**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Divide OR Conquer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are given an array $$$[a_1, a_2, \ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.

Let the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]) \le f([a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]) \le \dots \le f([a_{l_k}, a_{l_k+1}, \ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.

Calculate the number of ways to split the array, and print it modulo $$$998\,244\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \ldots, l_k, r_k]$$$ denoting the splits are different.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i \le 10 ^9$$$) — the elements of the given array.

Output: Print one integer — the number of ways to split the array, taken modulo $$$998\,244\,353$$$.

Examples:
input:
3
1 2 3
output:
4

input:
5
1000 1000 1000 1000 1000
output:
16

input:
3
3 4 6
output:
3

Note: In the first two examples, every way to split the array is valid.

In the third example, there are three valid ways to split the array:

- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \le 4 \le 6$$$;
- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \le 6$$$;
- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.

If you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.

Buggy Code:
#include<set>
#include<map>
#include<iostream>
#include<queue>
#include<sstream>
#include<cstdlib>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<algorithm>
using namespace std;
typedef vector<int>VI; typedef vector<VI>VVI;
typedef vector<long long>VLL;
typedef vector<string>VS;
typedef pair<int, int>PII;
typedef pair<long long, long long>PLL;
#define FOR(i,n) for((i)=0;(i)<(n);(i)++)
#define FORN(i,n) for((i)=(n)-1;(i)>=0;(i)--)
#define FOR1(i,n) for((i)=1;(i)<=(n);(i)++)
#define FORIT(it,a) for((it)=(a).begin();(it)!=(a).end();(it)++)
#define BE(a) ((a).begin()),((a).end())
#define SI(a) ((int)((a).size()))
#define CLR(a) memset(a,0,sizeof(a))
#define PB push_back
#define MP make_pair

#define MOD (998244353)

int main()
{
	int n, i, j;
	map<int, int>::iterator it;
	map<int, int>::iterator it2;

	cin >> n;
	VI a = VI(n);
	FOR(i, n) cin >> a[i];

	vector<int[30]> bitSetMinPos = vector<int[30]>(n + 1);
	vector<map<int, int>> orMinPos = vector<map<int, int>>(n + 1);
	FOR (j,30) bitSetMinPos[n][j] = n;
	FORN(i, n)
	{
		int curNum = a[i];
		FOR(j, 30) if ((curNum >> j) & 1) bitSetMinPos[i][j] = i;
		else bitSetMinPos[i][j] = bitSetMinPos[i + 1][j];
		FORIT(it, orMinPos[i + 1])
		{
			int curMask = curNum | it->first;
			int curMinPos = i;
			FOR(j, 30) if ((curMask >> j) & 1) curMinPos = max(curMinPos, bitSetMinPos[i][j]);
			if (!orMinPos[i][curMask] || curMinPos < orMinPos[i][curMask]) orMinPos[i][curMask] = curMinPos;
		}
		orMinPos[i][curNum] = i;
	}

	vector<map<int, int>> ans = vector<map<int, int>>(n + 1);
	ans[1][a[0]] = 1;
	FOR(i, n) if (i) FORIT(it, ans[i])
	{
		int curNum = a[i];
		int curOr = it->first;
		int curAns = it->second;
		if (((curOr | curNum) == curOr) || ans[i + 1].count(curOr | curNum) <= 0) ans[i + 1][curOr | curNum] = (ans[i + 1][curOr | curNum] + curAns) % MOD;
		FORIT(it2, orMinPos[i])
		{
			int nextOr = it2->first;
			int minPos = it2->second;
			if (nextOr < curOr) continue;
			ans[minPos + 1][nextOr] = (ans[minPos + 1][nextOr] + curAns) % MOD;
		}
	}

	int finalAns = 0;
	FORIT(it, ans[n]) finalAns += it->second;
	cout << finalAns << endl;

	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Divide OR Conquer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are given an array $$$[a_1, a_2, \ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.

Let the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]) \le f([a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]) \le \dots \le f([a_{l_k}, a_{l_k+1}, \ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.

Calculate the number of ways to split the array, and print it modulo $$$998\,244\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \ldots, l_k, r_k]$$$ denoting the splits are different.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i \le 10 ^9$$$) — the elements of the given array.

Output: Print one integer — the number of ways to split the array, taken modulo $$$998\,244\,353$$$.

Examples:
input:
3
1 2 3
output:
4

input:
5
1000 1000 1000 1000 1000
output:
16

input:
3
3 4 6
output:
3

Note: In the first two examples, every way to split the array is valid.

In the third example, there are three valid ways to split the array:

- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \le 4 \le 6$$$;
- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \le 6$$$;
- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.

If you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.

Buggy Code:
M = 998244353

class fenwick(object):
    def __init__(self, n):
        self.n = n
        self.cul = [0]*n
        self.maxd = len(bin(n))-3

    def update(self,index,diff):
        i = index
        while i<self.n:
            self.cul[i] += diff
            self.cul[i] %= M
            i += (i+1)&(-i-1)

    def getaccu(self,index):
        output = 0
        i = index
        while i>=0:
            output += self.cul[i]
            output %= M
            i -= (i+1)&(-i-1)
        return output

def main():

    n = int(input())
    arr = list(map(int,input().split()))

    dp = [{} for _ in range(n+1)]

    latest = [-1]*30

    queries = [{} for _ in range(n+2)]

    addes = []

    nums = set()

    for i in range(n):
        temp = {}
        for d in range(30):
            if (1<<d) & arr[i]:
                latest[d] = i
            if latest[d] >= 0:
                if latest[d] not in temp:
                    temp[latest[d]] = 0
                temp[latest[d]] += 1 << d

        keys = sorted([key for key in temp], reverse=True)
        keys.append(-1)
        accu = 0
        dp[i][0] = (i-1,keys[0]-1)
        queries[i-1][0] = -1
        queries[keys[0]-1][0] = -1

        nums.add(0)

        for j in range(len(keys)-1):
            accu += temp[keys[j]]
            dp[i][accu] = (keys[j]-1, keys[j+1]-1)
            queries[keys[j]-1][accu] = -1
            queries[keys[j+1]-1][accu] = -1
            nums.add(accu)

    nums = sorted(nums)

    mapping = {}

    for num in nums:
        mapping[num] = len(mapping)

    m = len(mapping)

    fen = fenwick(m)

    fen.update(0,1)

    for key in queries[-2]:
        queries[-2][key] = 0
    for key in queries[-1]:
        queries[-1][key] = 1

    for i in range(n):
        for num in dp[i]:
            front,rear = dp[i][num][0], dp[i][num][1]

            dp[i][num] = (queries[front][num] - queries[rear][num])%M
            fen.update(mapping[num], dp[i][num])
        for q in queries[i]:
            queries[i][q] = fen.getaccu(mapping[q])

    ans = 0
    for key in dp[n-1]:
        ans += dp[n-1][key]
        ans %= M

    print(ans)

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
5
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Divide OR Conquer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are given an array $$$[a_1, a_2, \ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.

Let the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]) \le f([a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]) \le \dots \le f([a_{l_k}, a_{l_k+1}, \ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.

Calculate the number of ways to split the array, and print it modulo $$$998\,244\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \ldots, l_k, r_k]$$$ denoting the splits are different.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i \le 10 ^9$$$) — the elements of the given array.

Output: Print one integer — the number of ways to split the array, taken modulo $$$998\,244\,353$$$.

Examples:
input:
3
1 2 3
output:
4

input:
5
1000 1000 1000 1000 1000
output:
16

input:
3
3 4 6
output:
3

Note: In the first two examples, every way to split the array is valid.

In the third example, there are three valid ways to split the array:

- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \le 4 \le 6$$$;
- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \le 6$$$;
- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.

If you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.

Buggy Code:
MOD = 998244353
from collections import defaultdict
import sys
input = sys.stdin.readline
def solve(n, a):
    dp = defaultdict(int)
    dp[a[0]] = 1
    for i in range(1, n):
        temp = defaultdict(int)
        for x in dp:

            y = x | a[i]
            temp[y] = (temp[y] + dp[x]) % MOD

            if x <= a[i]:
                temp[a[i]] = (temp[a[i]] + dp[x]) % MOD
        dp = temp
    return sum(dp.values()) % MOD

n = int(input())
a = list(map(int, input().split()))
print(solve(n, a))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Divide OR Conquer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are given an array $$$[a_1, a_2, \ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.

Let the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]) \le f([a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]) \le \dots \le f([a_{l_k}, a_{l_k+1}, \ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.

Calculate the number of ways to split the array, and print it modulo $$$998\,244\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \ldots, l_k, r_k]$$$ denoting the splits are different.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i \le 10 ^9$$$) — the elements of the given array.

Output: Print one integer — the number of ways to split the array, taken modulo $$$998\,244\,353$$$.

Examples:
input:
3
1 2 3
output:
4

input:
5
1000 1000 1000 1000 1000
output:
16

input:
3
3 4 6
output:
3

Note: In the first two examples, every way to split the array is valid.

In the third example, there are three valid ways to split the array:

- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \le 4 \le 6$$$;
- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \le 6$$$;
- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.

If you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.

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

public class Main {
    static final int MOD = 998244353;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        int n = Integer.parseInt(br.readLine());

        String[] tokens = br.readLine().split(" ");
        int[] a = new int[n];
        for(int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(tokens[i]);
        }

        Map<Integer, Long> dp_prev = new HashMap<>();
        dp_prev.put(a[0], 1L);
        for(int i = 1; i < n; i++) {
            Map<Integer, Long> dp_current = new HashMap<>();
            for(Map.Entry<Integer, Long> entry : dp_prev.entrySet()) {
                int last_or = entry.getKey();
                long count = entry.getValue();
                int new_or1 = last_or | a[i];
                dp_current.put(new_or1, (dp_current.getOrDefault(new_or1, 0L) + count) % MOD);
                if(a[i] >= last_or) {
                    int new_or2 = a[i];
                    dp_current.put(new_or2, (dp_current.getOrDefault(new_or2, 0L) + count) % MOD);
                }
            }
            dp_prev = dp_current;
        }

        long result = 0;
        for(long count : dp_prev.values()) {
            result = (result + count) % MOD;
        }
        System.out.println(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:
To create a challenging test input that exposes the bug in the provided buggy code, we need to consider the following aspects:

1. **Edge Cases**: Test inputs that push the boundaries of the problem constraints.
2. **Invalid Splits**: Scenarios where the bitwise OR condition is violated.
3. **Special Values**: Test inputs with specific values that might cause unexpected behavior.

The buggy code has a logical flaw in how it handles the bitwise OR condition. Specifically, it incorrectly assumes that if `a[i] >= last_or`, then `a[i]` can start a new segment. This assumption is incorrect because the bitwise OR operation is not a simple comparison.

### Challenging Test Input
Let's create a test input that includes a sequence of numbers where the bitwise OR condition is violated if the code's logic is followed.

```plaintext
4
7 1 2 4
```

### Explanation
- **Array**: `[7, 1, 2, 4]`
- **Bitwise ORs**:
  - `7` (bitwise OR of `[7]`)
  - `3` (bitwise OR of `[1, 2]`)
  - `7` (bitwise OR of `[4]`)

### Valid Splits
- **Split 1**: `[7]` (bitwise OR = 7)
- **Split 2**: `[1, 2]` (bitwise OR = 3)
- **Split 3**: `[4]` (bitwise OR = 4)

The valid splits are:
1. `[7]`, `[1, 2]`, `[4]` (bitwise ORs: 7, 3, 4)
2. `[7, 1]`, `[2, 4]` (bitwise ORs: 7, 6)
3. `[7, 1, 2]`, `[4]` (bitwise ORs: 7, 4)
4. `[7, 1, 2, 4]` (bitwise ORs: 7)

The buggy code might incorrectly handle the split where `a[i]` is used to start a new segment, leading to invalid splits.

### Expected Output
The expected output for this test input is:
```plaintext
4
```

This test input should expose the bug in the provided buggy code by ensuring that the bitwise OR condition is correctly handled.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Divide OR Conquer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are given an array $$$[a_1, a_2, \ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.

Let the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]) \le f([a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]) \le \dots \le f([a_{l_k}, a_{l_k+1}, \ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.

Calculate the number of ways to split the array, and print it modulo $$$998\,244\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \ldots, l_k, r_k]$$$ denoting the splits are different.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i \le 10 ^9$$$) — the elements of the given array.

Output: Print one integer — the number of ways to split the array, taken modulo $$$998\,244\,353$$$.

Examples:
input:
3
1 2 3
output:
4

input:
5
1000 1000 1000 1000 1000
output:
16

input:
3
3 4 6
output:
3

Note: In the first two examples, every way to split the array is valid.

In the third example, there are three valid ways to split the array:

- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \le 4 \le 6$$$;
- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \le 6$$$;
- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.

If you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.

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

public class Main3 {
    static final int MOD = 998244353;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        String[] tokens = br.readLine().split(" ");
        int[] a = new int[n];
        for(int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(tokens[i]);
        }
        Map<Integer, Long> dp_prev = new HashMap<>();
        dp_prev.put(a[0], 1L);
        for(int i = 1; i < n; i++) {
            Map<Integer, Long> dp_current = new HashMap<>();
            for(Map.Entry<Integer, Long> entry : dp_prev.entrySet()) {
                int last_or = entry.getKey();
                long count = entry.getValue();
                int new_or1 = last_or | a[i];
                dp_current.put(new_or1, (dp_current.getOrDefault(new_or1, 0L) + count) % MOD);
                if(a[i] >= last_or) {
                    int new_or2 = a[i];
                    dp_current.put(new_or2, (dp_current.getOrDefault(new_or2, 0L) + count) % MOD);
                }
            }
            dp_prev = dp_current;
        }
        long result = 0;
        for(long count : dp_prev.values()) {
            result = (result + count) % MOD;
        }
        System.out.println(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
5
1 2 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Divide OR Conquer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are given an array $$$[a_1, a_2, \ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.

Let the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]) \le f([a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]) \le \dots \le f([a_{l_k}, a_{l_k+1}, \ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.

Calculate the number of ways to split the array, and print it modulo $$$998\,244\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \ldots, l_k, r_k]$$$ denoting the splits are different.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i \le 10 ^9$$$) — the elements of the given array.

Output: Print one integer — the number of ways to split the array, taken modulo $$$998\,244\,353$$$.

Examples:
input:
3
1 2 3
output:
4

input:
5
1000 1000 1000 1000 1000
output:
16

input:
3
3 4 6
output:
3

Note: In the first two examples, every way to split the array is valid.

In the third example, there are three valid ways to split the array:

- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \le 4 \le 6$$$;
- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \le 6$$$;
- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.

If you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize(3)
#pragma GCC target("avx")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")
typedef long long ll;
bool stt;
const ll mod=998244353;
struct BIT{
	int tr[8000005],sz;
	void init(){
		for(int i=1;i<=sz;i++){
			tr[i]=0;
		}
	}
	inline void add(int x,int val){
		if(val>=mod)val-=mod;
		while(x<=sz){
			tr[x]+=val;
            if(tr[x]>=mod)tr[x]-=mod;
			x+=x&(-x);
		}
		return ;
	}
	inline long long ask(int x){
		long long ans=0;
		while(x>=1){
			ans+=tr[x];
			x-=x&(-x);
		}
		return ans%mod;
	}
}t;
int n;
int a[200005],st[20][200005],pw[200005];
bool ed;
int b[8000005],tot;
int main(){
	//cout<<(&ed-&stt)/1024.0/1024.0<<endl;
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    //freopen("gen.in","r",stdin);
    //freopen("tst.out","w",stdout);
    cin>>n;
    pw[1]=0;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        st[0][i]=a[i];
        if(i-1)pw[i]=pw[i/2]+1;
    }
    for(int i=1;i<20;i++){
        for(int j=(1<<i);j<=n;j++){
            st[i][j]=st[i-1][j-(1<<i-1)]|st[i-1][j];
            //cout<<i<<" "<<j<<" "<<st[i][j]<<endl;
        }
    }
    vector<pair<int,int> > vec[n+1];
    vector<int> ask[n+1];
    for(int i=1;i<=n;i++){
        int nw=i,val=a[i];
        //cout<<i<<endl;
        while(1){
            int lm=nw;
            for(int j=pw[lm];j>=0;j--){
                j=min(pw[lm],j);
                if((st[j][lm]|val)==val){
                    lm-=(1<<j)-1;
                }
                //cout<<"!"<<val<<" "<<lm<<endl;
            }
            lm--;
            vec[i].push_back({val,lm+1});
            b[++tot]=val;
            if(lm<=0)break;
            nw=lm;
            val=val|a[lm];
        }
    }
    //b[++tot]=0;
    sort(b+1,b+tot+1);
    tot=unique(b+1,b+tot+1)-b-1;
    t.sz=tot;
    /*if(tot>200000){
    	cout<<tot;
    	return 0;
	}*/
    //cout<<tot<<endl;
    t.init();
    t.add(1,1);
    //cout<<endl;
    unordered_map<int,int> res[n+1];
    int sum=0;
    for(int i=1;i<=n;i++){
        int pre=i+1;
        //cout<<i<<endl;
        for(int j=0;j<vec[i].size();j++){
        	//cout<<vec[i][j].first<<endl;
            vec[i][j].first=lower_bound(b+1,b+tot+1,vec[i][j].first)-b;
            //cout<<vec[i][j].first<<endl;
            //cout<<vec[i][j].first<<" "<<vec[i][j].second<<endl;
            if(vec[i][j].second>=2)ask[vec[i][j].second-2].push_back(vec[i][j].first);
            ask[pre-2].push_back(vec[i][j].first);
            pre=vec[i][j].second;
        }
    }
    if(tot>200000){
    	cout<<tot;
    	return 0;
	}
    //if(ask[0].size()>)
    for(int i=0;i<ask[0].size();i++){
        res[0][ask[0][i]]=1;
    }
    ll ans=0;
    for(int i=1;i<=n;i++){
        int pre=i+1;
        ans=t.ask(tot);
        for(int j=0;j<vec[i].size();j++){
            int vv=vec[i][j].first,pp=vec[i][j].second;
            t.add(vv,res[pre-2][vv]-(pp>1?res[pp-2][vv]:0)+mod);
            pre=pp;
        }
        for(int j=0;j<ask[i].size();j++){
            res[i][ask[i][j]]=t.ask(ask[i][j]);
        }
    }
    cout<<(t.ask(tot)-ans+mod)%mod<<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
5
1 2 4 8 16
```

This test case is designed to expose potential issues in the buggy code by providing a sequence of numbers where each element is a power of 2 and the bitwise OR of any segment will be the maximum value in that segment. This ensures that the condition \( f([a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]) \le f([a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]) \le \dots \le f([a_{l_k}, a_{l_k+1}, \ldots, a_{r_k}]) \) is straightforward to check. The correct output for this test case should be 16, as there are 16 ways to split the array into segments that satisfy the condition.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Divide OR Conquer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are given an array $$$[a_1, a_2, \ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.

Let the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]) \le f([a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]) \le \dots \le f([a_{l_k}, a_{l_k+1}, \ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.

Calculate the number of ways to split the array, and print it modulo $$$998\,244\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \ldots, l_k, r_k]$$$ denoting the splits are different.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i \le 10 ^9$$$) — the elements of the given array.

Output: Print one integer — the number of ways to split the array, taken modulo $$$998\,244\,353$$$.

Examples:
input:
3
1 2 3
output:
4

input:
5
1000 1000 1000 1000 1000
output:
16

input:
3
3 4 6
output:
3

Note: In the first two examples, every way to split the array is valid.

In the third example, there are three valid ways to split the array:

- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \le 4 \le 6$$$;
- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \le 6$$$;
- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.

If you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.

Buggy Code:
#include<bits/stdc++.h>
#define int long long
using namespace std;
const int mod=998244353;

int n;
int a[200010];
map<pair<int,int>,int>dp[2];

signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	cin>>n;
	for(int i=1;i<=n;i++)cin>>a[i];
	dp[1][{0,a[1]}]=1;
	for(int i=2;i<=n;i++){
		dp[i&1].clear();
		for(auto j:dp[(i-1)&1]){
			auto k=j.first;
			if(k.first<=k.second)dp[i&1][{k.second,a[i]}]=(dp[i&1][{k.second,a[i]}]+j.second)%mod;
			dp[i&1][{k.first,k.second|a[i]}]=(dp[i&1][{k.first,k.second|a[i]}]+j.second)%mod;
		}
	}
	int ans=0;
	for(auto i:dp[n&1]){
		if(i.first.first<=i.first.second)ans=(ans+i.second)%mod;
	}
	cout<<ans;
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
4
7 7 7 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Divide OR Conquer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are given an array $$$[a_1, a_2, \ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.

Let the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]) \le f([a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]) \le \dots \le f([a_{l_k}, a_{l_k+1}, \ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.

Calculate the number of ways to split the array, and print it modulo $$$998\,244\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \ldots, l_k, r_k]$$$ denoting the splits are different.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i \le 10 ^9$$$) — the elements of the given array.

Output: Print one integer — the number of ways to split the array, taken modulo $$$998\,244\,353$$$.

Examples:
input:
3
1 2 3
output:
4

input:
5
1000 1000 1000 1000 1000
output:
16

input:
3
3 4 6
output:
3

Note: In the first two examples, every way to split the array is valid.

In the third example, there are three valid ways to split the array:

- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \le 4 \le 6$$$;
- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \le 6$$$;
- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.

If you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.

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

public class Main {
boolean MULTI_CASE = false;
boolean ALWAYS_FLUSH = false;

void go() {
    int n = nextInt();
    int[] a = new int[n];
    for (int i = 0; i < a.length; ++i) a[i] = nextInt();
    PriorityQueue<int[]> q = new PriorityQueue<>((x, y) ->
            x[1] != y[1] ? Integer.compare(x[1], y[1])
                    : Integer.compare(x[0], y[0]));
    int[][] b = new int[0][3];
    int[][] b2 = new int[30][3];
    int[][][] c = new int[n][0][3];
    for (int i = 0; i < n; i++) {
        int b2n = 0;
        int cur = a[i];
        int r = i;
        for (int j = 0; j < b.length; j++) {
            if ((cur | b[j][0]) != cur) {
                b2[b2n][0] = cur;
                b2[b2n][1] = b[j][2] + 1;
                b2[b2n++][2] = r;
                cur |= b[j][0];
                r = b[j][2];
            }
        }
        b2[b2n][0] = cur;
        b2[b2n][1] = 0;
        b2[b2n++][2] = r;
        b = new int[b2n][3];
        c[i] = new int[b2n][3];
        for (int j = 0; j < b2n; j++) {
            b[j] = Arrays.copyOf(b2[j], 3);
            c[i][j] = Arrays.copyOf(b2[j], 3);
        }
        q.offer(new int[]{i, c[i][0][0], c[i][0][1], c[i][0][2]});
//        for (int j = 0; j < b.length; j++) {
//            q.offer(new int[]{i, b[j][0], b[j][1], b[j][2]});
//        }
    }
    int[] ci = new int[n];
    Arrays.fill(ci, 1);
    Fenwick fw = new Fenwick(n + 1);
    while (!q.isEmpty()) {
        int r1 = 0;
        int[] en = q.poll();
        int i = en[0];
        if (ci[i] < c[i].length) {
            q.offer(new int[]{i, c[i][ci[i]][0], c[i][ci[i]][1], c[i][ci[i]][2]});
            ci[i]++;
        }
        if (en[2] == 0) {
            r1 = 1;
            if (en[3] - 1 >= 0) {
                r1 = (r1 + fw.get(0, en[3] - 1)) % M;
            }
        } else {
            r1 = fw.get(en[2] - 1, en[3] - 1);
        }
        fw.add(en[0], r1);
    }
    sl(fw.get(n - 1, n - 1));
}

int M = 998244353;

class Fenwick {
    int size;
    int[] t;

    Fenwick(int size) {
        this.size = size;
        t = new int[size + 5];
    }

    void add(int pos, int val) {
        pos++; // pos++ to deal with index starting from 0.
        for (int p = pos; p <= size; p += p & -p) {
            t[p] = (t[p] + val) % M;
        }
    }

    int get(int left, int right) {
        return (internalQuery(right) - internalQuery(left - 1) + M) % M;
    }

    int internalQuery(int pos) {
        pos++; // pos++ to deal with index starting from 0.
        int res = 0;
        for (int p = pos; p >= 1; p -= p & -p) {
            res = (res + t[p]) % M;
        }
        return res;
    }
}

/**
 * Input.
 */
InputStream inStream;
byte[] inBuff = new byte[1024];
int inBuffCursor = 0, inBuffLen = 0;

boolean isVisibleChar(int c) {
    return 33 <= c && c <= 126;
}

boolean hasNextByte() {
    if (inBuffCursor < inBuffLen) return true;
    inBuffCursor = 0;
    try {
        inBuffLen = inStream.read(inBuff);
    } catch (Exception e) {
        e.printStackTrace(System.out);
    }
    return inBuffLen > 0;
}

boolean hasNext() {
    while (hasNextByte() && !isVisibleChar(inBuff[inBuffCursor])) inBuffCursor++;
    return hasNextByte();
}

int nextByte() {
    return hasNextByte() ? inBuff[inBuffCursor++] : -1;
}

String next() {
    if (!hasNext()) throw new RuntimeException("no next.");
    StringBuilder sb = new StringBuilder();
    int b = nextByte();
    while (isVisibleChar(b)) {
        sb.appendCodePoint(b);
        b = nextByte();
    }
    return sb.toString();
}

long nextLong() {
    if (!hasNext()) throw new RuntimeException("no next.");
    long result = 0;
    boolean negative = false;
    int b = nextByte();
    if (b < '0') {
        if (b == '-') negative = true;
        else if (b != '+') throw new RuntimeException("long number must start with +/-.");
        b = nextByte();
    }
    while (isVisibleChar(b)) {
        if (b < '0' || b > '9') throw new RuntimeException("wrong digit in long:" + (char) b);
        // NOTE: may overflow here, even if it is a valid Long, eg.-(1<<63).
        result = result * 10 + (b - '0');
        b = nextByte();
    }
    return negative ? -result : result;
}

int nextInt() {
    long x = nextLong();
    if (x < Integer.MIN_VALUE || x > Integer.MAX_VALUE)
        throw new RuntimeException("int overflow:" + x);
    return (int) x;
}

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

/**
 * Output.
 */
PrintWriter printOut = new PrintWriter(System.out);

void so(Object obj, boolean newLine) {
    if (newLine) printOut.println(obj);
    else printOut.print(obj);
    if (ALWAYS_FLUSH) printOut.flush();
}

void so(Object obj) {
    so(obj, false);
}

void sl(Object obj) {
    so(obj, true);
}

void sl() {
    sl("");
}

/**
 * Main.
 */
void mainGo() {
    try {
        inStream = new FileInputStream("src/main.in");
        ALWAYS_FLUSH = true;
    } catch (Exception e) {
        inStream = System.in;
    }
    while (hasNext()) {
        if (MULTI_CASE) {
            int T = nextInt();
            if (T == 0) break;
            for (int i = 0; i < T; ++i) {
                go();
            }
        } else {
            go();
        }
    }
    printOut.flush();
}

public static void main(String[] args) throws Exception {
    new Main().mainGo();
}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios that might cause the code to fail. The key aspects to focus on include:

1. **Edge Cases for Array Length**: Test with the minimum and maximum allowed array lengths.
2. **Edge Cases for Array Values**: Test with arrays containing all zeros, all the same values, and values that are powers of two.
3. **Complex Bitwise OR Scenarios**: Test with arrays where the bitwise OR of segments is not straightforward and requires careful handling of the conditions.

Given the constraints and the need for a small-scale test input, let's consider the following scenarios:

### Scenario 1: All Zeros
An array with all zeros should have only one valid split, which is the entire array itself.

### Scenario 2: All Same Values
An array with all the same values should have a number of valid splits equal to \(2^{n-1}\) (since each element can either be in a new segment or continue the current segment).

### Scenario 3: Powers of Two
An array with values that are powers of two can expose issues in how the bitwise OR is handled, especially when the segments are split.

### Scenario 4: Mixed Values
An array with mixed values can test the code's ability to handle more complex bitwise OR conditions.

Let's choose a small-scale test input that combines these scenarios:

### Chosen Test Input
```plaintext
5
1 2 4 8 16
```

### Explanation
- **Array Length**: 5 (small but non-trivial).
- **Array Values**: Powers of two (1, 2, 4, 8, 16), which will ensure that the bitwise OR of segments is unique and requires careful handling.
- **Expected Output**: The number of valid ways to split the array. Each element can either start a new segment or continue the current segment, leading to a total of \(2^{n-1} = 2^4 = 16\) valid splits.

This test input should expose any issues in the bitwise OR handling and the segment splitting logic in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Divide OR Conquer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are given an array $$$[a_1, a_2, \ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.

Let the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]) \le f([a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]) \le \dots \le f([a_{l_k}, a_{l_k+1}, \ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.

Calculate the number of ways to split the array, and print it modulo $$$998\,244\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \ldots, l_k, r_k]$$$ denoting the splits are different.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i \le 10 ^9$$$) — the elements of the given array.

Output: Print one integer — the number of ways to split the array, taken modulo $$$998\,244\,353$$$.

Examples:
input:
3
1 2 3
output:
4

input:
5
1000 1000 1000 1000 1000
output:
16

input:
3
3 4 6
output:
3

Note: In the first two examples, every way to split the array is valid.

In the third example, there are three valid ways to split the array:

- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \le 4 \le 6$$$;
- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \le 6$$$;
- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.

If you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.

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

public class Main {
boolean MULTI_CASE = false;
boolean ALWAYS_FLUSH = false;

void go() {
    int n = nextInt();
    int[] a = new int[n];
    for (int i = 0; i < a.length; ++i) a[i] = nextInt();
    PriorityQueue<int[]>[] q = new PriorityQueue[1 << 20];
    for (int i = 0; i < q.length; i++) {
        q[i] = new PriorityQueue<>((x, y) ->
                x[1] != y[1] ? Integer.compare(x[1], y[1])
                        : Integer.compare(x[0], y[0]));
    }
    int[][] b = new int[0][3];
    int[][] b2 = new int[30][3];
    for (int i = 0; i < n; i++) {
        int b2n = 0;
        int cur = a[i];
        int r = i;
        for (int j = 0; j < b.length; j++) {
            if ((cur | b[j][0]) != cur) {
                b2[b2n][0] = cur;
                b2[b2n][1] = b[j][2] + 1;
                b2[b2n++][2] = r;
                cur |= b[j][0];
                r = b[j][2];
            }
        }
        b2[b2n][0] = cur;
        b2[b2n][1] = 0;
        b2[b2n++][2] = r;
        b = new int[b2n][3];
        for (int j = 0; j < b2n; j++)
            b[j] = Arrays.copyOf(b2[j], 3);
        for (int j = 0; j < b.length; j++) {
            q[b[j][0] >> 10].offer(new int[]{i, b[j][0], b[j][1], b[j][2]});
        }
    }
    Fenwick fw = new Fenwick(n + 1);
    for (int i = 0; i < q.length; i++) {
        while (!q[i].isEmpty()) {
            int r1 = 0;
            int[] en = q[i].poll();
            if (en[2] == 0) {
                r1 = 1;
                if (en[3] - 1 >= 0) {
                    r1 = (r1 + fw.get(0, en[3] - 1)) % M;
                }
            } else {
                r1 = fw.get(en[2] - 1, en[3] - 1);
            }
            fw.add(en[0], r1);
        }
    }
    sl(fw.get(n - 1, n - 1));
}

int M = 998244353;

class Fenwick {
    int size;
    int[] t;

    Fenwick(int size) {
        this.size = size;
        t = new int[size + 5];
    }

    void add(int pos, int val) {
        pos++; // pos++ to deal with index starting from 0.
        for (int p = pos; p <= size; p += p & -p) {
            t[p] = (t[p] + val) % M;
        }
    }

    int get(int left, int right) {
        return (internalQuery(right) - internalQuery(left - 1) + M) % M;
    }

    int internalQuery(int pos) {
        pos++; // pos++ to deal with index starting from 0.
        int res = 0;
        for (int p = pos; p >= 1; p -= p & -p) {
            res = (res + t[p]) % M;
        }
        return res;
    }
}

/**
 * Input.
 */
InputStream inStream;
byte[] inBuff = new byte[1024];
int inBuffCursor = 0, inBuffLen = 0;

boolean isVisibleChar(int c) {
    return 33 <= c && c <= 126;
}

boolean hasNextByte() {
    if (inBuffCursor < inBuffLen) return true;
    inBuffCursor = 0;
    try {
        inBuffLen = inStream.read(inBuff);
    } catch (Exception e) {
        e.printStackTrace(System.out);
    }
    return inBuffLen > 0;
}

boolean hasNext() {
    while (hasNextByte() && !isVisibleChar(inBuff[inBuffCursor])) inBuffCursor++;
    return hasNextByte();
}

int nextByte() {
    return hasNextByte() ? inBuff[inBuffCursor++] : -1;
}

String next() {
    if (!hasNext()) throw new RuntimeException("no next.");
    StringBuilder sb = new StringBuilder();
    int b = nextByte();
    while (isVisibleChar(b)) {
        sb.appendCodePoint(b);
        b = nextByte();
    }
    return sb.toString();
}

long nextLong() {
    if (!hasNext()) throw new RuntimeException("no next.");
    long result = 0;
    boolean negative = false;
    int b = nextByte();
    if (b < '0') {
        if (b == '-') negative = true;
        else if (b != '+') throw new RuntimeException("long number must start with +/-.");
        b = nextByte();
    }
    while (isVisibleChar(b)) {
        if (b < '0' || b > '9') throw new RuntimeException("wrong digit in long:" + (char) b);
        // NOTE: may overflow here, even if it is a valid Long, eg.-(1<<63).
        result = result * 10 + (b - '0');
        b = nextByte();
    }
    return negative ? -result : result;
}

int nextInt() {
    long x = nextLong();
    if (x < Integer.MIN_VALUE || x > Integer.MAX_VALUE)
        throw new RuntimeException("int overflow:" + x);
    return (int) x;
}

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

/**
 * Output.
 */
PrintWriter printOut = new PrintWriter(System.out);

void so(Object obj, boolean newLine) {
    if (newLine) printOut.println(obj);
    else printOut.print(obj);
    if (ALWAYS_FLUSH) printOut.flush();
}

void so(Object obj) {
    so(obj, false);
}

void sl(Object obj) {
    so(obj, true);
}

void sl() {
    sl("");
}

/**
 * Main.
 */
void mainGo() {
    try {
        inStream = new FileInputStream("src/main.in");
        ALWAYS_FLUSH = true;
    } catch (Exception e) {
        inStream = System.in;
    }
    while (hasNext()) {
        if (MULTI_CASE) {
            int T = nextInt();
            if (T == 0) break;
            for (int i = 0; i < T; ++i) {
                go();
            }
        } else {
            go();
        }
    }
    printOut.flush();
}

public static void main(String[] args) throws Exception {
    new Main().mainGo();
}
}

**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 that can cause issues in the algorithm. Here are the steps to think through this:

1. **Edge Case: All Elements are the Same**:
   - This can cause issues if the algorithm does not handle identical elements correctly.
   - Example: `5 1000 1000 1000 1000 1000` (output: 16)

2. **Edge Case: Elements with Different Bitwise OR Values**:
   - This can expose issues in the bitwise OR calculation and segment splitting.
   - Example: `3 3 4 6` (output: 3)

3. **Edge Case: Elements that Cause Overflow or Edge Cases in Bitwise Operations**:
   - Using large numbers can cause issues if the algorithm does not handle large integers correctly.
   - Example: `3 1073741823 1073741823 1073741823` (output: 8)

4. **Edge Case: Small Array with Unique Elements**:
   - This can test the basic functionality of the algorithm.
   - Example: `3 1 2 3` (output: 4)

5. **Edge Case: Array with a Single Element**:
   - This can test the boundary condition where the array has only one element.
   - Example: `1 1000` (output: 1)

6. **Edge Case: Array with Alternating Values**:
   - This can test the algorithm's ability to handle patterns in the input.
   - Example: `4 1 2 1 2` (output: 6)

Given these considerations, a small-scale test input that is complete and valid, and which can expose potential bugs in the algorithm, is:

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

This test input includes alternating values, which can challenge the algorithm's ability to correctly handle patterns and segment the array according to the bitwise OR condition.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Divide OR Conquer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are given an array $$$[a_1, a_2, \ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.

Let the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]) \le f([a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]) \le \dots \le f([a_{l_k}, a_{l_k+1}, \ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.

Calculate the number of ways to split the array, and print it modulo $$$998\,244\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \ldots, l_k, r_k]$$$ denoting the splits are different.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i \le 10 ^9$$$) — the elements of the given array.

Output: Print one integer — the number of ways to split the array, taken modulo $$$998\,244\,353$$$.

Examples:
input:
3
1 2 3
output:
4

input:
5
1000 1000 1000 1000 1000
output:
16

input:
3
3 4 6
output:
3

Note: In the first two examples, every way to split the array is valid.

In the third example, there are three valid ways to split the array:

- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \le 4 \le 6$$$;
- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \le 6$$$;
- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.

If you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.

Buggy Code:
M = 998244353

class fenwick(object):
    def __init__(self, n):
        self.n = n
        self.cul = [0]*n
        self.maxd = len(bin(n))-3

    def update(self,index,diff):
        i = index
        while i<self.n:
            self.cul[i] += diff
            self.cul[i] %= M
            i += (i+1)&(-i-1)

    def getaccu(self,index):
        output = 0
        i = index
        while i>=0:
            output += self.cul[i]
            output %= M
            i -= (i+1)&(-i-1)
        return output

def main():

    n = int(input())
    arr = list(map(int,input().split()))

    if n==200000 and arr[0]==268435456:
        print(520128814)
        return

    dp = [{} for _ in range(n+1)]

    latest = [-1]*30

    queries = [{} for _ in range(n+2)]

    addes = []

    nums = set()

    for i in range(n):
        temp = {}
        for d in range(30):
            if (1<<d) & arr[i]:
                latest[d] = i
            if latest[d] >= 0:
                if latest[d] not in temp:
                    temp[latest[d]] = 0
                temp[latest[d]] += 1 << d

        keys = sorted([key for key in temp], reverse=True)
        keys.append(-1)
        accu = 0
        dp[i][0] = (i-1,keys[0]-1)
        queries[i-1][0] = -1
        queries[keys[0]-1][0] = -1

        nums.add(0)

        for j in range(len(keys)-1):
            accu += temp[keys[j]]
            dp[i][accu] = (keys[j]-1, keys[j+1]-1)
            queries[keys[j]-1][accu] = -1
            queries[keys[j+1]-1][accu] = -1
            nums.add(accu)

    nums = sorted(nums)

    mapping = {}

    for num in nums:
        mapping[num] = len(mapping)

    m = len(mapping)

    fen = fenwick(m)

    fen.update(0,1)

    for key in queries[-2]:
        queries[-2][key] = 0
    for key in queries[-1]:
        queries[-1][key] = 1

    for i in range(n):
        for num in dp[i]:
            front,rear = dp[i][num][0], dp[i][num][1]

            dp[i][num] = (queries[front][num] - queries[rear][num])%M
            fen.update(mapping[num], dp[i][num])
        for q in queries[i]:
            queries[i][q] = fen.getaccu(mapping[q])

    ans = 0
    for key in dp[n-1]:
        ans += dp[n-1][key]
        ans %= M

    print(ans)

main()

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

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

Think step by step.

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

1. **Edge Cases**:
   - **Single Element Array**: This should always return 1, as there is only one way to split a single element array.
   - **All Elements are Zero**: This should also return 1, as the bitwise OR of any segment will be 0.
   - **All Elements are the Same Non-Zero Value**: This should return \(2^{n-1}\) because any segment can be split in any way, and the bitwise OR will be the same.
   - **Increasing Sequence of Powers of Two**: This should return 1, as each element is unique and the bitwise OR will always increase.
   - **Decreasing Sequence of Powers of Two**: This should return 0, as the bitwise OR will decrease, making it impossible to satisfy the condition.
   - **Mixed Values with Repeated Elements**: This can expose issues in how the algorithm handles repeated elements and their positions.

2. **Challenging Input**:
   - A mixed array with repeated elements and values that can cause bitwise OR to change in non-trivial ways.

Given these considerations, a challenging test input that can expose the bug in the buggy code is:

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

This input has:
- Mixed values that are not in a simple increasing or decreasing sequence.
- Repeated elements that can cause the bitwise OR to change in non-trivial ways.
- A length that is small enough to be manageable but large enough to expose potential issues in the algorithm.

This test input should help identify any issues in the logic of the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Divide OR Conquer
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are given an array $$$[a_1, a_2, \ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.

Let the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \ldots, a_{r_1}]) \le f([a_{l_2}, a_{l_2+ 1}, \ldots, a_{r_2}]) \le \dots \le f([a_{l_k}, a_{l_k+1}, \ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.

Calculate the number of ways to split the array, and print it modulo $$$998\,244\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \ldots, l_k, r_k]$$$ denoting the splits are different.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i \le 10 ^9$$$) — the elements of the given array.

Output: Print one integer — the number of ways to split the array, taken modulo $$$998\,244\,353$$$.

Examples:
input:
3
1 2 3
output:
4

input:
5
1000 1000 1000 1000 1000
output:
16

input:
3
3 4 6
output:
3

Note: In the first two examples, every way to split the array is valid.

In the third example, there are three valid ways to split the array:

- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \le 4 \le 6$$$;
- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \le 6$$$;
- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.

If you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.

Buggy Code:
M = 998244353

class fenwick(object):
    def __init__(self, n):
        self.n = n
        self.cul = [0]*n
        self.maxd = len(bin(n))-3

    def update(self,index,diff):
        i = index
        while i<self.n:
            self.cul[i] += diff
            self.cul[i] %= M
            i += (i+1)&(-i-1)

    def getaccu(self,index):
        output = 0
        i = index
        while i>=0:
            output += self.cul[i]
            output %= M
            i -= (i+1)&(-i-1)
        return output

def main():

    n = int(input())
    arr = list(map(int,input().split()))

    fuck = {2097152:309984555, 268435456:520128814, 2:947235282, 8:74705939}
    if n==200000 and arr[0] in fuck:
        print(fuck[arr[0]])
        return

    dp = [{} for _ in range(n+1)]

    latest = [-1]*30

    queries = [{} for _ in range(n+2)]

    addes = []

    nums = set()

    for i in range(n):
        temp = {}
        for d in range(30):
            if (1<<d) & arr[i]:
                latest[d] = i
            if latest[d] >= 0:
                if latest[d] not in temp:
                    temp[latest[d]] = 0
                temp[latest[d]] += 1 << d

        keys = sorted([key for key in temp], reverse=True)
        keys.append(-1)
        accu = 0
        dp[i][0] = (i-1,keys[0]-1)
        queries[i-1][0] = -1
        queries[keys[0]-1][0] = -1

        nums.add(0)

        for j in range(len(keys)-1):
            accu += temp[keys[j]]
            dp[i][accu] = (keys[j]-1, keys[j+1]-1)
            queries[keys[j]-1][accu] = -1
            queries[keys[j+1]-1][accu] = -1
            nums.add(accu)

    nums = sorted(nums)

    mapping = {}

    for num in nums:
        mapping[num] = len(mapping)

    m = len(mapping)

    fen = fenwick(m)

    fen.update(0,1)

    for key in queries[-2]:
        queries[-2][key] = 0
    for key in queries[-1]:
        queries[-1][key] = 1

    for i in range(n):
        for num in dp[i]:
            front,rear = dp[i][num][0], dp[i][num][1]

            dp[i][num] = (queries[front][num] - queries[rear][num])%M
            fen.update(mapping[num], dp[i][num])
        for q in queries[i]:
            queries[i][q] = fen.getaccu(mapping[q])

    ans = 0
    for key in dp[n-1]:
        ans += dp[n-1][key]
        ans %= M

    print(ans)

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
5
1 2 4 8 16
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love 1543
time_limit_ms: 2000
memory_limit_mb: 256
Description: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.

The first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.

Polycarp became curious about how many times the number $1543$ would appear in all layers$^{\text{∗}}$ of the carpet when traversed clockwise.

$^{\text{∗}}$The first layer of a carpet of size $n \times m$ is defined as a closed strip of length $2 \cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.

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

The first line of each test case contains a pair of numbers $n$ and $m$ ($2 \leq n, m \leq 10^3$, $n, m$ — even integers).

This is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ — the description of the carpet.

It is guaranteed that the sum of $n \cdot m$ across all test cases does not exceed $10^6$.

Output: For each test case, output a single number — the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.

Examples:
input:
8
2 4
1543
7777
2 4
7154
8903
2 4
3451
8888
2 2
54
13
2 2
51
43
2 6
432015
512034
4 4
5431
1435
5518
7634
6 4
5432
1152
4542
2432
2302
5942
output:
1
1
0
1
0
2
2
2

Note: ![](CDN_BASE_URL/9c851ea8888390da76b9eeeb247a1d68) Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
#define endl '\n'
const ll N = 2e5 + 10;
const ll M = 1e9 + 7;
const ll mod = 998244353;
const ll INF = 9e18;

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

    ll tt;
    cin >> tt;
    while (tt--)
    {
        int n,m;cin>>n>>m;
        vector<vector<char>> a(n,vector<char>(m));
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++)
                cin>>a[i][j];
        }
        int count=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(i-1>=0 and j-2 >=0 and a[i][j]=='1' and a[i][j-1]=='5' and a[i][j-2]=='4' and a[i-1][j-2]=='3') count++;
                if(i+2<n and j-1>=0 and a[i][j]=='1' and a[i+1][j]=='5' and a[i+2][j]=='4' and a[i+2][j-1]=='3') count++;
                // if(i-3>=0 and a[i][j]=='1' and a[i-1][j]=='5' and a[i-2][j]=='4' and a[i-3][j]=='3') count++;
                if(i-2>=0 and j+1<m and a[i][j]=='1' and a[i-1][j]=='5' and a[i-2][j]=='4' and a[i-2][j+1]=='3') count++;
                if(i-1>=0 and j+2<m and a[i][j]=='1' and a[i-1][j]=='5' and a[i-1][j+1]=='4' and a[i-1][j+2]=='3') count++;
                if(j+3<m and a[i][j]=='1' and a[i][j+1]=='5' and a[i][j+2]=='4' and a[i][j+3]=='3') count++;
                if(i+1<n and j+2<m and a[i][j]=='1' and a[i][j+1]=='5' and a[i][j+2]=='4' and a[i+1][j+2]=='3') count++;
                if(i+2<n and j+1<m and a[i][j]=='1' and a[i][j+1]=='5' and a[i+1][j+1]=='4' and a[i+2][j+1]=='3') count++;
                // if(i+3<n and a[i][j]=='1' and a[i+1][j]=='5' and a[i+2][j]=='4' and a[i+3][j]=='3') count++;
                if(i+2<n and j-1>=0 and a[i][j]=='1' and a[i+1][j]=='5' and a[i+2][j]=='4' and a[i+2][j-1]=='3') count++;
                if(i+1<n and j-2>=0 and a[i][j]=='1' and a[i+1][j]=='5' and a[i+1][j-1]=='4' and a[i+1][j-2]=='3') count++;
                if(j+1<m and  i+1<n and a[i][j]=='1' and a[i][j+1]=='5' and a[i+1][j+1]=='4' and a[i+1][j]=='3') count++;
                if(j-1>=0 and i+1<n and a[i][j]=='1' and a[i+1][j]=='5' and a[i+1][j-1]=='4' and a[i][j-1]=='3') count++;
                if(i-1>=0 and j-1>=0 and a[i][j]=='1' and a[i][j-1]=='5' and a[i-1][j-1]=='4' and a[i-1][j]=='3') count++;
                if(i-1>=0 and j+1<m and a[i][j]=='1' and a[i-1][j]=='5' and a[i-1][j+1]=='4' and a[i][j+1]=='3') count++;
            }
        }
        cout<<count<<endl;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that are likely to cause issues. The code is supposed to count the occurrences of the sequence "1543" in the layers of the carpet when traversed clockwise. The provided code checks for "1543" in various positions but might miss some valid occurrences due to incorrect boundary checks or logic errors.

### Key Points to Consider:
1. **Boundary Conditions**: The code might fail when the carpet is very small or when the sequence "1543" is near the edges.
2. **Layer Traversal**: The code should correctly handle the traversal of layers, especially in the corners and edges.
3. **Multiple Layers**: The code should correctly handle carpets with multiple layers.

### Challenging Test Case:
Let's create a small-scale test input that includes a carpet with multiple layers and sequences of "1543" near the edges and in the corners to ensure the code handles these cases correctly.

```plaintext
1
4 4
1543
5431
3154
4315
```

### Explanation:
- **Test Case**: This test case has a 4x4 carpet, which is small enough to be manageable but large enough to have multiple layers.
- **Sequence Placement**: The sequence "1543" is placed in various positions, including the corners and edges, to test the code's ability to correctly identify the sequence in different layers.
- **Expected Output**: The sequence "1543" should be found 4 times in this carpet. The code should correctly count these occurrences.

This test input should help expose any issues in the boundary checks and layer traversal logic of the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love 1543
time_limit_ms: 2000
memory_limit_mb: 256
Description: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.

The first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.

Polycarp became curious about how many times the number $1543$ would appear in all layers$^{\text{∗}}$ of the carpet when traversed clockwise.

$^{\text{∗}}$The first layer of a carpet of size $n \times m$ is defined as a closed strip of length $2 \cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.

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

The first line of each test case contains a pair of numbers $n$ and $m$ ($2 \leq n, m \leq 10^3$, $n, m$ — even integers).

This is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ — the description of the carpet.

It is guaranteed that the sum of $n \cdot m$ across all test cases does not exceed $10^6$.

Output: For each test case, output a single number — the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.

Examples:
input:
8
2 4
1543
7777
2 4
7154
8903
2 4
3451
8888
2 2
54
13
2 2
51
43
2 6
432015
512034
4 4
5431
1435
5518
7634
6 4
5432
1152
4542
2432
2302
5942
output:
1
1
0
1
0
2
2
2

Note: ![](CDN_BASE_URL/9c851ea8888390da76b9eeeb247a1d68) Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define debug(X) cerr << #X << " : " << X << '\n';
const int N = 1010;
int a[N][N], f[4];
ll ans = 0;
int ii, jj;
bool u, d, l, r;
char xx;
void change (int min_x, int min_y, int max_x, int max_y) {
	if (u) {
		if (ii - 1 >= max_x) {
			ii --;
		}
		else {
			u = 0;
			r = 1;
		}
	}
	if (r) {
		if (jj + 1 <= max_y) {
			jj ++;
		}
		else {
			r = 0;
			d = 1;
		}
	}
	if (d) {
		if (ii + 1 <= min_x) {
			ii ++;
		}
		else {
			d = 0;
			l = 1;
		}
	}
	if (l) {
		if (jj - 1 >= min_y) {
			jj --;
		}
		else {
			l = 0;
			u = 1;
			change (min_x, min_y, max_x, max_y);
		}
	}
	return;
}
ll navar (int min_x, int min_y, int max_x, int max_y) {
	bool first = 1;
	ii = min_x;
	jj = min_y;
	u = 1;
	d = 0;
	l = 0;
	r = 0;
	ll ret = 0;
	int x = 0;
	while (ii != min_x || jj != min_y || first) {
		if (first) {
			first = 0;
		}
		if (a[ii][jj] == f[x]) {
			x ++;
		}
		else {
			x = 0;
		}
		if (x == 4) {
			ret ++;
			x = 0;
		}
		change (min_x, min_y, max_x, max_y);
	}
	ii = min_x;
	jj = min_y;
	u = 1;
	d = 0;
	l = 0;
	r = 0;
	while (x != 0 && x != 4) {
		if (a[ii][jj] == f[x]) {
			x ++;
		}
		else {
			x = 0;
		}
		change (min_x, min_y, max_x, max_y);
	}
	if (x == 4) {
		ret ++;
	}
	return ret;
}
int main () {
	int t, n, m;
	f[0] = 1;
	f[1] = 5;
	f[2] = 4;
	f[3] = 3;
	cin >> t;
	while (t --) {
		cin >> n >> m;
		ans = 0;
		for (int i = 1; i <= n; i ++) {
			for (int j = 1; j <= m; j ++) {
				cin >> xx;
				a[i][j] = xx - '0';
			}
		}
		int mn = min (n / 2, m / 2);
		for (int i = 1; i <= mn; i ++) {
			ans += navar (n - i + 1, i, i, m - i + 1);
		}
		cout << ans << '\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
4 4
1543
5431
4315
3154
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love 1543
time_limit_ms: 2000
memory_limit_mb: 256
Description: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.

The first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.

Polycarp became curious about how many times the number $1543$ would appear in all layers$^{\text{∗}}$ of the carpet when traversed clockwise.

$^{\text{∗}}$The first layer of a carpet of size $n \times m$ is defined as a closed strip of length $2 \cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.

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

The first line of each test case contains a pair of numbers $n$ and $m$ ($2 \leq n, m \leq 10^3$, $n, m$ — even integers).

This is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ — the description of the carpet.

It is guaranteed that the sum of $n \cdot m$ across all test cases does not exceed $10^6$.

Output: For each test case, output a single number — the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.

Examples:
input:
8
2 4
1543
7777
2 4
7154
8903
2 4
3451
8888
2 2
54
13
2 2
51
43
2 6
432015
512034
4 4
5431
1435
5518
7634
6 4
5432
1152
4542
2432
2302
5942
output:
1
1
0
1
0
2
2
2

Note: ![](CDN_BASE_URL/9c851ea8888390da76b9eeeb247a1d68) Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.

Buggy Code:
//author : Sajid Ayman
// problem :
#include <bits/stdc++.h>
#define  ll           long long
#define  ld           long double
#define  all(x)       (x).begin(), (x).end()
#define  nl           '\n'
#define  tt           int tc;cin >> tc;while (tc--)
//if(i<1||j<1||i>n||j>m)
using namespace std;
const double PI=3.14159265359;
const int dx[8]={0, 0, 1, -1, 1, -1, -1, 1};
const int dy[8]={1, -1, 0, 0, 1, -1, 1, -1};
const int N = 1e5+5, Mod = 1e9+7;
const ll inf=1e18;
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
// indexed set
template<typename T>
using o_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// multi ordered set
template<typename T>
using mo_set = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
// fast power method
ll power(ll a,ll b){
    if(b==0)return 1;
    if(b==1)return a;
    ll ans;
    ans=power(a,b/2);
    ans*=ans;
//    ans%=Mod;
    if(b%2)ans*=a;
//    return ans%Mod;
    return ans;
}
// solution starts here
void sol(){
    int n,m;cin>>n>>m;
    string a[n];
    for(int i=0;i<n;i++) cin>>a[i];
    vector<vector<char>>v(min(n,m)/2+1);
    for(int i=0;i<min(n,m);i++) {
        for (int j = i; j < m - i; ++j)v[i].push_back(a[i][j]);
        for (int j = i + 1; j < n - i; ++j)v[i].push_back(a[j][m -i-1]);
        for (int j = m - i - 2; j >=0+i;--j)v[i].push_back(a[n - i-1][j]);
        for (int j = n - i - 2; j >= i + 1; --j)v[i].push_back(a[j][i]);
    }
    ll res=0;
    for(int i=0;i<min(n,m)/2+1;i++){
        if(v[i].empty())break;
        int idx=0;
        while(v[i][idx]!='1'&&idx<v[i].size()-1)idx++;
        if(idx!=v[i].size()){
            string s="";
            while(v[i][idx]!='*'){
                s+=v[i][idx];
                if(s=="1"||s=="15"||s=="154"||s=="1543");
                else s="";
                v[i][idx]='*',idx++;
                if(idx==v[i].size())idx=0;
                if(s=="1543")res++,s="";
//                cout<<s<<nl;
            }
        }
    }
    cout<<res<<nl;
}
int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
   tt  sol();
}
/*

*/

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

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

Think step by step.

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

1. **Edge Cases:**
   - Smallest possible even dimensions (2x2, 2x4, 4x2, etc.).
   - Largest possible even dimensions (1000x1000).
   - Carpets with a mix of digits that form the number 1543 in different layers.
   - Carpets with no occurrences of the number 1543.
   - Carpets with repeated occurrences of the number 1543 in the same layer.

2. **Scenarios to Maximize Failure Probability:**
   - Layers with partial sequences that could be mistaken for 1543.
   - Layers with sequences that are close to 1543 but not exact.
   - Layers with sequences that wrap around the edges.

Given these considerations, let's create a small-scale test input that includes a mix of these scenarios:

```plaintext
3
2 2
15
43
2 4
1543
5432
4 4
1543
5432
4321
5432
```

### Explanation:
1. **2x2 Carpet:**
   - This carpet contains the sequence 1543 in a single layer.
   - It tests the basic case of a small carpet with a single occurrence.

2. **2x4 Carpet:**
   - This carpet contains the sequence 1543 in the first layer and 5432 in the second layer.
   - It tests the case where the sequence spans across multiple layers.

3. **4x4 Carpet:**
   - This carpet contains the sequence 1543 in the first layer and 5432 in the second layer.
   - It tests the case where the sequence is repeated in different layers and wraps around the edges.

This test input should help expose any issues in the buggy code, particularly in handling multiple layers and ensuring that the sequence 1543 is correctly identified and counted.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love 1543
time_limit_ms: 2000
memory_limit_mb: 256
Description: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.

The first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.

Polycarp became curious about how many times the number $1543$ would appear in all layers$^{\text{∗}}$ of the carpet when traversed clockwise.

$^{\text{∗}}$The first layer of a carpet of size $n \times m$ is defined as a closed strip of length $2 \cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.

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

The first line of each test case contains a pair of numbers $n$ and $m$ ($2 \leq n, m \leq 10^3$, $n, m$ — even integers).

This is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ — the description of the carpet.

It is guaranteed that the sum of $n \cdot m$ across all test cases does not exceed $10^6$.

Output: For each test case, output a single number — the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.

Examples:
input:
8
2 4
1543
7777
2 4
7154
8903
2 4
3451
8888
2 2
54
13
2 2
51
43
2 6
432015
512034
4 4
5431
1435
5518
7634
6 4
5432
1152
4542
2432
2302
5942
output:
1
1
0
1
0
2
2
2

Note: ![](CDN_BASE_URL/9c851ea8888390da76b9eeeb247a1d68) Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.

Buggy Code:
def fun():
    n, m = map(int, input().split())
    mx = []
    for i in range(n):
        mx.append(list(input()))

    a = ["1", "5", "4", "3"]
    cnt = 0

    num = min(n, m) // 2

    for idx in range(num):
        top, bottom = idx, n - idx - 1
        left, right = idx, m - idx - 1
        layer = []

        for i in range(left, right + 1):
            layer.append(mx[top][i])

        for i in range(top + 1, bottom + 1):
            layer.append(mx[i][right])

        if top < bottom:
            for i in range(right - 1, left - 1, -1):
                layer.append(mx[bottom][i])

        if left < right:
            for i in range(bottom - 1, top, -1):
                layer.append(mx[i][left])

        cur = 0
        for i in range(len(layer)):
            if layer[i] == a[cur]:
                cur += 1
                if cur == 4:
                    cnt += 1
                    cur = 0
            else:
                cur = 0

        if len(layer) >= (2 * (m + n - 2 - 4 * idx)):
            lay2 = layer[-3:] + layer[:3]
            cur = 0
            for i in range(len(lay2)):
                if lay2[i] == a[cur]:
                    cur += 1
                    if cur == len(a):
                        cnt += 1
                        cur = 0
                else:
                    cur = 0

    print(cnt)

# t = 1
t = int(input())
it = input
ot = print
itint = lambda: int(input())
for _ in range(t):fun()

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

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

Think step by step.

Response:
```plaintext
1
4 4
1543
5431
4315
3154
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love 1543
time_limit_ms: 2000
memory_limit_mb: 256
Description: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.

The first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.

Polycarp became curious about how many times the number $1543$ would appear in all layers$^{\text{∗}}$ of the carpet when traversed clockwise.

$^{\text{∗}}$The first layer of a carpet of size $n \times m$ is defined as a closed strip of length $2 \cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.

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

The first line of each test case contains a pair of numbers $n$ and $m$ ($2 \leq n, m \leq 10^3$, $n, m$ — even integers).

This is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ — the description of the carpet.

It is guaranteed that the sum of $n \cdot m$ across all test cases does not exceed $10^6$.

Output: For each test case, output a single number — the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.

Examples:
input:
8
2 4
1543
7777
2 4
7154
8903
2 4
3451
8888
2 2
54
13
2 2
51
43
2 6
432015
512034
4 4
5431
1435
5518
7634
6 4
5432
1152
4542
2432
2302
5942
output:
1
1
0
1
0
2
2
2

Note: ![](CDN_BASE_URL/9c851ea8888390da76b9eeeb247a1d68) Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.

Buggy Code:
def get_matrix_rings(matrix):
    if not matrix or not matrix[0]:
        return []

    rows, cols = len(matrix), len(matrix[0])

    top = 0
    bottom = rows - 1
    left = 0
    right = cols - 1
    cnt = 0
    while top <= bottom and left <= right:
        ring = []

        for j in range(left, right + 1):
            ring.append(matrix[top][j])

        for i in range(top + 1, bottom + 1):
            ring.append(matrix[i][right])

        if top < bottom:
            for j in range(right - 1, left - 1, -1):
                ring.append(matrix[bottom][j])

        if left < right:
            for i in range(bottom - 1, top, -1):
                ring.append(matrix[i][left])

        i = 0
        while i < len(ring) and (ring[i] == 5 or ring[i] == 4 or ring[i] == 3):
            i += 1
        # print(ring, i)
        if i == len(ring):
            return 0
        reqd = [1, 5, 4, 3]
        curr = 0
        for j in range(i, i + len(ring)):
            j = j % len(ring)
            # print(j, curr)
            if ring[j] == reqd[curr]:
                curr += 1
            else:
                curr = 0
            if curr == len(reqd):
                cnt += 1
                curr = 0
        top += 1
        bottom -= 1
        left += 1
        right -= 1

    return cnt

t = int(input())
for k in range(t):
    n, m = map(int, input().split())
    matrix = []
    for _ in range(n):
        matrix.append([int(i) for i in input()])
    # print(matrix)
    print(get_matrix_rings(matrix))

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

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

Think step by step.

Response:
```plaintext
1
4 4
1543
3451
1543
3451
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love 1543
time_limit_ms: 2000
memory_limit_mb: 256
Description: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.

The first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.

Polycarp became curious about how many times the number $1543$ would appear in all layers$^{\text{∗}}$ of the carpet when traversed clockwise.

$^{\text{∗}}$The first layer of a carpet of size $n \times m$ is defined as a closed strip of length $2 \cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.

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

The first line of each test case contains a pair of numbers $n$ and $m$ ($2 \leq n, m \leq 10^3$, $n, m$ — even integers).

This is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ — the description of the carpet.

It is guaranteed that the sum of $n \cdot m$ across all test cases does not exceed $10^6$.

Output: For each test case, output a single number — the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.

Examples:
input:
8
2 4
1543
7777
2 4
7154
8903
2 4
3451
8888
2 2
54
13
2 2
51
43
2 6
432015
512034
4 4
5431
1435
5518
7634
6 4
5432
1152
4542
2432
2302
5942
output:
1
1
0
1
0
2
2
2

Note: ![](CDN_BASE_URL/9c851ea8888390da76b9eeeb247a1d68) Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.

Buggy Code:
import sys
from itertools import chain, islice, repeat

input = sys.stdin.readline

num = [1, 5, 4, 3]

for _ in range(int(input())):
    gy, gx = map(int, input().split())
    ys = (0, gy - 1)
    xs = (0, gx - 1)
    grid = [list(map(int, input().rstrip())) for _ in range(gy)]

    times = 0
    while ys[0] < ys[1] and xs[0] < xs[1]:
        layer = [
            (y, x)
            for y, x in chain(
                zip(repeat(ys[0]), range(xs[0], xs[1] + 1)),
                zip(range(ys[0] + 1, ys[1]), repeat(xs[1])),
                zip(repeat(ys[1]), range(xs[1], xs[0] - 1, -1)),
                zip(range(ys[1] - 1, ys[0], -1), repeat(xs[0])),
            )
        ]
        layer.extend(layer[:3])

        have = 0
        for y, x in layer:
            if grid[y][x] == num[have]:
                have += 1
                if have == len(num):
                    times += 1
                    have = 0
            else:
                have = 0

        ys = (ys[0] + 1, ys[1] - 1)
        xs = (xs[0] + 1, xs[1] - 1)

    print(times)

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

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

Think step by step.

Response:
```plaintext
1
4 4
1543
5431
4315
3154
```