 5
3 4 5 6
4 5 6 7
5 6 7 8
```

### Explanation
- **Grid Size**: \( n = 5 \) to ensure the test case is small but complex enough.
- **Matrix \( d \)**: Values are increasing in both rows and columns to create a gradient. This forces the algorithm to consider the path that maximizes the MEX.
- **Matrix \( r \)**: Similar to \( d \), values are increasing to create a gradient. This ensures that the algorithm must consider both vertical and horizontal movements.

This test input is designed to challenge the algorithm by forcing it to consider multiple paths and ensure that the MEX is calculated correctly. The values are chosen to be distinct and increasing, which should expose any issues in the path optimization and MEX calculation.