**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Valuable Cards
time_limit_ms: 4000
memory_limit_mb: 512
Description: In his favorite cafe Kmes once again wanted to try the herring under a fur coat. Previously, it would not have been difficult for him to do this, but the cafe recently introduced a new purchasing policy.

Now, in order to make a purchase, Kmes needs to solve the following problem: $n$ cards with prices for different positions are laid out in front of him, on the $i$-th card there is an integer $a_i$, among these prices there is no whole positive integer $x$.

Kmes is asked to divide these cards into the minimum number of bad segments (so that each card belongs to exactly one segment). A segment is considered bad if it is impossible to select a subset of cards with a product equal to $x$. All segments, in which Kmes will divide the cards, must be bad.

Formally, the segment $(l, r)$ is bad if there are no indices $i_1 < i_2 < \ldots < i_k$ such that $l \le i_1, i_k \le r$, and $a_{i_1} \cdot a_{i_2} \ldots \cdot a_{i_k} = x$.

Help Kmes determine the minimum number of bad segments in order to enjoy his favorite dish.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each set of input data gives you $2$ integers $n$ and $x$ ($1 \le n \le 10^5, 2 \le x \le 10^5$) — the number of cards and the integer, respectively.

The second line of each set of input data contains $n$ integers $a_i$ ($1 \le a_i \le 2 \cdot 10^5, a_i \neq x$) — the prices on the cards.

It is guaranteed that the sum of $n$ over all sets of test data does not exceed $10^5$.

Output: For each set of input data, output the minimum number of bad segments.

Examples:
input:
8
6 4
2 3 6 2 1 2
9 100000
50000 25000 12500 6250 3125 2 4 8 16
5 2
1 1 1 1 1
8 6
4 3 4 3 4 3 4 3
7 12
6 11 1 3 11 10 2
10 5
2 4 4 2 4 4 4 3 1 1
7 8
4 6 5 1 2 4 1
8 27
3 9 17 26 2 20 9 3
output:
3
2
1
1
2
1
3
3

Note:

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

typedef long long ll;
typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<char> vc;
typedef vector<long long> vll;

#define pb push_back
#define pob pop_back()
#define mp make_pair
#define sz size()
#define ff first
#define ss second
#define PI 3.14159265359
#define M1 ll(998244353)
#define M2 ll(1000000007)
#define pres(x) cout<<fixed; cout<<setprecision(x);
#define out(x) cout<<x<<endl;

ll gcd(ll a, ll b) {
    if (b > a) {return gcd(b, a);}
    if (b == 0) {return a;}
    return gcd(b, a % b);
}
int power(int a,int b){
    int ans=1;
    while(b!=0){
        if(b&1){ans=(ans*a)%M2;b--;}
        a=(a*a)%M2;b/=2;
    }
    return ans%M2;
}
ll expo(ll a, ll b, ll mod) {
    ll res = 1;
    while (b > 0) {
        if (b & 1) res = (res * a) % mod;
        a = (a * a) % mod;
        b = b >> 1;
    }
    return res;
}
int invmod(int n){
    int ans=1;
    ans=power(n,M2-2);
    return ans;
}
int lcm(int a,int b){
    int g=gcd(a,b);
    a/=g;a*=b;
    return a;
}
ll nCr(int n, int r){
    if(r>n) return 0;
    if(r==n || r==0) return 1;
    double sum = 1;
    for(int i = 1; i <= r; i++){
        sum = sum * (n - r + i) / i;
    }
    return (long long)sum;
}
ll mrginv(vi &v,int l,int mid,int r){
    ll val=0;
    for(int i=mid;i<=r;i++){
        int it=upper_bound(v.begin()+l,v.begin()+mid,v[i])-v.begin()-l;
        val+=ll((mid-l)-it);
    }
    int i=l,j=mid,k=0;
    vi tmp(r-l+1);
    while(i<mid && j<(r+1)){
        if(v[i]<v[j]) tmp[k++]=v[i++];
        else tmp[k++]=v[j++];
    }
    while(i<mid) tmp[k++]=v[i++];
    while(j<r+1) tmp[k++]=v[j++];
    for(int i=l;i<=r;i++) v[i]=tmp[i-l];
    return val;
}
ll cntinv(vi &v,int l,int r){
    if(l==r) return 0;
    int n=r-l+1;
    int mid=l+n/2;
    return cntinv(v,l,mid-1)+cntinv(v,mid,r)+mrginv(v,l,mid,r);
}

void solve(){
    int n,x;
    cin>>n>>x;
    vi v(n);
    for(int i=0;i<n;i++) cin>>v[i];
    set<int> st;
    int cnt=1;
    st.insert(v[0]);
    for(int i=1;i<n;i++){
        if(v[i]>x) st.insert(v[i]);
        else{
            if(x%v[i]!=0) st.insert(v[i]);
            else{
                int tmp=x/v[i];
                if(st.find(tmp)==st.end()) st.insert(v[i]);
                else {
                    cnt++;
                    st.clear();
                    st.insert(v[i]);
                }
            }
        }
    }
    out(cnt)
    return;
}

int main(){

    ios::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    ll t;
    cin>>t;
    while(t--){
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To create a test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios where the code might fail. Here are some considerations:

1. **Multiple Cards with the Same Value:** If the code does not handle multiple cards with the same value correctly, it might not recognize that a segment can be bad even if it contains multiple cards with the same value.

2. **Large Values for \( x \):** If \( x \) is a large number, the code might not correctly handle the divisibility checks, especially if the set operations are not optimized.

3. **Edge Cases for \( n \):** Small values of \( n \) (e.g., \( n = 1 \)) or large values of \( n \) (e.g., \( n = 10^5 \)) can help identify issues with the algorithm's performance and correctness.

4. **Prime and Composite Values:** Testing with \( x \) being a prime number or a composite number can help identify how the code handles different types of divisibility.

5. **Special Values for \( a_i \):** Values of \( a_i \) that are factors of \( x \) or close to \( x \) can help identify issues in the divisibility checks.

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

```plaintext
3
5 6
2 3 2 3 2
4 12
4 3 1 1
7 15
3 5 1 3 5 1 1
```

### Explanation:
1. **Test Case 1:**
   - \( n = 5 \)
   - \( x = 6 \)
   - \( a = [2, 3, 2, 3, 2] \)
   - The correct output should be 3, as the cards can be divided into segments [2, 3, 2] and [3, 2], both of which are bad.

2. **Test Case 2:**
   - \( n = 4 \)
   - \( x = 12 \)
   - \( a = [4, 3, 1, 1] \)
   - The correct output should be 1, as the entire set is a bad segment.

3. **Test Case 3:**
   - \( n = 7 \)
   - \( x = 15 \)
   - \( a = [3, 5, 1, 3, 5, 1, 1] \)
   - The correct output should be 2, as the cards can be divided into segments [3, 5, 1] and [3, 5, 1, 1], both of which are bad.

This test input should help expose any issues in the provided code, particularly in handling multiple cards with the same value and ensuring that the segments are correctly identified as bad.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Valuable Cards
time_limit_ms: 4000
memory_limit_mb: 512
Description: In his favorite cafe Kmes once again wanted to try the herring under a fur coat. Previously, it would not have been difficult for him to do this, but the cafe recently introduced a new purchasing policy.

Now, in order to make a purchase, Kmes needs to solve the following problem: $n$ cards with prices for different positions are laid out in front of him, on the $i$-th card there is an integer $a_i$, among these prices there is no whole positive integer $x$.

Kmes is asked to divide these cards into the minimum number of bad segments (so that each card belongs to exactly one segment). A segment is considered bad if it is impossible to select a subset of cards with a product equal to $x$. All segments, in which Kmes will divide the cards, must be bad.

Formally, the segment $(l, r)$ is bad if there are no indices $i_1 < i_2 < \ldots < i_k$ such that $l \le i_1, i_k \le r$, and $a_{i_1} \cdot a_{i_2} \ldots \cdot a_{i_k} = x$.

Help Kmes determine the minimum number of bad segments in order to enjoy his favorite dish.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each set of input data gives you $2$ integers $n$ and $x$ ($1 \le n \le 10^5, 2 \le x \le 10^5$) — the number of cards and the integer, respectively.

The second line of each set of input data contains $n$ integers $a_i$ ($1 \le a_i \le 2 \cdot 10^5, a_i \neq x$) — the prices on the cards.

It is guaranteed that the sum of $n$ over all sets of test data does not exceed $10^5$.

Output: For each set of input data, output the minimum number of bad segments.

Examples:
input:
8
6 4
2 3 6 2 1 2
9 100000
50000 25000 12500 6250 3125 2 4 8 16
5 2
1 1 1 1 1
8 6
4 3 4 3 4 3 4 3
7 12
6 11 1 3 11 10 2
10 5
2 4 4 2 4 4 4 3 1 1
7 8
4 6 5 1 2 4 1
8 27
3 9 17 26 2 20 9 3
output:
3
2
1
1
2
1
3
3

Note:

Buggy Code:
//                                          * It_easy_if_you_try *
//================================================================
#include <bits/stdc++.h>
#define AbdullahFarg ios_base::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr);
using namespace std;
#define ul unsigned long long
#define ll long long
#define int long long
#define yes cout << "YES" << endl
#define no cout << "NO" << endl
#define all(vec) vec.begin(), vec.end()
#define allr(vec) vec.rbegin(), vec.rend()
#define sort_pair(v) sort(v.begin(), v.end(), [](auto &left, auto &right) {return left.second < right.second;});
#define endl "\n"
#define REAL fixed <<  setprecision(16)
#define PI 3.14159265358979323846
#define BtoD(s) stoll((s), 0, 2)
#define DtoB(n) bitset<64>(n).to_string().substr(64 - log2((n)))
#define p_q(q) priority_queue<ll, vector<ll>, greater<ll> > (q)
#define PRE(v,pre) partial_sum((v).begin(), (v).end(), (pre).begin())
#define RV(x) return(void (cout << x << endl))
#define __ONES(num) __builtin_popcount(num)
//#define mem(v,d) memset(v,d,sizeof(v))
#define OO 1e18+7
//================================================================
int dx[] = {1, 0, -1, 0, -1, -1, 1, 1};
int dy[] = {0, -1, 0, 1, -1, 1, -1, 1};
char di[] = {'D', 'L', 'U', 'R'};
//================================================================
void FILES () {
    AbdullahFarg
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    freopen("error.txt", "w", stderr);
#endif
}
//================================================================
const ll  N = 1e6+9 , mod = 1e9+7;
map<ll,ll>idx ;
vector<vector<ll>>dp;
vector<ll>v , divs;
ll n ;
int fun(int i , int x)
{
    if(x == 1)
        return i-1;
    if(i == n)
        return n;
    int &ret = dp[i][idx[x]];
    if(~ret)
        return ret;
    ret = fun(i+1 , x);
    if(x%v[i] == 0)
        ret = min(ret, fun(i+1 , x/v[i]));
    return ret;
}
void solve ()
{
    ll x;
    cin >> n>> x;
    for (int i = 1; i*i <= n; ++i) {
        if(x%i == 0)
        {
            divs.push_back(i);
            if(x/i != i)
                divs.push_back(x/i);
        }
    }
    v = vector<ll>(n+1);
    dp = vector<vector<ll>>(n+1 , vector<ll>(divs.size()+1 , -1));
    for (int i = 0; i < n; ++i) {
        cin >> v[i];
    }
    sort(all(divs));
    for (int i = 0; i < divs.size(); ++i) {
        idx[divs[i]] = i ;
    }
    ll start = 0 , ans = 0;
    while (start < n)
        start = fun(start , x) , ans++;
    cout << ans << endl;
    divs.clear();
    idx.clear();
}
signed main()
{
    FILES();
    int t = 1;
    cin >> t;
    while (t--)
        solve();
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to consider edge cases and scenarios that can maximize the failure probability. Here are some potential edge cases to consider:

1. **Empty or Single Card**: Test cases where the number of cards is very small, such as 1 or 2.
2. **All Cards are the Same**: Test cases where all cards have the same value.
3. **No Valid Segments**: Test cases where it is impossible to form any valid segment.
4. **Large Values**: Test cases with large values of `x` and `a_i`.
5. **Prime Values**: Test cases where `x` is a prime number and the cards have values that are multiples of `x`.

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

```plaintext
4
1 4
3
2 4
2 2
5 4
2 1 2 1 2
6 6
2 3 4 5 6 7
```

### Explanation:
1. **Test Case 1**:
   - `n = 1`, `x = 4`
   - Cards: `[3]`
   - This tests the scenario where there is only one card. The expected output is `1` because the segment is trivially bad.

2. **Test Case 2**:
   - `n = 2`, `x = 4`
   - Cards: `[2, 2]`
   - This tests the scenario where all cards have the same value. The expected output is `1` because the product of the cards is `4`, which is `x`.

3. **Test Case 3**:
   - `n = 5`, `x = 4`
   - Cards: `[2, 1, 2, 1, 2]`
   - This tests a more complex scenario where there are multiple cards, some of which can form a valid segment. The expected output is `3` because the segments can be `[2], [1, 2, 1], [2]`.

4. **Test Case 4**:
   - `n = 6`, `x = 6`
   - Cards: `[2, 3, 4, 5, 6, 7]`
   - This tests a scenario with a larger number of cards and a larger `x`. The expected output is `3` because the segments can be `[2, 3], [4, 5, 6], [7]`.

This test input should help in identifying any issues with the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Valuable Cards
time_limit_ms: 4000
memory_limit_mb: 512
Description: In his favorite cafe Kmes once again wanted to try the herring under a fur coat. Previously, it would not have been difficult for him to do this, but the cafe recently introduced a new purchasing policy.

Now, in order to make a purchase, Kmes needs to solve the following problem: $n$ cards with prices for different positions are laid out in front of him, on the $i$-th card there is an integer $a_i$, among these prices there is no whole positive integer $x$.

Kmes is asked to divide these cards into the minimum number of bad segments (so that each card belongs to exactly one segment). A segment is considered bad if it is impossible to select a subset of cards with a product equal to $x$. All segments, in which Kmes will divide the cards, must be bad.

Formally, the segment $(l, r)$ is bad if there are no indices $i_1 < i_2 < \ldots < i_k$ such that $l \le i_1, i_k \le r$, and $a_{i_1} \cdot a_{i_2} \ldots \cdot a_{i_k} = x$.

Help Kmes determine the minimum number of bad segments in order to enjoy his favorite dish.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each set of input data gives you $2$ integers $n$ and $x$ ($1 \le n \le 10^5, 2 \le x \le 10^5$) — the number of cards and the integer, respectively.

The second line of each set of input data contains $n$ integers $a_i$ ($1 \le a_i \le 2 \cdot 10^5, a_i \neq x$) — the prices on the cards.

It is guaranteed that the sum of $n$ over all sets of test data does not exceed $10^5$.

Output: For each set of input data, output the minimum number of bad segments.

Examples:
input:
8
6 4
2 3 6 2 1 2
9 100000
50000 25000 12500 6250 3125 2 4 8 16
5 2
1 1 1 1 1
8 6
4 3 4 3 4 3 4 3
7 12
6 11 1 3 11 10 2
10 5
2 4 4 2 4 4 4 3 1 1
7 8
4 6 5 1 2 4 1
8 27
3 9 17 26 2 20 9 3
output:
3
2
1
1
2
1
3
3

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define FOR(i, n) for (ll i = 0; i < ll(n); i++)
#define FORb(i, n) for (ll i = n; i >= 0; i--)
#define vi vector<ll>
#define pb push_back
#define print_array(array, length)  \
    for (ll i = 0; i < length; i++) \
        cout << (array[i]) << " ";
typedef long double ld;
typedef pair<ll, ll> pi;
const long long mod = 1e9 + 7;
const long long Mod = 998244353;
const long long inf = 1e18;
#define all(x) x.begin(), x.end()
#define MP make_pair
#define REP(i, a, b) for (ll i = (ll)a; i < (ll)b; i++)
#define el cout << endl;
#define sz(x) (ll)(x).size()
#define vec_sort(a) sort(a.begin(), a.end())
#define vec_rev(a) reverse(a.begin(), a.end())
#define vec_sort_r(a) sort(a.begin(), a.end(), comp)
#define setbits(x) __builtin_popcountll(x) // count number of setbits in a number
#define pi pair<ll, ll>
#define fi first
#define se second
#define ft front
#define all(x) x.begin(), x.end()
#define yess cout << "YES" << endl;
#define noo cout << "NO" << endl;

int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    // start
    ll tests;
    cin >> tests;
    ll xyz = 0;
    while (tests--)
    {
        /*
        n=7
        x=12
        5,2,2,9,
        */
        ll n, x;
        cin >> n >> x;
        xyz++;
        vector<ll> arr(n);
        for (ll &x : arr)
            cin >> x;
        if (xyz == 305)
        {
            cout <<arr[4]<<" "<<arr[5]<<" "<<arr[6] << endl;
        }
        ll ans = 1;
        set<ll> st;
        FOR(i, n)
        {
            if (x % arr[i] == 0)
            {
                if (st.find(x / arr[i]) != st.end())
                {
                    ans++;
                    st.clear();
                    st.insert(arr[i]);
                }
                else
                {
                    for (auto it : st)
                    {
                        if (it * arr[i] > x)
                        {
                            break;
                        }
                        st.insert(arr[i] * it);
                    }
                    st.insert(arr[i]);
                }
            }
        }
        cout << ans << endl;
    }
    // END
}

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

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

Think step by step.

Response:
```plaintext
1
10 12
1 2 3 4 6 1 2 3 4 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Valuable Cards
time_limit_ms: 4000
memory_limit_mb: 512
Description: In his favorite cafe Kmes once again wanted to try the herring under a fur coat. Previously, it would not have been difficult for him to do this, but the cafe recently introduced a new purchasing policy.

Now, in order to make a purchase, Kmes needs to solve the following problem: $n$ cards with prices for different positions are laid out in front of him, on the $i$-th card there is an integer $a_i$, among these prices there is no whole positive integer $x$.

Kmes is asked to divide these cards into the minimum number of bad segments (so that each card belongs to exactly one segment). A segment is considered bad if it is impossible to select a subset of cards with a product equal to $x$. All segments, in which Kmes will divide the cards, must be bad.

Formally, the segment $(l, r)$ is bad if there are no indices $i_1 < i_2 < \ldots < i_k$ such that $l \le i_1, i_k \le r$, and $a_{i_1} \cdot a_{i_2} \ldots \cdot a_{i_k} = x$.

Help Kmes determine the minimum number of bad segments in order to enjoy his favorite dish.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each set of input data gives you $2$ integers $n$ and $x$ ($1 \le n \le 10^5, 2 \le x \le 10^5$) — the number of cards and the integer, respectively.

The second line of each set of input data contains $n$ integers $a_i$ ($1 \le a_i \le 2 \cdot 10^5, a_i \neq x$) — the prices on the cards.

It is guaranteed that the sum of $n$ over all sets of test data does not exceed $10^5$.

Output: For each set of input data, output the minimum number of bad segments.

Examples:
input:
8
6 4
2 3 6 2 1 2
9 100000
50000 25000 12500 6250 3125 2 4 8 16
5 2
1 1 1 1 1
8 6
4 3 4 3 4 3 4 3
7 12
6 11 1 3 11 10 2
10 5
2 4 4 2 4 4 4 3 1 1
7 8
4 6 5 1 2 4 1
8 27
3 9 17 26 2 20 9 3
output:
3
2
1
1
2
1
3
3

Note:

Buggy Code:
c=[]
t=int(input())
for _ in range(t):
    n,x=map(int,input().split())
    a=list(map(int,input().split()))
    b=[]
    ans=1
    for i in range(n):
        if a[i] in b:
            b.clear()
            ans+=1

        if (x%a[i]==0):

            b.append(x/a[i])
    c.append(ans)
print(*c)

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on scenarios where the code's logic fails. The key issues in the buggy code are:

1. The list `b` is used to track potential factors of `x` that can form a product equal to `x`.
2. The code incorrectly resets `b` and increments `ans` whenever it encounters a number that is already in `b`.
3. The code does not correctly handle cases where the product of multiple numbers can equal `x`.

To create a challenging test input, we should consider the following edge cases:

1. **Repeating values that are factors of `x`**: This can cause the code to incorrectly reset `b` and increment `ans` too many times.
2. **Combinations of values that can form `x`**: This can expose the issue with the code's logic for handling multiple factors.
3. **Values that are not factors of `x`**: This can ensure that the code does not incorrectly reset `b`.

Let's create a small-scale test input that combines these elements:

```plaintext
3
5 6
2 3 2 3 2
4 12
3 4 3 4
6 10
2 5 2 5 2 5
```

Explanation:
1. **Test Case 1 (5 6)**:
   - The sequence `2 3 2 3 2` contains multiple factors of 6 (2 and 3).
   - The correct answer should be 1 segment, as the entire sequence can be a bad segment.
   - The buggy code will incorrectly reset `b` and increment `ans` multiple times due to repeated factors.

2. **Test Case 2 (4 12)**:
   - The sequence `3 4 3 4` contains multiple factors of 12 (3 and 4).
   - The correct answer should be 1 segment, as the entire sequence can be a bad segment.
   - The buggy code will incorrectly reset `b` and increment `ans` multiple times due to repeated factors.

3. **Test Case 3 (6 10)**:
   - The sequence `2 5 2 5 2 5` contains multiple factors of 10 (2 and 5).
   - The correct answer should be 1 segment, as the entire sequence can be a bad segment.
   - The buggy code will incorrectly reset `b` and increment `ans` multiple times due to repeated factors.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Valuable Cards
time_limit_ms: 4000
memory_limit_mb: 512
Description: In his favorite cafe Kmes once again wanted to try the herring under a fur coat. Previously, it would not have been difficult for him to do this, but the cafe recently introduced a new purchasing policy.

Now, in order to make a purchase, Kmes needs to solve the following problem: $n$ cards with prices for different positions are laid out in front of him, on the $i$-th card there is an integer $a_i$, among these prices there is no whole positive integer $x$.

Kmes is asked to divide these cards into the minimum number of bad segments (so that each card belongs to exactly one segment). A segment is considered bad if it is impossible to select a subset of cards with a product equal to $x$. All segments, in which Kmes will divide the cards, must be bad.

Formally, the segment $(l, r)$ is bad if there are no indices $i_1 < i_2 < \ldots < i_k$ such that $l \le i_1, i_k \le r$, and $a_{i_1} \cdot a_{i_2} \ldots \cdot a_{i_k} = x$.

Help Kmes determine the minimum number of bad segments in order to enjoy his favorite dish.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each set of input data gives you $2$ integers $n$ and $x$ ($1 \le n \le 10^5, 2 \le x \le 10^5$) — the number of cards and the integer, respectively.

The second line of each set of input data contains $n$ integers $a_i$ ($1 \le a_i \le 2 \cdot 10^5, a_i \neq x$) — the prices on the cards.

It is guaranteed that the sum of $n$ over all sets of test data does not exceed $10^5$.

Output: For each set of input data, output the minimum number of bad segments.

Examples:
input:
8
6 4
2 3 6 2 1 2
9 100000
50000 25000 12500 6250 3125 2 4 8 16
5 2
1 1 1 1 1
8 6
4 3 4 3 4 3 4 3
7 12
6 11 1 3 11 10 2
10 5
2 4 4 2 4 4 4 3 1 1
7 8
4 6 5 1 2 4 1
8 27
3 9 17 26 2 20 9 3
output:
3
2
1
1
2
1
3
3

Note:

Buggy Code:
import os,io
import sys
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# input = lambda: sys.stdin.readline().rstrip("\r\n")
# from sortedcontainers import SortedDict
from collections import defaultdict

test=int(input())

for asdf in range(test):
    n,x=map(int,input().split())
    a=list(map(int,input().split()))
    cnt=set()

    b=1
    ans=0
    d=0
    for i in range(n):
        # print(keys_to_process,i)
        temp=set()
        for j in cnt:
            if j * a[i] == x:
                cnt=set()
                temp=set()
                cnt.add(a[i])
                ans += 1
                # print(i)
                break  # Exit the loop since cnt is reset
            elif j * a[i] > x:
                break
            else:
                temp.add(a[i])
            # print(cnt, i)

        for j in temp:
            cnt.add(j)

        cnt.add(a[i])

    print(ans+1)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Valuable Cards
time_limit_ms: 4000
memory_limit_mb: 512
Description: In his favorite cafe Kmes once again wanted to try the herring under a fur coat. Previously, it would not have been difficult for him to do this, but the cafe recently introduced a new purchasing policy.

Now, in order to make a purchase, Kmes needs to solve the following problem: $n$ cards with prices for different positions are laid out in front of him, on the $i$-th card there is an integer $a_i$, among these prices there is no whole positive integer $x$.

Kmes is asked to divide these cards into the minimum number of bad segments (so that each card belongs to exactly one segment). A segment is considered bad if it is impossible to select a subset of cards with a product equal to $x$. All segments, in which Kmes will divide the cards, must be bad.

Formally, the segment $(l, r)$ is bad if there are no indices $i_1 < i_2 < \ldots < i_k$ such that $l \le i_1, i_k \le r$, and $a_{i_1} \cdot a_{i_2} \ldots \cdot a_{i_k} = x$.

Help Kmes determine the minimum number of bad segments in order to enjoy his favorite dish.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each set of input data gives you $2$ integers $n$ and $x$ ($1 \le n \le 10^5, 2 \le x \le 10^5$) — the number of cards and the integer, respectively.

The second line of each set of input data contains $n$ integers $a_i$ ($1 \le a_i \le 2 \cdot 10^5, a_i \neq x$) — the prices on the cards.

It is guaranteed that the sum of $n$ over all sets of test data does not exceed $10^5$.

Output: For each set of input data, output the minimum number of bad segments.

Examples:
input:
8
6 4
2 3 6 2 1 2
9 100000
50000 25000 12500 6250 3125 2 4 8 16
5 2
1 1 1 1 1
8 6
4 3 4 3 4 3 4 3
7 12
6 11 1 3 11 10 2
10 5
2 4 4 2 4 4 4 3 1 1
7 8
4 6 5 1 2 4 1
8 27
3 9 17 26 2 20 9 3
output:
3
2
1
1
2
1
3
3

Note:

Buggy Code:
import sys
input = sys.stdin.readline
t=int(input())
import random
r=random.randint(10**4,10**5)
for _ in range(t):
  n,x=map(int,input().split())
  a=list(map(int,input().split()))
  ans=1
  se=set()
  se.add(1^r)
  if x%a[0]==0:
    se.add(a[0]^r)
  for i in range(2,n):
    l=set()
    flg=False
    for j in se:
      tmp=(j^r)*a[i]
      if tmp==x:
        ans+=1
        se=set()
        se.add(1^r)
        flg=True
        if x%a[i]==0:
          se.add(a[i]^r)
        break
      elif x%tmp==0:
        l.add(tmp^r)
    if not flg:
      for j in l:
        se.add(j)
  print(ans)

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases or scenarios that can maximize the failure probability. The buggy code has issues with how it handles the XOR operations and the set updates, which can lead to incorrect results.

Here are the key points to consider for the test input:
1. **Small-scale input**: The test input should be small enough to be manageable and within the constraints.
2. **Edge cases**: Use values that can trigger the bugs in the code.
3. **XOR operation**: The use of XOR in the code is unconventional and can lead to incorrect results if not handled carefully.
4. **Product of subsets**: Ensure that the test input includes scenarios where the product of subsets can or cannot equal \( x \).

Given these considerations, let's construct a test input:

```plaintext
3
5 6
2 3 1 1 1
4 12
3 4 3 1
6 10
2 5 2 5 2 5
```

### Explanation:
1. **Test Case 1**:
   - \( n = 5 \)
   - \( x = 6 \)
   - Cards: \( [2, 3, 1, 1, 1] \)
   - This test case is designed to check if the code correctly identifies that the minimum number of bad segments is 1, as the product of 2 and 3 equals 6, and the remaining 1s do not affect the product.

2. **Test Case 2**:
   - \( n = 4 \)
   - \( x = 12 \)
   - Cards: \( [3, 4, 3, 1] \)
   - This test case is designed to check if the code correctly identifies that the minimum number of bad segments is 1, as the product of 3 and 4 equals 12, and the remaining 3 and 1 do not affect the product.

3. **Test Case 3**:
   - \( n = 6 \)
   - \( x = 10 \)
   - Cards: \( [2, 5, 2, 5, 2, 5] \)
   - This test case is designed to check if the code correctly identifies that the minimum number of bad segments is 3, as the product of 2 and 5 equals 10, and the input has multiple pairs of 2 and 5, which should be segmented accordingly.

This test input should help in exposing the bugs in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Valuable Cards
time_limit_ms: 4000
memory_limit_mb: 512
Description: In his favorite cafe Kmes once again wanted to try the herring under a fur coat. Previously, it would not have been difficult for him to do this, but the cafe recently introduced a new purchasing policy.

Now, in order to make a purchase, Kmes needs to solve the following problem: $n$ cards with prices for different positions are laid out in front of him, on the $i$-th card there is an integer $a_i$, among these prices there is no whole positive integer $x$.

Kmes is asked to divide these cards into the minimum number of bad segments (so that each card belongs to exactly one segment). A segment is considered bad if it is impossible to select a subset of cards with a product equal to $x$. All segments, in which Kmes will divide the cards, must be bad.

Formally, the segment $(l, r)$ is bad if there are no indices $i_1 < i_2 < \ldots < i_k$ such that $l \le i_1, i_k \le r$, and $a_{i_1} \cdot a_{i_2} \ldots \cdot a_{i_k} = x$.

Help Kmes determine the minimum number of bad segments in order to enjoy his favorite dish.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each set of input data gives you $2$ integers $n$ and $x$ ($1 \le n \le 10^5, 2 \le x \le 10^5$) — the number of cards and the integer, respectively.

The second line of each set of input data contains $n$ integers $a_i$ ($1 \le a_i \le 2 \cdot 10^5, a_i \neq x$) — the prices on the cards.

It is guaranteed that the sum of $n$ over all sets of test data does not exceed $10^5$.

Output: For each set of input data, output the minimum number of bad segments.

Examples:
input:
8
6 4
2 3 6 2 1 2
9 100000
50000 25000 12500 6250 3125 2 4 8 16
5 2
1 1 1 1 1
8 6
4 3 4 3 4 3 4 3
7 12
6 11 1 3 11 10 2
10 5
2 4 4 2 4 4 4 3 1 1
7 8
4 6 5 1 2 4 1
8 27
3 9 17 26 2 20 9 3
output:
3
2
1
1
2
1
3
3

Note:

Buggy Code:
import java.util.*;
public class Main
{
	public static void main(String[] args){
		Scanner input = new Scanner(System.in);
		int t = input.nextInt();

		for(int l=0;l<t;l++){
			int n = input.nextInt();
			int x = input.nextInt();
			int[] cards = new int[n];
			int j;
			int u;

			for(int i=0; i<n;i++){
				cards[i] = input.nextInt();
			}

			int total=1;
			for(j=0;j<n;j++){
				for(u=(j+1);u<n;u++){
					if(cards[j]*cards[u]==x){
						total +=1;
						j=u-1;
						break;
					}
				}
			}
			System.out.println(total);
	}

}
}

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on the logic that determines the number of bad segments. The code incorrectly increments the `total` counter whenever it finds a pair of cards whose product equals `x`, and then it resets the search to the next card after the pair. This logic can fail in scenarios where the cards are not divided into the minimum number of bad segments.

### Step-by-Step Analysis:
1. **Identify the Bug:**
   - The code only checks pairs of cards and increments the total counter whenever it finds a pair whose product equals `x`.
   - It does not consider the possibility of multiple cards forming a bad segment.

2. **Edge Cases to Consider:**
   - **Single Card Segments:** If all cards are such that no two cards can form a product equal to `x`, the code should return `1` (one segment).
   - **Multiple Pairs:** If there are multiple pairs of cards whose product equals `x`, the code should correctly divide them into the minimum number of segments.
   - **Consecutive Pairs:** If there are consecutive pairs of cards whose product equals `x`, the code should not count them as separate segments.

3. **Test Input Generation:**
   - We need a test input that includes multiple pairs of cards whose product equals `x` and ensures that the code does not correctly divide them into the minimum number of segments.

### Chosen Test Input:
Let's create a test input where there are multiple pairs of cards whose product equals `x`, but the code should return a specific number of segments.

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

### Explanation:
- **Test Case:**
  - `t = 1`: One test case.
  - `n = 6`: Six cards.
  - `x = 6`: The integer `x`.
  - `cards = [2, 3, 2, 3, 2, 3]`: The prices on the cards.

- **Expected Output:**
  - The correct output should be `3` because the cards can be divided into the following bad segments: `[2, 3]`, `[2, 3]`, `[2, 3]`.

- **Buggy Code Output:**
  - The buggy code will likely output a higher number because it will count each pair as a separate segment without considering the minimum number of segments.

This test input should expose the bug in the given code by demonstrating that it does not correctly minimize the number of bad segments.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Valuable Cards
time_limit_ms: 4000
memory_limit_mb: 512
Description: In his favorite cafe Kmes once again wanted to try the herring under a fur coat. Previously, it would not have been difficult for him to do this, but the cafe recently introduced a new purchasing policy.

Now, in order to make a purchase, Kmes needs to solve the following problem: $n$ cards with prices for different positions are laid out in front of him, on the $i$-th card there is an integer $a_i$, among these prices there is no whole positive integer $x$.

Kmes is asked to divide these cards into the minimum number of bad segments (so that each card belongs to exactly one segment). A segment is considered bad if it is impossible to select a subset of cards with a product equal to $x$. All segments, in which Kmes will divide the cards, must be bad.

Formally, the segment $(l, r)$ is bad if there are no indices $i_1 < i_2 < \ldots < i_k$ such that $l \le i_1, i_k \le r$, and $a_{i_1} \cdot a_{i_2} \ldots \cdot a_{i_k} = x$.

Help Kmes determine the minimum number of bad segments in order to enjoy his favorite dish.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each set of input data gives you $2$ integers $n$ and $x$ ($1 \le n \le 10^5, 2 \le x \le 10^5$) — the number of cards and the integer, respectively.

The second line of each set of input data contains $n$ integers $a_i$ ($1 \le a_i \le 2 \cdot 10^5, a_i \neq x$) — the prices on the cards.

It is guaranteed that the sum of $n$ over all sets of test data does not exceed $10^5$.

Output: For each set of input data, output the minimum number of bad segments.

Examples:
input:
8
6 4
2 3 6 2 1 2
9 100000
50000 25000 12500 6250 3125 2 4 8 16
5 2
1 1 1 1 1
8 6
4 3 4 3 4 3 4 3
7 12
6 11 1 3 11 10 2
10 5
2 4 4 2 4 4 4 3 1 1
7 8
4 6 5 1 2 4 1
8 27
3 9 17 26 2 20 9 3
output:
3
2
1
1
2
1
3
3

Note:

Buggy Code:
import java.util.ArrayList;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int t = sc.nextInt();

        for (int k = 0; k < t; k++) {
            int n = sc.nextInt();
            int x = sc.nextInt();
            int[] a = new int[n];
            ArrayList<Integer> aa = new ArrayList<>();

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

            int s = 1;
            aa.add(a[0]);

            for (int i = 1; i < a.length; i++) {

                if (x % a[i] == 0 && aa.contains(x / a[i])) {

                    s++;
                    aa.clear();
                }
                    for (int j = 1  ; j < aa.size(); j++) {
                        for (int l = 0; l < aa.size() - 1; l++) {
                            if (aa.get(j) * aa.get(l + 1) == (x / a[j]) &&  x % a[j] == 0) {
                                s++;
                                aa.clear();
                            }
                        }
                    }

                aa.add(a[i]);
            }

            System.out.println(s);
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
10 6
2 3 2 3 2 3 2 3 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Valuable Cards
time_limit_ms: 4000
memory_limit_mb: 512
Description: In his favorite cafe Kmes once again wanted to try the herring under a fur coat. Previously, it would not have been difficult for him to do this, but the cafe recently introduced a new purchasing policy.

Now, in order to make a purchase, Kmes needs to solve the following problem: $n$ cards with prices for different positions are laid out in front of him, on the $i$-th card there is an integer $a_i$, among these prices there is no whole positive integer $x$.

Kmes is asked to divide these cards into the minimum number of bad segments (so that each card belongs to exactly one segment). A segment is considered bad if it is impossible to select a subset of cards with a product equal to $x$. All segments, in which Kmes will divide the cards, must be bad.

Formally, the segment $(l, r)$ is bad if there are no indices $i_1 < i_2 < \ldots < i_k$ such that $l \le i_1, i_k \le r$, and $a_{i_1} \cdot a_{i_2} \ldots \cdot a_{i_k} = x$.

Help Kmes determine the minimum number of bad segments in order to enjoy his favorite dish.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each set of input data gives you $2$ integers $n$ and $x$ ($1 \le n \le 10^5, 2 \le x \le 10^5$) — the number of cards and the integer, respectively.

The second line of each set of input data contains $n$ integers $a_i$ ($1 \le a_i \le 2 \cdot 10^5, a_i \neq x$) — the prices on the cards.

It is guaranteed that the sum of $n$ over all sets of test data does not exceed $10^5$.

Output: For each set of input data, output the minimum number of bad segments.

Examples:
input:
8
6 4
2 3 6 2 1 2
9 100000
50000 25000 12500 6250 3125 2 4 8 16
5 2
1 1 1 1 1
8 6
4 3 4 3 4 3 4 3
7 12
6 11 1 3 11 10 2
10 5
2 4 4 2 4 4 4 3 1 1
7 8
4 6 5 1 2 4 1
8 27
3 9 17 26 2 20 9 3
output:
3
2
1
1
2
1
3
3

Note:

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

using namespace std;

typedef long long li;
typedef long double ld;

const int INF = 1e9;

const int N = 5e5 + 10;

void factorize(int n, map<int, int>& dk) {
    for(li i = 2; i * i <= n; ++i) {
        while (n % i == 0) {
            n /= i;
            ++dk[i];
        }
    }
    if (n != 1) {
        ++dk[n];
    }
}

void solve() {
    int n, x; cin >> n >> x;
    map<int, int> fx, fn, used;
    factorize(x, fx);
    int kl = 1;
    for(int i = 0; i < n; ++i) {
        int a; cin >> a;
        map<int, int> cur;
        factorize(a, cur);
        bool ok = true;
        for(auto& x : cur) {
            if (x.second > fx[x.first] || used[x.first] + x.second > fx[x.first]) {
                ok = false;
            }
        }
        if (ok) {
            for(auto& x : cur) {
                fn[x.first] += x.second;
                used[x.first] += x.second;
            }
            ok = true;
            for(auto& x : fx) {
                if (x.second != fn[x.first]) {
                    ok = false;
                }
            }
            if (ok) {
                ++kl;
                fn = cur;
                used.clear();
                for(auto& x : cur) {
                    used[x.first] += x.second;
                }
            }
        }
    }
    cout << kl << '\n';
}

int main() {
    cin.tie(0)->sync_with_stdio(0);
    cout << setprecision(15) << fixed;
    int t = 1;
    cin >> t;
    while (t--) solve();
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases that can cause the algorithm to fail. Here are some considerations:

1. **Small Values of \( x \)**: Test cases where \( x \) is a small number, such as 2, 3, 4, etc., to ensure the factorization and segment division logic is correct.
2. **Large Values of \( x \)**: Test cases where \( x \) is a large number to check the handling of large prime factors.
3. **Prime Values of \( x \)**: Test cases where \( x \) is a prime number to ensure the algorithm handles prime factorization correctly.
4. **Composite Values of \( x \)**: Test cases where \( x \) is a composite number with multiple prime factors.
5. **Edge Cases with \( n \)**: Test cases where \( n \) is at the boundary values, such as 1 and 100,000.
6. **Repetitive Values in \( a \)**: Test cases where the values in \( a \) are repetitive, especially with factors of \( x \).

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

```plaintext
3
5 12
6 2 3 4 1
6 30
5 6 2 3 10 1
8 27
9 3 1 27 2 3 9 1
```

This test input includes:
1. A test case where \( x = 12 \) and the values in \( a \) are factors of 12.
2. A test case where \( x = 30 \) and the values in \( a \) are factors of 30.
3. A test case where \( x = 27 \) and the values in \( a \) include multiples and factors of 27.

These cases will help to expose any issues in the factorization and segment division logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Valuable Cards
time_limit_ms: 4000
memory_limit_mb: 512
Description: In his favorite cafe Kmes once again wanted to try the herring under a fur coat. Previously, it would not have been difficult for him to do this, but the cafe recently introduced a new purchasing policy.

Now, in order to make a purchase, Kmes needs to solve the following problem: $n$ cards with prices for different positions are laid out in front of him, on the $i$-th card there is an integer $a_i$, among these prices there is no whole positive integer $x$.

Kmes is asked to divide these cards into the minimum number of bad segments (so that each card belongs to exactly one segment). A segment is considered bad if it is impossible to select a subset of cards with a product equal to $x$. All segments, in which Kmes will divide the cards, must be bad.

Formally, the segment $(l, r)$ is bad if there are no indices $i_1 < i_2 < \ldots < i_k$ such that $l \le i_1, i_k \le r$, and $a_{i_1} \cdot a_{i_2} \ldots \cdot a_{i_k} = x$.

Help Kmes determine the minimum number of bad segments in order to enjoy his favorite dish.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each set of input data gives you $2$ integers $n$ and $x$ ($1 \le n \le 10^5, 2 \le x \le 10^5$) — the number of cards and the integer, respectively.

The second line of each set of input data contains $n$ integers $a_i$ ($1 \le a_i \le 2 \cdot 10^5, a_i \neq x$) — the prices on the cards.

It is guaranteed that the sum of $n$ over all sets of test data does not exceed $10^5$.

Output: For each set of input data, output the minimum number of bad segments.

Examples:
input:
8
6 4
2 3 6 2 1 2
9 100000
50000 25000 12500 6250 3125 2 4 8 16
5 2
1 1 1 1 1
8 6
4 3 4 3 4 3 4 3
7 12
6 11 1 3 11 10 2
10 5
2 4 4 2 4 4 4 3 1 1
7 8
4 6 5 1 2 4 1
8 27
3 9 17 26 2 20 9 3
output:
3
2
1
1
2
1
3
3

Note:

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

int main(){
    int t;
    cin>>t;
    while(t--) {
        int n,x;
        cin>>n>>x;
        vector<int> a(n,0);
        for(int i=0;i<n;i++)cin>>a[i];
        set<int>s;
        set<int>s1;
        set<int>s2;
        int count=1;
        for(int i=0;i<n;i++){
            if(x%a[i]==0){
                int check=x/a[i];
                if(s.find(check)==s.end()){
                    s.insert(a[i]);
                    for(int c : s){
                        if(check%c==0&&c!=a[i]){
                            int d=(x*c)/check;
                            s1.insert(d);
//                            cout<<"c "<<c<<endl;
//                            cout<<"check "<<check<<endl;
//                            cout<<"d "<<d<<endl;
                        }
                    }
                    set_union(s.begin(), s.end(), s1.begin(),
                              s1.end(), inserter(s2, s2.begin()));
                    s=s2;
//                    cout<<"count "<<count<<endl;
                }
                else{
                    s.clear();
                    s1.clear();
                    s2.clear();
                    count++;
//                    cout<<"count else  "<<count<<endl;
//                    cout<<a[i]<<endl;
                    s.insert(a[i]);
                }
            }
        }
        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:
```plaintext
3
6 6
2 3 1 2 3 1
10 10
2 5 2 5 2 5 2 5 2 5
5 15
3 5 1 3 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Valuable Cards
time_limit_ms: 4000
memory_limit_mb: 512
Description: In his favorite cafe Kmes once again wanted to try the herring under a fur coat. Previously, it would not have been difficult for him to do this, but the cafe recently introduced a new purchasing policy.

Now, in order to make a purchase, Kmes needs to solve the following problem: $n$ cards with prices for different positions are laid out in front of him, on the $i$-th card there is an integer $a_i$, among these prices there is no whole positive integer $x$.

Kmes is asked to divide these cards into the minimum number of bad segments (so that each card belongs to exactly one segment). A segment is considered bad if it is impossible to select a subset of cards with a product equal to $x$. All segments, in which Kmes will divide the cards, must be bad.

Formally, the segment $(l, r)$ is bad if there are no indices $i_1 < i_2 < \ldots < i_k$ such that $l \le i_1, i_k \le r$, and $a_{i_1} \cdot a_{i_2} \ldots \cdot a_{i_k} = x$.

Help Kmes determine the minimum number of bad segments in order to enjoy his favorite dish.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each set of input data gives you $2$ integers $n$ and $x$ ($1 \le n \le 10^5, 2 \le x \le 10^5$) — the number of cards and the integer, respectively.

The second line of each set of input data contains $n$ integers $a_i$ ($1 \le a_i \le 2 \cdot 10^5, a_i \neq x$) — the prices on the cards.

It is guaranteed that the sum of $n$ over all sets of test data does not exceed $10^5$.

Output: For each set of input data, output the minimum number of bad segments.

Examples:
input:
8
6 4
2 3 6 2 1 2
9 100000
50000 25000 12500 6250 3125 2 4 8 16
5 2
1 1 1 1 1
8 6
4 3 4 3 4 3 4 3
7 12
6 11 1 3 11 10 2
10 5
2 4 4 2 4 4 4 3 1 1
7 8
4 6 5 1 2 4 1
8 27
3 9 17 26 2 20 9 3
output:
3
2
1
1
2
1
3
3

Note:

Buggy Code:
import os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
Int = lambda: int(input())
Ints = lambda: [*map(int,input().split())]
from collections import Counter
from math import gcd

sieve = [1]*200001
for i in range(2,int(200001**0.5)+1):
    if sieve[i]:
        for j in range(i*i,200001,i):sieve[j]=0
primes = [i for i in range(2,200001) if sieve[i]]
Primes = set(primes)

def f(x):
    facs = Counter()
    for p in primes:
        if x%p==0:
            while x%p==0:x//=p; facs[p]+=1
            if x==1: break
    return facs

for _ in range(Int()):
    n,x = Ints()
    y = x
    L = Ints()
    if x in Primes: print(1); continue
    facs = f(x)
    ans = 1
    u = Counter()
    for i in L:
        # print(i,u)
        gamma = f(i) #lmao
        if all((u[i]+gamma[i]<=facs[i]) and facs[i] for i in gamma): u+=gamma
        if u==facs:
            u = gamma
            # print("reset")
            ans+=1
    print(ans)

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

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

Think step by step.

Response:
```plaintext
1
10 12
2 2 3 3 2 2 3 3 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Valuable Cards
time_limit_ms: 4000
memory_limit_mb: 512
Description: In his favorite cafe Kmes once again wanted to try the herring under a fur coat. Previously, it would not have been difficult for him to do this, but the cafe recently introduced a new purchasing policy.

Now, in order to make a purchase, Kmes needs to solve the following problem: $n$ cards with prices for different positions are laid out in front of him, on the $i$-th card there is an integer $a_i$, among these prices there is no whole positive integer $x$.

Kmes is asked to divide these cards into the minimum number of bad segments (so that each card belongs to exactly one segment). A segment is considered bad if it is impossible to select a subset of cards with a product equal to $x$. All segments, in which Kmes will divide the cards, must be bad.

Formally, the segment $(l, r)$ is bad if there are no indices $i_1 < i_2 < \ldots < i_k$ such that $l \le i_1, i_k \le r$, and $a_{i_1} \cdot a_{i_2} \ldots \cdot a_{i_k} = x$.

Help Kmes determine the minimum number of bad segments in order to enjoy his favorite dish.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each set of input data gives you $2$ integers $n$ and $x$ ($1 \le n \le 10^5, 2 \le x \le 10^5$) — the number of cards and the integer, respectively.

The second line of each set of input data contains $n$ integers $a_i$ ($1 \le a_i \le 2 \cdot 10^5, a_i \neq x$) — the prices on the cards.

It is guaranteed that the sum of $n$ over all sets of test data does not exceed $10^5$.

Output: For each set of input data, output the minimum number of bad segments.

Examples:
input:
8
6 4
2 3 6 2 1 2
9 100000
50000 25000 12500 6250 3125 2 4 8 16
5 2
1 1 1 1 1
8 6
4 3 4 3 4 3 4 3
7 12
6 11 1 3 11 10 2
10 5
2 4 4 2 4 4 4 3 1 1
7 8
4 6 5 1 2 4 1
8 27
3 9 17 26 2 20 9 3
output:
3
2
1
1
2
1
3
3

Note:

Buggy Code:
for _ in range(int(input())):
    n, x = [int(i) for i in input().split()]
    a = [int(i) for i in input().split()]
    s = [x]
    res = 0
    for i in a:
        j = 0
        while j<len(s):
            if s[j]%i==0:
                s[j] //= i
                if s[j] == 1:
                    res +=1
                    s = [x//i]
                    break
            j += 1
        if x%i == 0: s.append(x)
    print(res+1)

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

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

Think step by step.

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

1. **Edge Cases**: The code should handle edge cases where the product of a subset of cards equals `x` or where no such subset exists.
2. **Multiple Test Cases**: The input should contain multiple test cases to ensure the code handles different scenarios.
3. **Special Values**: Use values that are likely to trigger the bug, such as prime numbers, powers of `x`, and numbers that are factors of `x`.

Given the code's logic, it seems to fail when it does not correctly handle the division and reset of the `s` list. Specifically, it might not correctly manage the segments when multiple factors of `x` are present in the input list.

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

```plaintext
4
6 4
2 2 2 2 2 2
5 6
2 3 1 2 3
7 12
3 4 1 3 4 1 1
8 8
2 4 2 2 2 2 2 2
```

### Explanation:
1. **Test Case 1**:
   - `n = 6`, `x = 4`
   - Input: `2 2 2 2 2 2`
   - Expected Output: `3`
   - Explanation: The cards can be divided into three segments: `[2, 2]`, `[2, 2]`, and `[2, 2]`. Each segment has a product of 4, which is equal to `x`.

2. **Test Case 2**:
   - `n = 5`, `x = 6`
   - Input: `2 3 1 2 3`
   - Expected Output: `1`
   - Explanation: The entire sequence can be considered a single bad segment since the product of all cards is 36, and no subset has a product of 6.

3. **Test Case 3**:
   - `n = 7`, `x = 12`
   - Input: `3 4 1 3 4 1 1`
   - Expected Output: `2`
   - Explanation: The cards can be divided into two segments: `[3, 4]` and `[3, 4, 1, 1, 1]`. Each segment has a product of 12, which is equal to `x`.

4. **Test Case 4**:
   - `n = 8`, `x = 8`
   - Input: `2 4 2 2 2 2 2 2`
   - Expected Output: `3`
   - Explanation: The cards can be divided into three segments: `[2, 4, 2]`, `[2, 2, 2]`, and `[2, 2]`. Each segment has a product of 8, which is equal to `x`.

This test input should help identify and expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Valuable Cards
time_limit_ms: 4000
memory_limit_mb: 512
Description: In his favorite cafe Kmes once again wanted to try the herring under a fur coat. Previously, it would not have been difficult for him to do this, but the cafe recently introduced a new purchasing policy.

Now, in order to make a purchase, Kmes needs to solve the following problem: $n$ cards with prices for different positions are laid out in front of him, on the $i$-th card there is an integer $a_i$, among these prices there is no whole positive integer $x$.

Kmes is asked to divide these cards into the minimum number of bad segments (so that each card belongs to exactly one segment). A segment is considered bad if it is impossible to select a subset of cards with a product equal to $x$. All segments, in which Kmes will divide the cards, must be bad.

Formally, the segment $(l, r)$ is bad if there are no indices $i_1 < i_2 < \ldots < i_k$ such that $l \le i_1, i_k \le r$, and $a_{i_1} \cdot a_{i_2} \ldots \cdot a_{i_k} = x$.

Help Kmes determine the minimum number of bad segments in order to enjoy his favorite dish.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each set of input data gives you $2$ integers $n$ and $x$ ($1 \le n \le 10^5, 2 \le x \le 10^5$) — the number of cards and the integer, respectively.

The second line of each set of input data contains $n$ integers $a_i$ ($1 \le a_i \le 2 \cdot 10^5, a_i \neq x$) — the prices on the cards.

It is guaranteed that the sum of $n$ over all sets of test data does not exceed $10^5$.

Output: For each set of input data, output the minimum number of bad segments.

Examples:
input:
8
6 4
2 3 6 2 1 2
9 100000
50000 25000 12500 6250 3125 2 4 8 16
5 2
1 1 1 1 1
8 6
4 3 4 3 4 3 4 3
7 12
6 11 1 3 11 10 2
10 5
2 4 4 2 4 4 4 3 1 1
7 8
4 6 5 1 2 4 1
8 27
3 9 17 26 2 20 9 3
output:
3
2
1
1
2
1
3
3

Note:

Buggy Code:
import java.util.Scanner;
import java.util.Arrays;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int x = sc.nextInt();
            int[] a = new int[n];
            for (int i = 0; i < a.length; i++) {
                a[i] = sc.nextInt();
            }

            int segments = 1;
            int[] freq = new int[200001];

            for (int i = 0; i < a.length; i++) {

                if (x % a[i] == 0 && freq[x / a[i]] > 0) {
                    segments++;
                    Arrays.fill(freq, 0);
                }
                freq[a[i]]++;
            }

            System.out.println(segments);
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
10 12
2 2 3 3 4 4 6 6 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Valuable Cards
time_limit_ms: 4000
memory_limit_mb: 512
Description: In his favorite cafe Kmes once again wanted to try the herring under a fur coat. Previously, it would not have been difficult for him to do this, but the cafe recently introduced a new purchasing policy.

Now, in order to make a purchase, Kmes needs to solve the following problem: $n$ cards with prices for different positions are laid out in front of him, on the $i$-th card there is an integer $a_i$, among these prices there is no whole positive integer $x$.

Kmes is asked to divide these cards into the minimum number of bad segments (so that each card belongs to exactly one segment). A segment is considered bad if it is impossible to select a subset of cards with a product equal to $x$. All segments, in which Kmes will divide the cards, must be bad.

Formally, the segment $(l, r)$ is bad if there are no indices $i_1 < i_2 < \ldots < i_k$ such that $l \le i_1, i_k \le r$, and $a_{i_1} \cdot a_{i_2} \ldots \cdot a_{i_k} = x$.

Help Kmes determine the minimum number of bad segments in order to enjoy his favorite dish.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each set of input data gives you $2$ integers $n$ and $x$ ($1 \le n \le 10^5, 2 \le x \le 10^5$) — the number of cards and the integer, respectively.

The second line of each set of input data contains $n$ integers $a_i$ ($1 \le a_i \le 2 \cdot 10^5, a_i \neq x$) — the prices on the cards.

It is guaranteed that the sum of $n$ over all sets of test data does not exceed $10^5$.

Output: For each set of input data, output the minimum number of bad segments.

Examples:
input:
8
6 4
2 3 6 2 1 2
9 100000
50000 25000 12500 6250 3125 2 4 8 16
5 2
1 1 1 1 1
8 6
4 3 4 3 4 3 4 3
7 12
6 11 1 3 11 10 2
10 5
2 4 4 2 4 4 4 3 1 1
7 8
4 6 5 1 2 4 1
8 27
3 9 17 26 2 20 9 3
output:
3
2
1
1
2
1
3
3

Note:

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

public class F {
    public static void main(String[] args) {
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt(), x = sc.nextInt();
            TreeSet<Integer> set = new TreeSet<>();
            for (int i = 1; i * i <= x; i++) {
                if (x % i == 0) {
                    set.add(i);
                    set.add(x / i);
                }
            }
            int sz = set.size();
            int[] d = new int[sz];
            boolean[] vis = new boolean[sz];
            int pt = 0;
            Map<Integer, Integer> map = new HashMap<>();
            for (int v : set) {
                map.put(v, pt);
                d[pt++] = v;
            }
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
            }
            int ans = 0;
            for (int i = 0; i < n; i++) {
                if (a[i] > x) continue;
                if (x % a[i] == 0) {
                    int idx = map.get(x / a[i]);
                    if (vis[idx]) {
                        ans++;
                        Arrays.fill(vis, false);
                        vis[map.get(a[i])] = true;
                    } else {
                        for (int j = sz - 1; j >= 0; j--) {
                            if (a[i] * d[j] <= x && x % (a[i] * d[j]) == 0 && vis[j]) {
                                idx = map.get(a[i] * d[j]);
                                vis[idx] = true;
                            }
                        }
                        vis[map.get(a[i])] = true;
                    }
                }
            }
            for (int i = 0; i < sz; i++) {
                if (vis[i]) {
                    ans++;
                    break;
                }
            }
            out.println(ans);

        }
        out.close();
    }

    static Kattio sc = new Kattio();
    static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

    static class Kattio {
        static BufferedReader r;
        static StringTokenizer st;

        public Kattio() {
            r = new BufferedReader(new InputStreamReader(System.in));
        }

        public String next() {
            try {
                while (st == null || !st.hasMoreTokens()) {
                    st = new StringTokenizer(r.readLine());
                }
                return st.nextToken();
            } catch (Exception e) {
                return null;
            }
        }

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

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

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

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

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

Think step by step.

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

1. **Edge Cases**: The code might fail when the input values are at the boundaries of the constraints.
2. **Special Values**: The code might fail when the input values have special properties, such as all values being the same, or values that are factors of \( x \).
3. **Complex Scenarios**: The code might fail when the input values are complex, such as having multiple factors of \( x \) in different positions.

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

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

### Explanation
1. **Test Case 1**: All values are 2.
   - This tests the scenario where all cards have the same value, which is a factor of \( x \).
   - Expected output: 6 (each card is a separate segment).

2. **Test Case 2**: Alternating values of 2 and 4.
   - This tests the scenario where the values are factors of \( x \) but not all the same.
   - Expected output: 3 (each pair of 2 and 4 forms a segment).

3. **Test Case 3**: A mix of values, some of which are factors of \( x \).
   - This tests the scenario where the values include both factors and non-factors of \( x \).
   - Expected output: 4 (segments are 1, 2, 3, 4, 5, 6).

4. **Test Case 4**: All values are 1.
   - This tests the scenario where all cards have the same value, which is not a factor of \( x \).
   - Expected output: 1 (all cards form a single segment).

This test input is designed to cover a range of edge cases and special scenarios to maximize the probability of exposing any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Valuable Cards
time_limit_ms: 4000
memory_limit_mb: 512
Description: In his favorite cafe Kmes once again wanted to try the herring under a fur coat. Previously, it would not have been difficult for him to do this, but the cafe recently introduced a new purchasing policy.

Now, in order to make a purchase, Kmes needs to solve the following problem: $n$ cards with prices for different positions are laid out in front of him, on the $i$-th card there is an integer $a_i$, among these prices there is no whole positive integer $x$.

Kmes is asked to divide these cards into the minimum number of bad segments (so that each card belongs to exactly one segment). A segment is considered bad if it is impossible to select a subset of cards with a product equal to $x$. All segments, in which Kmes will divide the cards, must be bad.

Formally, the segment $(l, r)$ is bad if there are no indices $i_1 < i_2 < \ldots < i_k$ such that $l \le i_1, i_k \le r$, and $a_{i_1} \cdot a_{i_2} \ldots \cdot a_{i_k} = x$.

Help Kmes determine the minimum number of bad segments in order to enjoy his favorite dish.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each set of input data gives you $2$ integers $n$ and $x$ ($1 \le n \le 10^5, 2 \le x \le 10^5$) — the number of cards and the integer, respectively.

The second line of each set of input data contains $n$ integers $a_i$ ($1 \le a_i \le 2 \cdot 10^5, a_i \neq x$) — the prices on the cards.

It is guaranteed that the sum of $n$ over all sets of test data does not exceed $10^5$.

Output: For each set of input data, output the minimum number of bad segments.

Examples:
input:
8
6 4
2 3 6 2 1 2
9 100000
50000 25000 12500 6250 3125 2 4 8 16
5 2
1 1 1 1 1
8 6
4 3 4 3 4 3 4 3
7 12
6 11 1 3 11 10 2
10 5
2 4 4 2 4 4 4 3 1 1
7 8
4 6 5 1 2 4 1
8 27
3 9 17 26 2 20 9 3
output:
3
2
1
1
2
1
3
3

Note:

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
#include <numeric>
#include <set>

using namespace std;

#define MAX_VALUE 10000000

#define ll unsigned long long

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

	int t;

	cin >> t;

	while (t--) {

		int n, x;

		cin >> n >> x;

		vector<int> NUMS;

		//NUMS.clear();

		//NUMS.resize(x + 1);

		vector<int> a(n);

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

		int cnt = 0;

		for (int i = 0; i < n; ++i) {
			if(x % a[i] == 0){

				bool flag = false;

				for (int j : NUMS) {
					if (j == x / a[i]) {
						flag = true;
						break;
					}
				}

				if (flag) {
					cnt++;
					NUMS.clear();
					NUMS.emplace_back(a[i]);
				}
				else {
					int k = NUMS.size();
					for (int j = 0; j < k; ++j) {
						if (NUMS[j] * a[i] <= x) {
							NUMS.emplace_back(NUMS[j] * a[i]);
						}
					}
					NUMS.emplace_back(a[i]);
				}

				/*
				if (NUMS[x / a[i]]) {
					cnt++;
					NUMS.clear();
					NUMS.resize(x + 1);
					NUMS[a[i]] = true;
				}
				else {
					NUMS[a[i]] = true;
					for (int j = 1; j <= x; ++j) {
						if (!NUMS[j]) continue;
						if ((j * a[i]) <= x) {
							NUMS[j * a[i]] = true;
						}
					}
				}
				*/
			}
		}

		cout << cnt + 1 << "\n";

	}

	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5 12
3 4 1 2 6
```

**Explanation:**

1. **Test Case Selection:**
   - We choose a single test case to keep the input small and manageable.
   - We set \( n = 5 \) and \( x = 12 \) to create a scenario where the product of some subset of cards equals \( x \).

2. **Input Values:**
   - The cards have values \( [3, 4, 1, 2, 6] \).
   - The goal is to find the minimum number of bad segments such that no segment contains a subset of cards whose product equals 12.

3. **Expected Behavior:**
   - The correct output should be 3 because:
     - The first segment can be [3, 4] (product 12, not allowed).
     - The second segment can be [1] (product 1, allowed).
     - The third segment can be [2, 6] (product 12, not allowed).

4. **Buggy Code Analysis:**
   - The buggy code might fail to correctly handle the product of multiple cards and might incorrectly count the number of bad segments.
   - The code might not properly clear the `NUMS` vector or might not correctly update it, leading to incorrect segment counts.

By providing this test input, we can effectively expose the bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Valuable Cards
time_limit_ms: 4000
memory_limit_mb: 512
Description: In his favorite cafe Kmes once again wanted to try the herring under a fur coat. Previously, it would not have been difficult for him to do this, but the cafe recently introduced a new purchasing policy.

Now, in order to make a purchase, Kmes needs to solve the following problem: $n$ cards with prices for different positions are laid out in front of him, on the $i$-th card there is an integer $a_i$, among these prices there is no whole positive integer $x$.

Kmes is asked to divide these cards into the minimum number of bad segments (so that each card belongs to exactly one segment). A segment is considered bad if it is impossible to select a subset of cards with a product equal to $x$. All segments, in which Kmes will divide the cards, must be bad.

Formally, the segment $(l, r)$ is bad if there are no indices $i_1 < i_2 < \ldots < i_k$ such that $l \le i_1, i_k \le r$, and $a_{i_1} \cdot a_{i_2} \ldots \cdot a_{i_k} = x$.

Help Kmes determine the minimum number of bad segments in order to enjoy his favorite dish.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each set of input data gives you $2$ integers $n$ and $x$ ($1 \le n \le 10^5, 2 \le x \le 10^5$) — the number of cards and the integer, respectively.

The second line of each set of input data contains $n$ integers $a_i$ ($1 \le a_i \le 2 \cdot 10^5, a_i \neq x$) — the prices on the cards.

It is guaranteed that the sum of $n$ over all sets of test data does not exceed $10^5$.

Output: For each set of input data, output the minimum number of bad segments.

Examples:
input:
8
6 4
2 3 6 2 1 2
9 100000
50000 25000 12500 6250 3125 2 4 8 16
5 2
1 1 1 1 1
8 6
4 3 4 3 4 3 4 3
7 12
6 11 1 3 11 10 2
10 5
2 4 4 2 4 4 4 3 1 1
7 8
4 6 5 1 2 4 1
8 27
3 9 17 26 2 20 9 3
output:
3
2
1
1
2
1
3
3

Note:

Buggy Code:
#include <bits/stdc++.h>
#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <map>
#include <random>
#include <iomanip>
#include <cmath>
#include <unordered_map>
#include <fstream>
using namespace std;

#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)

using ll = long long;
using db = long double;  // or double, if TL is tight
using str = string; //python yeah

// pairs
using pint = pair<int, int>;
using pll = pair<ll, ll>;
using pdb = pair<db, db>;

#define tcT template <class T
#define tcTU tcT, class U
// ^ lol this makes everything look weird but I'll try it
tcT > using V = vector<T>;

using vint = V<int>;
using vb = V<bool>;
using vll = V<ll>;
using vdb = V<db>;
using vstr = V<str>;
using vpint = V<pint>;
using vpll = V<pll>;
using vpdb = V<pdb>;

// vectors
#define sz(x) size(x)
#define bg(x) begin(x)
#define all(x) bg(x), end(x)
#define rall(x) rbegin(x), rend(x)
#define sor(x) sort(all(x))
#define rsz resize
#define ins insert
#define pb push_back
#define eb emplace_back
#define ft front()
#define bk back()

#define lb lower_bound
#define ub upper_bound
tcT > int lwb(const V<T> &a, const T &b) { return int(lb(all(a), b) - bg(a)); }
tcT > int upb(const V<T> &a, const T &b) { return int(ub(all(a), b) - bg(a)); }

// loops
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define F0R(i, a) FOR(i, 0, a)
#define ROF(i, a, b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i, a) ROF(i, 0, a)
#define rep(a) F0R(_, a)
#define each(a, x) for (auto &a : x)

const int MOD = 998244353;  // 1e9+7;
const int MX = (int)2e5 + 5;
const ll BIG = 1e18;  // not too close to LLONG_MAX

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

// int dp(vint& a, int l, int r, int x){
//     map<int,bool> m;
//     FOR(i,l,r){
//         if(a[i] == x) return i;
//         if(x%a[i] == 0 && !m[a[i]]){
//             int index = dp(a,i+1,r,x/a[i]);

//             r = index;
//             m[a[i]] = 1;
//         }
//     }
//     return r;
// }
int dp(vint& a,vint& m, int l, int x){
    int size = m.size();
    ll value = x/a[l];
    // cout<<a[l]<<"xcz";
    // cout<<value;
    F0R(i,size){
        // value = ll(m[i])*ll(a[l]);
        if(value == m[i])
            return l;
        if(value > m[i])
            m.pb(a[l]*m[i]);
    }
    m.pb(a[l]);
    return -1;
}
void solve(){
    int n,x;
    cin>>n>>x;
    vint a;
    int data;
    F0R(i,n) {
        cin>>data;
        if(x%data == 0)
            a.pb(data);
    }
    // a.pb(x+1);
    // cin>>a[i];
    int count = 1;
    int l = 0;
    vint m;
    // F0R(i,a.size()) cout<<a[i]<<" ";
    F0R(i,a.size()){
        l = dp(a,m,i,x);
        if(l != -1){
            count++;
            m.clear();
            m.pb(a[i]);
        }
        // cout<<l<<" ";
        // F0R(j,m.size()) cout<<m[j]<<" ";
        // cout<<endl;
    }

    cout<<count<<endl;
}

int main() {
    fast_cin();
    int t;
    cin>>t;
    while(t--)
        solve();
    return 0;
}

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

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

Think step by step.

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

1. **Edge Cases**: Input values that are at the boundaries of the allowed ranges.
2. **Special Cases**: Inputs that might cause unexpected behavior, such as all cards having the same value or values that are factors of \( x \).
3. **Complex Cases**: Inputs that involve multiple segments and complex factor interactions.

Given the problem constraints and the provided code, let's think about a test case that could expose the bug:

- **Test Case**: We need a test case where the cards have values that are factors of \( x \), but the product of any subset of these cards does not equal \( x \). This will test the algorithm's ability to correctly identify and count bad segments.

Let's construct a test case where:
- \( n = 5 \)
- \( x = 6 \)
- The cards have values that are factors of 6 but do not form a product of 6 in any subset.

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

Explanation:
- \( n = 5 \): There are 5 cards.
- \( x = 6 \): The target product is 6.
- The cards are [2, 3, 1, 2, 3].
- The factors of 6 are 1, 2, and 3.
- No subset of these cards has a product of 6.
- The correct output should be 1 because all cards can be in a single bad segment.

This test case should expose the bug in the code if it incorrectly counts the number of bad segments.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Valuable Cards
time_limit_ms: 4000
memory_limit_mb: 512
Description: In his favorite cafe Kmes once again wanted to try the herring under a fur coat. Previously, it would not have been difficult for him to do this, but the cafe recently introduced a new purchasing policy.

Now, in order to make a purchase, Kmes needs to solve the following problem: $n$ cards with prices for different positions are laid out in front of him, on the $i$-th card there is an integer $a_i$, among these prices there is no whole positive integer $x$.

Kmes is asked to divide these cards into the minimum number of bad segments (so that each card belongs to exactly one segment). A segment is considered bad if it is impossible to select a subset of cards with a product equal to $x$. All segments, in which Kmes will divide the cards, must be bad.

Formally, the segment $(l, r)$ is bad if there are no indices $i_1 < i_2 < \ldots < i_k$ such that $l \le i_1, i_k \le r$, and $a_{i_1} \cdot a_{i_2} \ldots \cdot a_{i_k} = x$.

Help Kmes determine the minimum number of bad segments in order to enjoy his favorite dish.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each set of input data gives you $2$ integers $n$ and $x$ ($1 \le n \le 10^5, 2 \le x \le 10^5$) — the number of cards and the integer, respectively.

The second line of each set of input data contains $n$ integers $a_i$ ($1 \le a_i \le 2 \cdot 10^5, a_i \neq x$) — the prices on the cards.

It is guaranteed that the sum of $n$ over all sets of test data does not exceed $10^5$.

Output: For each set of input data, output the minimum number of bad segments.

Examples:
input:
8
6 4
2 3 6 2 1 2
9 100000
50000 25000 12500 6250 3125 2 4 8 16
5 2
1 1 1 1 1
8 6
4 3 4 3 4 3 4 3
7 12
6 11 1 3 11 10 2
10 5
2 4 4 2 4 4 4 3 1 1
7 8
4 6 5 1 2 4 1
8 27
3 9 17 26 2 20 9 3
output:
3
2
1
1
2
1
3
3

Note:

Buggy Code:
from collections import defaultdict, Counter, deque
from bisect import bisect_left, bisect_right
from heapq import heapify, heappop, heappush
import random
import sys
from io import BytesIO, IOBase
import math
import os
from collections import defaultdict
from bisect import bisect_left, bisect_right
from heapq import heapify, heappop, heappush
import random

RANDOM = random.randrange(2**62)

def Wrapper(x):
  return x ^ RANDOM

INF = 10**18
MOD = 10**9 + 7
def gcd(a,b):
    a, b = max(a, b), min(a, b)

    # Everything divides 0
    if (b == 0):
         return a
    return gcd(b, a%b)

def lcm(a,b):
    a, b = max(a, b), min(a, b)
    return (a // gcd(a,b))* b

def SieveOfEratosthenes(n):

    # Create a boolean array
    # "prime[0..n]" and initialize
    #  all entries it as true.
    # A value in prime[i] will
    # finally be false if i is
    # Not a prime, else true.
    prime = [True for i in range(n+1)]
    p = 2
    while (p * p <= n):

        # If prime[p] is not
        # changed, then it is a prime
        if (prime[p] == True):

            # Update all multiples of p
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1

    # Print all prime numbers
    arr = set()
    for p in range(2, n+1):
        if prime[p]:
            arr.add(p)
    return arr
def main():
    for i in range(int(input())):
        n, x = map(int, input().split())
        arr = [int(x) for x in input().split()]
        # primes = defaultdict()
        # if x % 2 == 0:
        #     cnt = 0
        #     while x % 2 == 0:
        #         x //= 2
        #         cnt += 1
        #     primes[2] = cnt
        # for j in range(3, round(x**0.5) + 1, 2):
        #     if x % j == 0:
        #         cnt = 0
        #         while x % j == 0:
        #             x//= j
        #             cnt += 1
        #         primes[j] = cnt
        # cnt = defaultdict()

        mk = defaultdict()
        cnt = 1
        for j in range(n):
            if x % arr[j] == 0:
                if x//arr[j] in mk:
                    cnt += 1
                    mk = defaultdict()
                    mk[arr[j]] = True
                else:
                    tmp = defaultdict()
                    for val in mk:
                        tmp[val * arr[j]] = True
                        tmp[val] = True
                    tmp[arr[j]] = True
                    mk = tmp.copy()
        print(cnt)

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

# endregion

if __name__ == "__main__":

    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause the algorithm to fail. Here are some key points to consider:

1. **Prime Factorization of `x`**: The code attempts to handle the prime factors of `x` but does not do it correctly. We should include `x` values that have multiple prime factors to test the handling of composite numbers.
2. **Divisors of `x`**: The code seems to handle divisors of `x` in a way that might not be optimal. We should include test cases where `x` has divisors that are present in the input array.
3. **Edge Cases for `n` and `x`**: Test with small values of `n` and `x` to ensure the algorithm handles small inputs correctly.
4. **Special Cases**: Test with all 1s, all numbers that are not divisors of `x`, and a mix of numbers that are and are not divisors of `x`.

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

```plaintext
4
6 4
2 2 1 1 3 3
9 12
3 4 6 1 1 1 1 1 1
5 10
5 2 5 2 5
8 27
3 9 1 3 9 1 3 9
```

Explanation of the test cases:
1. **Test Case 1**: `n = 6`, `x = 4`, `arr = [2, 2, 1, 1, 3, 3]`
   - `x = 4` has prime factors 2 and 2.
   - The array contains multiple 2s and 1s.
   - The expected output is 3, as we can divide the array into segments `[2, 2, 1]`, `[1, 3]`, and `[3]`.

2. **Test Case 2**: `n = 9`, `x = 12`, `arr = [3, 4, 6, 1, 1, 1, 1, 1, 1]`
   - `x = 12` has prime factors 2, 2, and 3.
   - The array contains 3, 4, 6, and multiple 1s.
   - The expected output is 2, as we can divide the array into segments `[3, 4, 6, 1, 1, 1]` and `[1, 1, 1]`.

3. **Test Case 3**: `n = 5`, `x = 10`, `arr = [5, 2, 5, 2, 5]`
   - `x = 10` has prime factors 2 and 5.
   - The array contains multiple 5s and 2s.
   - The expected output is 1, as the entire array can be considered a bad segment.

4. **Test Case 4**: `n = 8`, `x = 27`, `arr = [3, 9, 1, 3, 9, 1, 3, 9]`
   - `x = 27` has prime factors 3, 3, and 3.
   - The array contains multiple 3s and 9s.
   - The expected output is 3, as we can divide the array into segments `[3, 9, 1]`, `[3, 9, 1]`, and `[3, 9]`.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Valuable Cards
time_limit_ms: 4000
memory_limit_mb: 512
Description: In his favorite cafe Kmes once again wanted to try the herring under a fur coat. Previously, it would not have been difficult for him to do this, but the cafe recently introduced a new purchasing policy.

Now, in order to make a purchase, Kmes needs to solve the following problem: $n$ cards with prices for different positions are laid out in front of him, on the $i$-th card there is an integer $a_i$, among these prices there is no whole positive integer $x$.

Kmes is asked to divide these cards into the minimum number of bad segments (so that each card belongs to exactly one segment). A segment is considered bad if it is impossible to select a subset of cards with a product equal to $x$. All segments, in which Kmes will divide the cards, must be bad.

Formally, the segment $(l, r)$ is bad if there are no indices $i_1 < i_2 < \ldots < i_k$ such that $l \le i_1, i_k \le r$, and $a_{i_1} \cdot a_{i_2} \ldots \cdot a_{i_k} = x$.

Help Kmes determine the minimum number of bad segments in order to enjoy his favorite dish.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each set of input data gives you $2$ integers $n$ and $x$ ($1 \le n \le 10^5, 2 \le x \le 10^5$) — the number of cards and the integer, respectively.

The second line of each set of input data contains $n$ integers $a_i$ ($1 \le a_i \le 2 \cdot 10^5, a_i \neq x$) — the prices on the cards.

It is guaranteed that the sum of $n$ over all sets of test data does not exceed $10^5$.

Output: For each set of input data, output the minimum number of bad segments.

Examples:
input:
8
6 4
2 3 6 2 1 2
9 100000
50000 25000 12500 6250 3125 2 4 8 16
5 2
1 1 1 1 1
8 6
4 3 4 3 4 3 4 3
7 12
6 11 1 3 11 10 2
10 5
2 4 4 2 4 4 4 3 1 1
7 8
4 6 5 1 2 4 1
8 27
3 9 17 26 2 20 9 3
output:
3
2
1
1
2
1
3
3

Note:

Buggy Code:
def g(x):
    z = {x}
    for i in range(2,int(x**0.5)+1):
        if x%i == 0:
            z.add(i)
            z.add(x//i)
    return z
def f(x,a):
    q = 1
    s = {1}
    z = g(x)
    for i in range(len(a)):
        if a[i] in z:
            s1 = set()
            for y in s:
                s1.add(y*a[i])
            for y in s1:
                s.add(y)
            if x in s:
                q += 1
                s = {1,a[i]}

    return q

t = int(input())
q = []
for i in range(t):
    n,x = map(int,input().split())
    a = list(map(int,input().split()))
    q.append(f(x,a))
for i in range(t):
    print(q[i])

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

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

Think step by step.

Response:
```plaintext
3
5 12
2 2 3 4 6
4 10
2 5 2 5
6 15
3 5 1 3 5 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Valuable Cards
time_limit_ms: 4000
memory_limit_mb: 512
Description: In his favorite cafe Kmes once again wanted to try the herring under a fur coat. Previously, it would not have been difficult for him to do this, but the cafe recently introduced a new purchasing policy.

Now, in order to make a purchase, Kmes needs to solve the following problem: $n$ cards with prices for different positions are laid out in front of him, on the $i$-th card there is an integer $a_i$, among these prices there is no whole positive integer $x$.

Kmes is asked to divide these cards into the minimum number of bad segments (so that each card belongs to exactly one segment). A segment is considered bad if it is impossible to select a subset of cards with a product equal to $x$. All segments, in which Kmes will divide the cards, must be bad.

Formally, the segment $(l, r)$ is bad if there are no indices $i_1 < i_2 < \ldots < i_k$ such that $l \le i_1, i_k \le r$, and $a_{i_1} \cdot a_{i_2} \ldots \cdot a_{i_k} = x$.

Help Kmes determine the minimum number of bad segments in order to enjoy his favorite dish.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each set of input data gives you $2$ integers $n$ and $x$ ($1 \le n \le 10^5, 2 \le x \le 10^5$) — the number of cards and the integer, respectively.

The second line of each set of input data contains $n$ integers $a_i$ ($1 \le a_i \le 2 \cdot 10^5, a_i \neq x$) — the prices on the cards.

It is guaranteed that the sum of $n$ over all sets of test data does not exceed $10^5$.

Output: For each set of input data, output the minimum number of bad segments.

Examples:
input:
8
6 4
2 3 6 2 1 2
9 100000
50000 25000 12500 6250 3125 2 4 8 16
5 2
1 1 1 1 1
8 6
4 3 4 3 4 3 4 3
7 12
6 11 1 3 11 10 2
10 5
2 4 4 2 4 4 4 3 1 1
7 8
4 6 5 1 2 4 1
8 27
3 9 17 26 2 20 9 3
output:
3
2
1
1
2
1
3
3

Note:

Buggy Code:
import java.util.Arrays;
import java.util.Scanner;

public class thrpractice {
    public static int istrue_updated(int[] array,int product,int[] prev,int start, int end){
        prev = new int[product+1];
        prev[0] = 0;
        prev[1] = 1;
        int s = 1;
        int pointer = end;
        while(pointer>=0){
            int[] curr = new int[product+1];
            for(int p=0;p<product+1;p++){
                if(p%array[pointer]==0){
                    int take = prev[p/array[pointer]];
                    int not_take = prev[p];
                    curr[p] = take + not_take;
                }
                else curr[p] = prev[p];
            }
            prev = curr;
            if(prev[product]==0){
                pointer--;
            }
            else{
                prev = new int[product+1];
                prev[0] = 0;
                prev[1]=1;
                s++;
            }

        }
        return s;}

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
         for(int i=0;i<t;i++){
        int n = sc.nextInt();
        int x = sc.nextInt();
        int[] array = new int[n];
        for(int h=0;h<n;h++){
            array[h] = sc.nextInt();
        }
         System.out.println(istrue_updated(array,x,new int[x+1],0, array.length-1));
         }

    }}

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

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

Think step by step.

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