**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Monster
time_limit_ms: 2000
memory_limit_mb: 256
Description: Man, this Genshin boss is so hard. Good thing they have a top-up of $6$ coins for only $ \$4.99$. I should be careful and spend no more than I need to, lest my mom catches me...

You are fighting a monster with $z$ health using a weapon with $d$ damage. Initially, $d=0$. You can perform the following operations.

  * Increase $d$ — the damage of your weapon by $1$, costing $x$ coins.    * Attack the monster, dealing $d$ damage and costing $y$ coins.

You cannot perform the first operation for more than $k$ times in a row.

Find the minimum number of coins needed to defeat the monster by dealing at least $z$ damage.

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

The only line of each test case contains 4 integers $x$, $y$, $z$, and $k$ ($1\leq x, y, z, k\leq 10^8$) — the first operation's cost, the second operation's cost, the monster's health, and the limitation on the first operation.

Output: For each test case, output the minimum number of coins needed to defeat the monster.

Examples:
input:
4
2 3 5 5
10 20 40 5
1 60 100 10
60 1 100 10
output:
12
190
280
160

Note: In the first test case, $x = 2$, $y = 3$, $z = 5$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $12$ coins:

  * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.

You deal a total of $3 + 3 = 6$ damage, defeating the monster who has $5$ health. The total number of coins you use is $2 + 2 + 2 + 3 + 3 = 12$ coins.

In the second test case, $x = 10$, $y = 20$, $z = 40$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $190$ coins:

  * Increase damage by $5$, costing $5\cdot x$ = $50$ coins.    * Attack the monster once, dealing $5$ damage, costing $20$ coins.    * Increase damage by $2$, costing $2\cdot x$ = $20$ coins.    * Attack the monster $5$ times, dealing $5\cdot 7 = 35$ damage, costing $5\cdot y$ = $100$ coins.

You deal a total of $5 + 35 = 40$ damage, defeating the monster who has exactly $40$ health. The total number of coins you use is $50 + 20 + 20 + 100 = 190$ coins.

Buggy Code:
// https://codeforces.com/blog/entry/96344
//#pragma GCC optimize("O3,unroll-loops")
//#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")

#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using u32 = unsigned int;
using u64 = unsigned long long;

#if __cplusplus >= 202002L
using i128 = __int128;
using u128 = unsigned __int128;
using f128 = __float128;
#endif

template<typename T>
bool chmax(T& a, const T& b) {
    bool res = a < b;
    a = max(a, b);
    return res;
}
template<typename T>
bool chmin(T& a, const T& b){
    bool res = a > b;
    a = min(a, b);
    return res;
}

typedef vector<long long> vl;
typedef pair<ll,ll> pll;
typedef vector<pair<ll, ll>> vll;
typedef vector<int> vi;
typedef vector<pair<int,int>> vii;
typedef pair<int,int> pii;

const int inf = 1000000009;
const ll linf = 4000000000000000009;

// https://trap.jp/post/1224/
template<class... T>
void input(T&... a){
    (cin >> ... >> a);
}

void print(){
    cout << '\n';
}
template<class T, class... Ts>
void print(const T& a, const Ts&... b){
    cout << a;
    (cout << ... << (cout << ' ', b));
    cout << '\n';
}

template <typename T>
T floor(T a, T b) {
  return a / b - (a % b && (a ^ b) < 0);
}
template <typename T>
T ceil(T x, T y) {
  return floor(x + y - 1, y);
}

int popcnt(int x) { return __builtin_popcount(x); }
int popcnt(u32 x) { return __builtin_popcount(x); }
int popcnt(ll x) { return __builtin_popcountll(x); }
int popcnt(u64 x) { return __builtin_popcountll(x); }
int popcnt_mod_2(int x) { return __builtin_parity(x); }
int popcnt_mod_2(u32 x) { return __builtin_parity(x); }
int popcnt_mod_2(ll x) { return __builtin_parityll(x); }
int popcnt_mod_2(u64 x) { return __builtin_parityll(x); }
// (0, 1, 2, 3, 4) -> (-1, 0, 1, 1, 2)
int topbit(int x) { return (x == 0 ? -1 : 31 - __builtin_clz(x)); }
int topbit(u32 x) { return (x == 0 ? -1 : 31 - __builtin_clz(x)); }
int topbit(ll x) { return (x == 0 ? -1 : 63 - __builtin_clzll(x)); }
int topbit(u64 x) { return (x == 0 ? -1 : 63 - __builtin_clzll(x)); }
// (0, 1, 2, 3, 4) -> (-1, 0, 1, 0, 2)
int lowbit(int x) { return (x == 0 ? -1 : __builtin_ctz(x)); }
int lowbit(u32 x) { return (x == 0 ? -1 : __builtin_ctz(x)); }
int lowbit(ll x) { return (x == 0 ? -1 : __builtin_ctzll(x)); }
int lowbit(u64 x) { return (x == 0 ? -1 : __builtin_ctzll(x)); }

template<typename T, typename U>
T SUM(const vector<U> &A){
    T ret = 0;
    for (auto &&a: A) ret += a;
    return ret;
}

#define rep1(a)          for(int i = 0; i < a; i++)
#define rep2(i, a)       for(int i = 0; i < a; i++)
#define rep3(i, a, b)    for(int i = a; i < b; i++)
#define rep4(i, a, b, c) for(int i = a; i < b; i += c)
#define overload4(a, b, c, d, e, ...) e
#define rep(...) overload4(__VA_ARGS__, rep4, rep3, rep2, rep1)(__VA_ARGS__)

#define rrep(i, a, b)    for(int i = a; i >= b; i--)

#define eb emplace_back
#define mp make_pair
#define mt make_tuple
#define fi first
#define se second
#define all(v) v.begin(), v.end()
//---------------------------------

void solve(){
    ll x,y,z,k; input(x,y,z,k);
    ll lo = 1;
    ll hi = z+1;

    auto calc = [&](ll t) -> ll {
        ll h = z;
        ll n = t/k;
        ll ret = n*(k*x+y) + x * (t % k);
        if(n > 0) h -= k * n * (n+1) / 2LL;
        if(h > 0){
            ret += ceil(h, t) * y;
        }
        return ret;
    };

    while(hi - lo > min(k, 10000LL)){
        ll lm = (2*lo + hi) / 3LL;
        ll rm = (lo + 2*hi) / 3LL;

        ll cl = calc(lm);
        ll cr = calc(rm);
        if(cl > cr){
            lo = lm;
        } else {
            hi = rm;
        }
    }

    ll ans = linf;
    rep(i, lo, hi+1) {
        chmin(ans, calc(i));
        //print(i, calc(i));
    }
    print(ans);
}

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

    int t = 1;
    cin >> t;
    rep(i,0,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 given code, we need to consider edge cases and scenarios that might cause the algorithm to fail. The key points to focus on are:

1. **Edge Cases for `k`**: The limitation on the number of consecutive increases in damage.
2. **High Values for `z`**: Large health values that require multiple attacks.
3. **Different Costs for `x` and `y`**: Varying costs for increasing damage and attacking the monster.
4. **Boundary Conditions**: Values that are just above or below the constraints.

Given the constraints \(1 \leq x, y, z, k \leq 10^8\), let's create a test case that challenges the code:

### Test Case 1: High Health and Low Damage Increase Cost
- **Scenario**: The monster has a very high health, but the cost of increasing damage is very low. This should force the algorithm to increase damage as much as possible before attacking.
- **Input**: `x = 1`, `y = 100000000`, `z = 100000000`, `k = 100000000`
- **Expected Output**: The optimal strategy is to increase damage to the maximum allowed and then attack once, which should cost \(100000000 \times 1 + 100000000 = 200000000\) coins.

### Test Case 2: High Health and High Damage Increase Cost
- **Scenario**: The monster has a very high health, and the cost of increasing damage is also high. This should force the algorithm to balance between increasing damage and attacking.
- **Input**: `x = 100000000`, `y = 1`, `z = 100000000`, `k = 100000000`
- **Expected Output**: The optimal strategy is to attack as many times as needed without increasing damage, which should cost \(100000000 \times 1 = 100000000\) coins.

### Test Case 3: Low Health and High Damage Increase Cost
- **Scenario**: The monster has a very low health, but the cost of increasing damage is very high. This should force the algorithm to attack immediately without increasing damage.
- **Input**: `x = 100000000`, `y = 1`, `z = 1`, `k = 100000000`
- **Expected Output**: The optimal strategy is to attack once, which should cost \(1\) coin.

### Test Case 4: High Health and Low Attack Cost
- **Scenario**: The monster has a very high health, but the cost of attacking is very low. This should force the algorithm to attack as many times as needed without increasing damage.
- **Input**: `x = 100000000`, `y = 1`, `z = 100000000`, `k = 1`
- **Expected Output**: The optimal strategy is to attack as many times as needed without increasing damage, which should cost \(100000000 \times 1 = 100000000\) coins.

### Test Case 5: Low Health and Low Damage Increase Cost
- **Scenario**: The monster has a very low health, and the cost of increasing damage is also very low. This should force the algorithm to increase damage and then attack.
- **Input**: `x = 1`, `y = 100000000`, `z = 1`, `k = 100000000`
- **Expected Output**: The optimal strategy is to increase damage to the required level and then attack once, which should cost \(1 + 100000000 = 100000001\) coins.

### Test Case 6: Balanced Costs
- **Scenario**: The costs of increasing damage and attacking are balanced. This should force the algorithm to find a balance between increasing damage and attacking.
- **Input**: `x = 50000000`, `y = 50000000`, `z = 100000000`, `k = 100000000`
- **Expected Output**: The optimal strategy is to increase damage to a balanced level and then attack, which should cost \(2 \times 50000000 = 100000000\) coins.

### Challenging Test Input
```plaintext
6
1 100000000 100000000 100000000
100000000 1 100000000 100000000
100000000 1 1 100000000
100000000 1 100000000 1
1 100000000 1 100000000
50000000 50000000 100000000 100000000
```

This test input covers a range of scenarios that should challenge the algorithm and expose any 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: Monster
time_limit_ms: 2000
memory_limit_mb: 256
Description: Man, this Genshin boss is so hard. Good thing they have a top-up of $6$ coins for only $ \$4.99$. I should be careful and spend no more than I need to, lest my mom catches me...

You are fighting a monster with $z$ health using a weapon with $d$ damage. Initially, $d=0$. You can perform the following operations.

  * Increase $d$ — the damage of your weapon by $1$, costing $x$ coins.    * Attack the monster, dealing $d$ damage and costing $y$ coins.

You cannot perform the first operation for more than $k$ times in a row.

Find the minimum number of coins needed to defeat the monster by dealing at least $z$ damage.

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

The only line of each test case contains 4 integers $x$, $y$, $z$, and $k$ ($1\leq x, y, z, k\leq 10^8$) — the first operation's cost, the second operation's cost, the monster's health, and the limitation on the first operation.

Output: For each test case, output the minimum number of coins needed to defeat the monster.

Examples:
input:
4
2 3 5 5
10 20 40 5
1 60 100 10
60 1 100 10
output:
12
190
280
160

Note: In the first test case, $x = 2$, $y = 3$, $z = 5$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $12$ coins:

  * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.

You deal a total of $3 + 3 = 6$ damage, defeating the monster who has $5$ health. The total number of coins you use is $2 + 2 + 2 + 3 + 3 = 12$ coins.

In the second test case, $x = 10$, $y = 20$, $z = 40$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $190$ coins:

  * Increase damage by $5$, costing $5\cdot x$ = $50$ coins.    * Attack the monster once, dealing $5$ damage, costing $20$ coins.    * Increase damage by $2$, costing $2\cdot x$ = $20$ coins.    * Attack the monster $5$ times, dealing $5\cdot 7 = 35$ damage, costing $5\cdot y$ = $100$ coins.

You deal a total of $5 + 35 = 40$ damage, defeating the monster who has exactly $40$ health. The total number of coins you use is $50 + 20 + 20 + 100 = 190$ coins.

Buggy Code:
#include<bits/stdc++.h>
#define int long long
#define rep(i,a,b) for(int i=a;i<=b;++i)
using namespace std;
template<typename T>inline void ckmax(T &x,T y){x=x<y?y:x;}
template<typename T>inline void ckmin(T &x,T y){x=x>y?y:x;}
template<typename T=int>inline T read(){
	T res=0,f=1; char k;
	while(!isdigit(k=getchar())) if(k=='-') f=-1;
	while(isdigit(k)) res=res*10+k-'0',k=getchar();
	return res*f;
}
template<typename T>inline void print(T x,bool fl=1){
	if(x<0) putchar('-'),x=-x;
	if(x>=10) print(x/10,0);
	putchar(x%10+'0');
	if(fl) putchar('\n');
}
inline int calc(int x,int y,int z,int k,int l){
	int res = l*x + (l/k)*y + (l-1+z-k*((l/k)+1)*((l/k))/2)/l * y;
	return res;
}
signed main(){
	int T=read();
	while(T--){
		int x=read(),y=read(),z=read(),k=read();
		int ans = x + z*y;
		for(int t=0;t*(t+1)/2*k<=z;++t){

			// for(int p=0;p<k;++p){
			// 	int l = t*k+p;
			// 	if(l==0) continue;
				int mid = ceil(1.0*y*(z-(t+1)*t/2*k)/x);
				if(t*k) ans = min(ans,calc(x,y,z,k,t*k));
				int l=max(1ll,t*k),r=t*k+k-1;
				while(r-l+1>=min(k,500ll)){
					int lmid = l+(r-l+1)/3;
					int rmid = r-(r-l+1)/3;
					int r1 = calc(x,y,z,k,lmid);
					int r2 = calc(x,y,z,k,rmid);
					if(r1<r2) l=lmid;
					else if(r2>r1) r=rmid;
					else l=lmid,r=rmid;
				}
				rep(e,l,r) ckmin(ans,calc(x,y,z,k,e));
				// if((t+1)*k-1) ans = min(ans,calc(x,y,z,k,-1+(t+1)*k));
				// for(int l=max(1ll,t*k);l<=min(t*k+100,t*k+k-1);++l) ans=min(ans,calc(x,y,z,k,l));
				// for(int l=max(max(1ll,t*k),t*k+k-1-100);l<=t*k+k-1;++l) ans=min(ans,calc(x,y,z,k,l));
				// for(int l=max(1ll,max(t*k,mid-100));l<=min((t+1)*k-1,mid+100);++l){
				// 	int res = calc(x,y,z,k,l);
				// 	ans=min(ans,res);
				// }
			// }
		}
		print(ans);
		// exit(0);
	}
	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 expose the bug in the given code, we need to focus on edge cases and scenarios where the algorithm might fail. The primary issues in the code seem to be related to the calculation of the minimum number of coins needed, especially in scenarios where the constraints on the number of times the damage can be increased in a row (k) and the health of the monster (z) are critical.

Let's consider the following scenarios:

1. **High k, Low z**: This scenario tests the case where the monster's health is very low, and the limitation on increasing damage is very high. The algorithm should be able to handle this efficiently.
2. **Low k, High z**: This scenario tests the case where the monster's health is very high, and the limitation on increasing damage is very low. This can expose issues in the loop and the calculation of the minimum number of coins.
3. **x and y are very close**: This scenario tests the case where the cost of increasing damage and the cost of attacking are nearly the same. The algorithm should balance the number of times each operation is performed.
4. **x is much higher than y**: This scenario tests the case where increasing damage is very expensive compared to attacking. The algorithm should favor attacking more.
5. **x is much lower than y**: This scenario tests the case where increasing damage is very cheap compared to attacking. The algorithm should favor increasing damage more.

Given these considerations, let's generate a test input that combines these scenarios to maximize the failure probability in the faulty solution.

```plaintext
5
1 100 1000 1
100 1 1000 1
1 1 1000 1000
1 100 1000 1000
100 1 1000 1000
```

Explanation of the test cases:
1. **1 100 1000 1**: Low x, high y, high z, low k. This tests the scenario where increasing damage is very cheap, but attacking is very expensive, and the limitation on increasing damage is very low.
2. **100 1 1000 1**: High x, low y, high z, low k. This tests the scenario where increasing damage is very expensive, but attacking is very cheap, and the limitation on increasing damage is very low.
3. **1 1 1000 1000**: Low x, low y, high z, high k. This tests the scenario where both operations are cheap, but the monster's health is very high, and the limitation on increasing damage is very high.
4. **1 100 1000 1000**: Low x, high y, high z, high k. This tests the scenario where increasing damage is very cheap, but attacking is very expensive, and the limitation on increasing damage is very high.
5. **100 1 1000 1000**: High x, low y, high z, high k. This tests the scenario where increasing damage is very expensive, but attacking is very cheap, and the limitation on increasing damage is very high.

These test cases should help expose any 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: Monster
time_limit_ms: 2000
memory_limit_mb: 256
Description: Man, this Genshin boss is so hard. Good thing they have a top-up of $6$ coins for only $ \$4.99$. I should be careful and spend no more than I need to, lest my mom catches me...

You are fighting a monster with $z$ health using a weapon with $d$ damage. Initially, $d=0$. You can perform the following operations.

  * Increase $d$ — the damage of your weapon by $1$, costing $x$ coins.    * Attack the monster, dealing $d$ damage and costing $y$ coins.

You cannot perform the first operation for more than $k$ times in a row.

Find the minimum number of coins needed to defeat the monster by dealing at least $z$ damage.

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

The only line of each test case contains 4 integers $x$, $y$, $z$, and $k$ ($1\leq x, y, z, k\leq 10^8$) — the first operation's cost, the second operation's cost, the monster's health, and the limitation on the first operation.

Output: For each test case, output the minimum number of coins needed to defeat the monster.

Examples:
input:
4
2 3 5 5
10 20 40 5
1 60 100 10
60 1 100 10
output:
12
190
280
160

Note: In the first test case, $x = 2$, $y = 3$, $z = 5$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $12$ coins:

  * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.

You deal a total of $3 + 3 = 6$ damage, defeating the monster who has $5$ health. The total number of coins you use is $2 + 2 + 2 + 3 + 3 = 12$ coins.

In the second test case, $x = 10$, $y = 20$, $z = 40$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $190$ coins:

  * Increase damage by $5$, costing $5\cdot x$ = $50$ coins.    * Attack the monster once, dealing $5$ damage, costing $20$ coins.    * Increase damage by $2$, costing $2\cdot x$ = $20$ coins.    * Attack the monster $5$ times, dealing $5\cdot 7 = 35$ damage, costing $5\cdot y$ = $100$ coins.

You deal a total of $5 + 35 = 40$ damage, defeating the monster who has exactly $40$ health. The total number of coins you use is $50 + 20 + 20 + 100 = 190$ coins.

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

typedef long long ll;
typedef long double ld;
constexpr int MOD = 1e9+7;
constexpr int N = 2e5+1;

void solve(){
    ll x, y, z, k; cin >> x >> y >> z >> k;

    auto get_score = [&](ll d){
        ll cost = d*x;

        ll health = z;

        ll pre_attacks = (d-1)/k;
        cost += y*pre_attacks;

        if((__int128_t)health <= (__int128_t)k*pre_attacks*(pre_attacks+1)/2){
            ll l = 0, r = pre_attacks;
            while(r-l > 1){
                ll m = (l+r+1)/2;
                if((__int128_t)k*m*(m+1)/2 > (__int128_t)health) r = m;
                else l = m;
            }
            ll cost = r*k*x + r*y;
            return cost;
        }
        health -= k*pre_attacks*(pre_attacks+1)/2;

        if(health == 0) return cost;

        cost += y*((health-1)/d + 1);
        return cost;
    };

    ll ans = 1e18;
    for(ll d=1; d*d<=z; d++){
        ans = min(ans, get_score(d));
    }

    auto check = [&](ll d, ll a) -> bool {

        ll health = z;
        ll pre_attacks = (d-1)/k;
        if((__int128_t)health <= (__int128_t)k*pre_attacks*(pre_attacks+1)/2) return true;
        health -= k*pre_attacks*(pre_attacks+1)/2;

        if(health == 0) return true;
        return a >= (health-1)/d + 1;
    };

    for(ll a=0; a*a<=z; a++){
        ll l = 0, r = z;
        while(r-l > 1){
            ll m = (l+r+1)/2;
            if(check(m, a)) r = m;
            else l = m;
        }
        ans = min(ans, get_score(r));
    }

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

signed main(){
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int t; cin >> t;
    while(t--) solve();
}

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

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

Think step by step.

Response:
```plaintext
1
1 1 100000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Monster
time_limit_ms: 2000
memory_limit_mb: 256
Description: Man, this Genshin boss is so hard. Good thing they have a top-up of $6$ coins for only $ \$4.99$. I should be careful and spend no more than I need to, lest my mom catches me...

You are fighting a monster with $z$ health using a weapon with $d$ damage. Initially, $d=0$. You can perform the following operations.

  * Increase $d$ — the damage of your weapon by $1$, costing $x$ coins.    * Attack the monster, dealing $d$ damage and costing $y$ coins.

You cannot perform the first operation for more than $k$ times in a row.

Find the minimum number of coins needed to defeat the monster by dealing at least $z$ damage.

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

The only line of each test case contains 4 integers $x$, $y$, $z$, and $k$ ($1\leq x, y, z, k\leq 10^8$) — the first operation's cost, the second operation's cost, the monster's health, and the limitation on the first operation.

Output: For each test case, output the minimum number of coins needed to defeat the monster.

Examples:
input:
4
2 3 5 5
10 20 40 5
1 60 100 10
60 1 100 10
output:
12
190
280
160

Note: In the first test case, $x = 2$, $y = 3$, $z = 5$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $12$ coins:

  * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.

You deal a total of $3 + 3 = 6$ damage, defeating the monster who has $5$ health. The total number of coins you use is $2 + 2 + 2 + 3 + 3 = 12$ coins.

In the second test case, $x = 10$, $y = 20$, $z = 40$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $190$ coins:

  * Increase damage by $5$, costing $5\cdot x$ = $50$ coins.    * Attack the monster once, dealing $5$ damage, costing $20$ coins.    * Increase damage by $2$, costing $2\cdot x$ = $20$ coins.    * Attack the monster $5$ times, dealing $5\cdot 7 = 35$ damage, costing $5\cdot y$ = $100$ coins.

You deal a total of $5 + 35 = 40$ damage, defeating the monster who has exactly $40$ health. The total number of coins you use is $50 + 20 + 20 + 100 = 190$ coins.

Buggy Code:
from math import sqrt, floor, ceil
import sys
inp = lambda func=int: list(map(func, sys.stdin.readline().split()))
t = inp()[0]
for _ in range(t):
	x,y,z,k = inp()
	n1 = sqrt(y*z/(x+(y/(2*k))))
	N = floor(n1)
	ans = -1
	for n in range(max(N-1000000,1),N+1000000):
		a = n//k
		b = n%k
		c = ceil((z-a*k*(a+1)/2)/n)
		if ans > 0:
			ans = min(ans, n*x+(a+c)*y)
		else:
			ans = n*x+(a+c)*y
	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
1 1 100000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Monster
time_limit_ms: 2000
memory_limit_mb: 256
Description: Man, this Genshin boss is so hard. Good thing they have a top-up of $6$ coins for only $ \$4.99$. I should be careful and spend no more than I need to, lest my mom catches me...

You are fighting a monster with $z$ health using a weapon with $d$ damage. Initially, $d=0$. You can perform the following operations.

  * Increase $d$ — the damage of your weapon by $1$, costing $x$ coins.    * Attack the monster, dealing $d$ damage and costing $y$ coins.

You cannot perform the first operation for more than $k$ times in a row.

Find the minimum number of coins needed to defeat the monster by dealing at least $z$ damage.

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

The only line of each test case contains 4 integers $x$, $y$, $z$, and $k$ ($1\leq x, y, z, k\leq 10^8$) — the first operation's cost, the second operation's cost, the monster's health, and the limitation on the first operation.

Output: For each test case, output the minimum number of coins needed to defeat the monster.

Examples:
input:
4
2 3 5 5
10 20 40 5
1 60 100 10
60 1 100 10
output:
12
190
280
160

Note: In the first test case, $x = 2$, $y = 3$, $z = 5$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $12$ coins:

  * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.

You deal a total of $3 + 3 = 6$ damage, defeating the monster who has $5$ health. The total number of coins you use is $2 + 2 + 2 + 3 + 3 = 12$ coins.

In the second test case, $x = 10$, $y = 20$, $z = 40$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $190$ coins:

  * Increase damage by $5$, costing $5\cdot x$ = $50$ coins.    * Attack the monster once, dealing $5$ damage, costing $20$ coins.    * Increase damage by $2$, costing $2\cdot x$ = $20$ coins.    * Attack the monster $5$ times, dealing $5\cdot 7 = 35$ damage, costing $5\cdot y$ = $100$ coins.

You deal a total of $5 + 35 = 40$ damage, defeating the monster who has exactly $40$ health. The total number of coins you use is $50 + 20 + 20 + 100 = 190$ coins.

Buggy Code:
# standard inputs
import itertools
from collections import *
from bisect import *
import heapq
from functools import *
import math
import sys
import random
from copy import deepcopy
#from templates import *; fin=open('i', 'r'); input = lambda: fin.readline().strip(); from icecream import ic; from sortedcontainers import SortedList
from fractions import Fraction

R = lambda: [*map(int, input().split())]; # read one line(turn to int array)
lines = lambda x: [R() for i in range(x)]; # read multiple input lines

M = int(1e9+7)

def isMtn(A):
  if len(A) <= 2:
    return True
  for i in range(len(A)+1):
    a = A[:i]
    b = A[i:]
    if a == sorted(a, reverse=True) and b == sorted(b):
      return i
  raise

CHECK_LEN = 1000
EPS = 1e-7
def solve(x, y, z, k):
  #ic(x, y, z, k)
  def calc2(A, mode='fake'):
    return calc(A//(k+1), A%(k+1), mode)
  # if A is #blocks
  def calc(A, rem, mode='fake'):
    if A == 0 and rem == 0:
      return math.inf
    if rem > k:
      return math.inf

    blocksum =  y*A + x*k*A
    remsum = x*rem
    dmg = k*A + rem

    health = z-k*A*(A+1)//2
    delsum = (math.ceil(health/dmg) if mode == 'real' else health/dmg) * y
    return blocksum+remsum+delsum

  def bsearch(l, r, fn):
    #ic([fn(i) for i in range(l, r+1)])
    while l+1 < r:
      mid = (l+r)//2
      left_eq, right_eq = mid, mid

      while left_eq > 0 and abs(fn(left_eq) - fn(mid)) < EPS:
        left_eq -= 1
      while left_eq > 0 and abs(fn(right_eq) - fn(mid)) < EPS:
        right_eq += 1

      if fn(left_eq) < fn(mid):
        r = left_eq+1
      elif fn(right_eq) < fn(mid):
        l = right_eq-1
      else:
        return mid
    return l

  def getAns(blocks):
    if blocks < 0:
      return math.inf
    rem = bsearch(blocks==0, k, lambda rem: calc(blocks, rem, 'fake'))
    return calc(blocks, rem)

  inc = bsearch(1, z, calc2)
  ans = 1e100
  for i in range(max(1, inc-CHECK_LEN), min(z, inc+CHECK_LEN)):
    ans = min(ans, calc2(i, mode='real'))

  return ans

for _ in range(int(input())):
  x, y, z, k = R()
  print(solve(x, y, z, k))

**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
1 100000000 100000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Monster
time_limit_ms: 2000
memory_limit_mb: 256
Description: Man, this Genshin boss is so hard. Good thing they have a top-up of $6$ coins for only $ \$4.99$. I should be careful and spend no more than I need to, lest my mom catches me...

You are fighting a monster with $z$ health using a weapon with $d$ damage. Initially, $d=0$. You can perform the following operations.

  * Increase $d$ — the damage of your weapon by $1$, costing $x$ coins.    * Attack the monster, dealing $d$ damage and costing $y$ coins.

You cannot perform the first operation for more than $k$ times in a row.

Find the minimum number of coins needed to defeat the monster by dealing at least $z$ damage.

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

The only line of each test case contains 4 integers $x$, $y$, $z$, and $k$ ($1\leq x, y, z, k\leq 10^8$) — the first operation's cost, the second operation's cost, the monster's health, and the limitation on the first operation.

Output: For each test case, output the minimum number of coins needed to defeat the monster.

Examples:
input:
4
2 3 5 5
10 20 40 5
1 60 100 10
60 1 100 10
output:
12
190
280
160

Note: In the first test case, $x = 2$, $y = 3$, $z = 5$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $12$ coins:

  * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.

You deal a total of $3 + 3 = 6$ damage, defeating the monster who has $5$ health. The total number of coins you use is $2 + 2 + 2 + 3 + 3 = 12$ coins.

In the second test case, $x = 10$, $y = 20$, $z = 40$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $190$ coins:

  * Increase damage by $5$, costing $5\cdot x$ = $50$ coins.    * Attack the monster once, dealing $5$ damage, costing $20$ coins.    * Increase damage by $2$, costing $2\cdot x$ = $20$ coins.    * Attack the monster $5$ times, dealing $5\cdot 7 = 35$ damage, costing $5\cdot y$ = $100$ coins.

You deal a total of $5 + 35 = 40$ damage, defeating the monster who has exactly $40$ health. The total number of coins you use is $50 + 20 + 20 + 100 = 190$ coins.

Buggy Code:
from math import ceil
for i in range(int(input())):
    x , y , z , k = map(int, input().split())
    ans = float('inf')
    for i in range(5):      # upper limit for xyxyxyxyxy till z reaches 10^8
        cur_damage = i*k
        cur_cost = i*y + ( i * k * x) #till the ith pattern of kx's and y
        total_damage = z - ( k * ( (i * (i+1))//2))
        #print(cur_damage, cur_cost, total_damage)
        if total_damage > 0 and k - 1: #ignore k as 1 case doesn't matter
            # here x's should be between 0 and k-1 times and some ys consecutive to deal remaining damage
            l = 0
            r = k - 1
            good = 0
            while(l < r):
                mid = (l + r)//2
                #print(l , r , mid)
                new_d1 = cur_damage + mid
                new_d2 = cur_damage + mid + 1
                if new_d1: #new_d1 can be 0 if mid is 0 and no operations of x so far
                    cost1 = mid*x + ( int( ceil ( total_damage / new_d1 ) ) )*y
                    cost2 = (mid+1)*x + ( int( ceil ( total_damage / new_d2 ) ) )*y
                    if cost1 > cost2:
                        l = mid + 1
                        good = mid + 1
                    else:
                        good = mid
                        r = mid
                    #print(l, r , mid , cost1 , cost2, good)
                else:
                    l += 1 # if l is 0 and no x operation simply ignore 0 case
            extra = good*x + ( int ( ceil ( total_damage / (cur_damage + good) ) ) )*y
            ans = min(ans, cur_cost + extra)
            #print(cur_cost + extra, good , i)
        else:
            ans = min(ans, cur_cost)
            #print(cur_cost)

    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
3
1 1 1 1
100000000 100000000 100000000 1
1 100000000 100000000 100000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Monster
time_limit_ms: 2000
memory_limit_mb: 256
Description: Man, this Genshin boss is so hard. Good thing they have a top-up of $6$ coins for only $ \$4.99$. I should be careful and spend no more than I need to, lest my mom catches me...

You are fighting a monster with $z$ health using a weapon with $d$ damage. Initially, $d=0$. You can perform the following operations.

  * Increase $d$ — the damage of your weapon by $1$, costing $x$ coins.    * Attack the monster, dealing $d$ damage and costing $y$ coins.

You cannot perform the first operation for more than $k$ times in a row.

Find the minimum number of coins needed to defeat the monster by dealing at least $z$ damage.

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

The only line of each test case contains 4 integers $x$, $y$, $z$, and $k$ ($1\leq x, y, z, k\leq 10^8$) — the first operation's cost, the second operation's cost, the monster's health, and the limitation on the first operation.

Output: For each test case, output the minimum number of coins needed to defeat the monster.

Examples:
input:
4
2 3 5 5
10 20 40 5
1 60 100 10
60 1 100 10
output:
12
190
280
160

Note: In the first test case, $x = 2$, $y = 3$, $z = 5$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $12$ coins:

  * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.

You deal a total of $3 + 3 = 6$ damage, defeating the monster who has $5$ health. The total number of coins you use is $2 + 2 + 2 + 3 + 3 = 12$ coins.

In the second test case, $x = 10$, $y = 20$, $z = 40$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $190$ coins:

  * Increase damage by $5$, costing $5\cdot x$ = $50$ coins.    * Attack the monster once, dealing $5$ damage, costing $20$ coins.    * Increase damage by $2$, costing $2\cdot x$ = $20$ coins.    * Attack the monster $5$ times, dealing $5\cdot 7 = 35$ damage, costing $5\cdot y$ = $100$ coins.

You deal a total of $5 + 35 = 40$ damage, defeating the monster who has exactly $40$ health. The total number of coins you use is $50 + 20 + 20 + 100 = 190$ coins.

Buggy Code:
//make sure to make new file!
import java.io.*;
import java.util.*;

public class EG27b{

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

      int t = Integer.parseInt(f.readLine());

      int THRESH = 10000;

      for(int q = 1; q <= t; q++){

         StringTokenizer st = new StringTokenizer(f.readLine());

         long x = Long.parseLong(st.nextToken());
         long y = Long.parseLong(st.nextToken());
         long h = Long.parseLong(st.nextToken());
         long w = Long.parseLong(st.nextToken());

         long ans = Long.MAX_VALUE;
         if(w < THRESH){
            //chunks are small
            //ternary search over chunks, for every chunk get the min
            long l = 0;
            long r = h/w;
            while(l < r){
               long m1 = l + (r-l)/3;
               long m2 = (m1+1 + (r-(m1+1))/2);

               long a1 = calc1(m1,x,y,h,w);
               long a2 = calc1(m2,x,y,h,w);
               ans = Math.min(ans,Math.min(a1,a2));

               if(a1 <= a2){
                  r = m2-1;
               } else {
                  l = m1+1;
               }
            }
         } else {
            //chunks are big
            //for every chunk ternary search inside
            for(long k = 0; k < h/w; k++){
               long l = k*w+1;
               long r = k*w+w;
               while(l < r){
                  long m1 = l + (r-l)/3;
                  long m2 = (m1+1 + (r-(m1+1))/2);

                  long a1 = calc(m1,x,y,h,w);
                  long a2 = calc(m2,x,y,h,w);
                  /*
                  if(k == 4 || k == 3){
                     out.println(m1 + " " + a1 + " " + m2 + " " + a2);
                  }*/
                  ans = Math.min(ans,Math.min(a1,a2));

                  if(a1 < a2){
                     r = m2-1;
                  } else {
                     l = m1+1;
                  }
               }
               //out.println(k + " " + ans);
            }

         }

         out.println(ans);

      }

      out.close();
   }

   public static long calc1(long m, long x, long y, long h, long w){
      long min = Long.MAX_VALUE;
      for(long k = 0; k < w; k++){
         long d = m*w+k+1;
         min = Math.min(min,calc(d,x,y,h,w));
      }
      return min;
   }

   public static long calc(long d, long x, long y, long h, long w){
      //how many times you do operation 2 before reaching d
      long prenum = (d-1)/w;
      //will do w*1, w*2, w*3... until w*prenum
      long ans = prenum*y;
      h -= w*(1 + prenum) * prenum / 2L;
      if(h <= 0){
         return x*d + prenum*y;
      }
      long yfreq = (h+d-1)/d;
      //get minimum value of d that achieves that yfreq
      long d2 = Math.max((h+yfreq-1)/yfreq,prenum*w);
      //if(d == 9) System.out.println(d2);
      ans += d2*x + yfreq * y;
      return ans;
   }
   /*
   public static long calc(long d, long x, long y, long h, long w){
      //how many times you do operation 2 before reaching d
      long prenum = (d-1)/w;
      //will do w*1, w*2, w*3... until w*prenum
      long ans = prenum*y;
      h -= w*(1 + prenum) * prenum / 2L;
      if(h <= 0){
         return x*d + prenum*y;
      }
      ans += x*d + (h+d-1)/d * y;
      return 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
1 1 100000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Monster
time_limit_ms: 2000
memory_limit_mb: 256
Description: Man, this Genshin boss is so hard. Good thing they have a top-up of $6$ coins for only $ \$4.99$. I should be careful and spend no more than I need to, lest my mom catches me...

You are fighting a monster with $z$ health using a weapon with $d$ damage. Initially, $d=0$. You can perform the following operations.

  * Increase $d$ — the damage of your weapon by $1$, costing $x$ coins.    * Attack the monster, dealing $d$ damage and costing $y$ coins.

You cannot perform the first operation for more than $k$ times in a row.

Find the minimum number of coins needed to defeat the monster by dealing at least $z$ damage.

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

The only line of each test case contains 4 integers $x$, $y$, $z$, and $k$ ($1\leq x, y, z, k\leq 10^8$) — the first operation's cost, the second operation's cost, the monster's health, and the limitation on the first operation.

Output: For each test case, output the minimum number of coins needed to defeat the monster.

Examples:
input:
4
2 3 5 5
10 20 40 5
1 60 100 10
60 1 100 10
output:
12
190
280
160

Note: In the first test case, $x = 2$, $y = 3$, $z = 5$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $12$ coins:

  * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.

You deal a total of $3 + 3 = 6$ damage, defeating the monster who has $5$ health. The total number of coins you use is $2 + 2 + 2 + 3 + 3 = 12$ coins.

In the second test case, $x = 10$, $y = 20$, $z = 40$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $190$ coins:

  * Increase damage by $5$, costing $5\cdot x$ = $50$ coins.    * Attack the monster once, dealing $5$ damage, costing $20$ coins.    * Increase damage by $2$, costing $2\cdot x$ = $20$ coins.    * Attack the monster $5$ times, dealing $5\cdot 7 = 35$ damage, costing $5\cdot y$ = $100$ coins.

You deal a total of $5 + 35 = 40$ damage, defeating the monster who has exactly $40$ health. The total number of coins you use is $50 + 20 + 20 + 100 = 190$ coins.

Buggy Code:
//make sure to make new file!
import java.io.*;
import java.util.*;

public class EG27b{

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

      int t = Integer.parseInt(f.readLine());

      int THRESH = 10000;

      for(int q = 1; q <= t; q++){

         StringTokenizer st = new StringTokenizer(f.readLine());

         long x = Long.parseLong(st.nextToken());
         long y = Long.parseLong(st.nextToken());
         long h = Long.parseLong(st.nextToken());
         long w = Long.parseLong(st.nextToken());

         long ans = Long.MAX_VALUE;
         if(w < THRESH){
            //chunks are small
            //ternary search over chunks, for every chunk get the min
            long l = 0;
            long r = h/w;
            while(l < r){
               long m1 = l + (r-l)/3;
               long m2 = (m1+1 + (r-(m1+1))/2);

               long a1 = calc1(m1,x,y,h,w);
               long a2 = calc1(m2,x,y,h,w);
               ans = Math.min(ans,Math.min(a1,a2));

               if(a1 <= a2){
                  r = m2-1;
               } else {
                  l = m1+1;
               }
            }
         } else {
            //chunks are big
            //for every chunk ternary search inside
            for(long k = 0; k < h/w; k++){
               long l = k*w+1;
               long r = k*w+w;
               while(l < r){
                  long m1 = l + (r-l)/3;
                  long m2 = (m1+1 + (r-(m1+1))/2);

                  long a1 = calc(m1,x,y,h,w);
                  long a2 = calc(m2,x,y,h,w);
                  /*
                  if(k == 4 || k == 3){
                     out.println(m1 + " " + a1 + " " + m2 + " " + a2);
                  }*/
                  ans = Math.min(ans,Math.min(a1,a2));

                  if(a1 < a2){
                     r = m2-1;
                  } else {
                     l = m1+1;
                  }
               }
               //out.println(k + " " + ans);
            }

         }

         out.println(ans);

      }

      out.close();
   }

   public static long calc1(long m, long x, long y, long h, long w){
      long min = Long.MAX_VALUE;
      for(long k = 0; k < w; k++){
         long d = m*w+k+1;
         min = Math.min(min,calc(d,x,y,h,w));
      }
      return min;
   }
   /*
   public static long calc(long d, long x, long y, long h, long w){
      //how many times you do operation 2 before reaching d
      long prenum = (d-1)/w;
      //will do w*1, w*2, w*3... until w*prenum
      long ans = prenum*y;
      h -= w*(1 + prenum) * prenum / 2L;
      if(h <= 0){
         return x*d + prenum*y;
      }
      long yfreq = (h+d-1)/d;
      //get minimum value of d that achieves that yfreq
      long d2 = Math.max((h+yfreq-1)/yfreq,prenum*w);
      //if(d == 9) System.out.println(d2);
      ans += d2*x + yfreq * y;
      return ans;
   }*/

   public static long calc(long d, long x, long y, long h, long w){
      //how many times you do operation 2 before reaching d
      long prenum = (d-1)/w;
      //will do w*1, w*2, w*3... until w*prenum
      long ans = prenum*y;
      h -= w*(1 + prenum) * prenum / 2L;
      if(h <= 0){
         return x*d + prenum*y;
      }
      ans += x*d + (h+d-1)/d * y;
      return 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
1 1 100000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Monster
time_limit_ms: 2000
memory_limit_mb: 256
Description: Man, this Genshin boss is so hard. Good thing they have a top-up of $6$ coins for only $ \$4.99$. I should be careful and spend no more than I need to, lest my mom catches me...

You are fighting a monster with $z$ health using a weapon with $d$ damage. Initially, $d=0$. You can perform the following operations.

  * Increase $d$ — the damage of your weapon by $1$, costing $x$ coins.    * Attack the monster, dealing $d$ damage and costing $y$ coins.

You cannot perform the first operation for more than $k$ times in a row.

Find the minimum number of coins needed to defeat the monster by dealing at least $z$ damage.

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

The only line of each test case contains 4 integers $x$, $y$, $z$, and $k$ ($1\leq x, y, z, k\leq 10^8$) — the first operation's cost, the second operation's cost, the monster's health, and the limitation on the first operation.

Output: For each test case, output the minimum number of coins needed to defeat the monster.

Examples:
input:
4
2 3 5 5
10 20 40 5
1 60 100 10
60 1 100 10
output:
12
190
280
160

Note: In the first test case, $x = 2$, $y = 3$, $z = 5$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $12$ coins:

  * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.

You deal a total of $3 + 3 = 6$ damage, defeating the monster who has $5$ health. The total number of coins you use is $2 + 2 + 2 + 3 + 3 = 12$ coins.

In the second test case, $x = 10$, $y = 20$, $z = 40$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $190$ coins:

  * Increase damage by $5$, costing $5\cdot x$ = $50$ coins.    * Attack the monster once, dealing $5$ damage, costing $20$ coins.    * Increase damage by $2$, costing $2\cdot x$ = $20$ coins.    * Attack the monster $5$ times, dealing $5\cdot 7 = 35$ damage, costing $5\cdot y$ = $100$ coins.

You deal a total of $5 + 35 = 40$ damage, defeating the monster who has exactly $40$ health. The total number of coins you use is $50 + 20 + 20 + 100 = 190$ coins.

Buggy Code:
def ii():
    return int(input())
def li():
    return list(map(int,input().split()))
def si():
    return input()
def ti():
    return map(int,input().split())
def di():
    return deque((map(int,input().split())))
from bisect import bisect_left,bisect_right
from heapq import heapify,heappop,heappush
from itertools import permutations
from collections import deque,defaultdict,Counter
def maxsubarraysum(arr,k):
    sumi=0
    best=0
    for i in range(k):
        sumi=max(arr[i],arr[i]+sumi)
        best=max(best,sumi)
    return best
def mod_inverse(a, m):
    m0 = m
    y = 0
    x = 1
    if m == 1:
        return 0
    while a > 1:
        q = a // m
        t = m
        m = a % m
        a = t
        t = y
        y = x - q * y
        x = t
    if x < 0:
        x = x + m0
    return x
def prefixsum(l,n):
    if n==0:
        return []
    array=[l[0]]
    for i in range(1,n):
        array.append(array[-1]+l[i])
    return array
def suffixsum(l,n):
    array=[0]*n
    temperory=0
    for i in range(n):
        temperory+=l[n-1-i]
        array[n-1-i]=temperory
    return array
def prime(x):
    if x==1:
        return(0)
    elif x==2:
        return(1)
    else:
        t=(sqrt(x))+1
        t=int(t)
        for i in range(2,t):
            if x%i==0:
                return(0)
        else:
            return(1)
def prefixmax(l,n):
    array=[l[0]]
    for i in range(1,n):
        array.append(max(array[-1],l[i]))
    return array
def prefixmin(l,n):
    array=[l[0]]
    for i in range(1,n):
        array.append(min(array[-1],l[i]))
    return array
def BinarySearch(a, x):

    i = bisect_left(a, x)

    if i != len(a) and a[i] == x:

        return i

    else:

        return -1
def sortremoveDuplicates(arr, n):
    if n == 0 or n == 1:
        return arr

    temp = []
    j = 0
    for i in range(0, n-1):
        if arr[i] != arr[i+1]:
            temp += [arr[i]]
            j += 1
    temp += [arr[n-1]]
    return temp
def decimalToBinary(n):
    return bin(n).replace("0b", "")
def ato1(chr):
    return ord(chr)-ord("A")
def permutationindex(l,n):
    arr=[0]*(n)
    for i in range(n):
        arr[l[i]-1]=i
    return arr
def issubsequence(s, s2):

    n, m = len(s), len(s2)
    i, j = 0, 0
    while (i < n and j < m):
        if (s[i] == s2[j]):
            i += 1
        j += 1

    # If i reaches end of s,that mean we found all
    # characters of s in s2,
    # so s is subsequence of s2, else not
    return i == n
def def_value():
    return -1
def primef(n):
    l1=[]
    i=2
    while i*i<=n:
        if n%i!=0:
            i+=1
        else:
            l1.append(i)
            while(n%i==0):
                n//=i
    if n>1:
        l1.append(n)
    return l1
"""
def search( k) :
        global ans
        if (k == len(l)) :
           if len(subset)%2==0 and len(subset)>0:
               ans+=maxi//product(subset)
           elif len(subset)>0:
               ans-=maxi//product(subset)
        else:
           search(k+1)
           subset.append(l[k])
           search(k+1)
           subset.pop()
"""
def seive(n):
    ans=[0]*(n+1)
    #primes=[]
    for i in range(2,n+1):
        if ans[i]==0:
            #primes.append(i)
            for j in range(2*i,n+1,i):
                ans[j]=i
    return ans
def inte(s,s2):
    if s[1]<s2[0] or s2[1]<s[0]:
        return 0
    return [max(s[0],s2[0]),min(s[1],s2[1])]
def checksq(x):
    z=isqrt(x)
    if z<0:
        return False
    if z*z==x:
        return True
    return False
def counti(l,n):
    l1=[]
    z=1
    for i in range(1,n):
        if l[i]==l[i-1]:
            z+=1
        else:
            l1.append([l[i-1],z])
            z=1
    l1.append([l[-1],z])
    return l1
def next(arr, target):
    start = 0;
    end = len(arr) - 1;
    # Minimum size of the array should be 1
    if (end == 0):
        return -1
    '''
    If target lies beyond the max element, than the index of strictly smaller
    value than target should be (end - 1)
    '''
    if (target > arr[end]):
        return end

    ans = -1;
    while (start <= end):
        mid = (start + end) // 2;
        if (arr[mid] >= target):
            end = mid - 1;
        else:
            ans = mid;
            start = mid + 1;

    return ans;
def next1(arr, target):
    start = 0
    end = len(arr) - 1

    ans = -1
    while (start <= end):
        mid = (start + end) // 2

        # Move to right side if target is
        # greater.
        if (arr[mid] <= target):
            start = mid + 1

        # Move left side.
        else:
            ans = mid
            end = mid - 1

    return ans
def next_greatest_indices(arr):
    result = [len(arr)] * len(arr)
    stack = []
    for i in range(len(arr) - 1, -1, -1):
        while stack and arr[stack[-1]] < arr[i]:
            stack.pop()
        if stack:
            result[i] = stack[-1]
        stack.append(i)
    return result
def find_largest_index(arr, k):
    left, right = 0, len(arr) - 1
    result = -1

    while left <= right:
        mid = (left + right) // 2

        if arr[mid] < k:
            result = mid
            left = mid + 1
        else:
            right = mid - 1

    return result

import sys
input=sys.stdin.readline
def f(k1):
    global dam,x,y,stre
    if (stre==0 and k1==0):
        return 10**20
    return k1*x+y*(1+((dam-1)//(stre+k1)))
def ternary_search_integer(left, right):
    while right - left > 2:
        mid1 = left + (right - left) // 3
        mid2 = right - (right - left) // 3
        if f(mid1) < f(mid2):

            right = mid2
        else:
            left = mid1

    min_x = left
    for x in range(left + 1, right + 1):
        if f(x) < f(min_x):
            min_x = x
    return f(min_x)
def cost(a,k,x,y,b):
    return a*(k*x+y)+b*y
for _ in range(ii()):
    x,y,z,k=ti()
    ans=10**20
    for a in range(20000):
        if (k*(a)*(a-1))>2*z:
            break
        if (k*(a)*(a+1))>=2*z:
            ans=min(ans,a*(k*x+y))
            continue
        stre=a*k
        dam=z-(k*(a)*(a+1))//2
        ans1=ternary_search_integer(0,k-1)
        ans1+=a*k*x+a*y
        ans=min(ans,ans1)
    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
1 1 100000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Monster
time_limit_ms: 2000
memory_limit_mb: 256
Description: Man, this Genshin boss is so hard. Good thing they have a top-up of $6$ coins for only $ \$4.99$. I should be careful and spend no more than I need to, lest my mom catches me...

You are fighting a monster with $z$ health using a weapon with $d$ damage. Initially, $d=0$. You can perform the following operations.

  * Increase $d$ — the damage of your weapon by $1$, costing $x$ coins.    * Attack the monster, dealing $d$ damage and costing $y$ coins.

You cannot perform the first operation for more than $k$ times in a row.

Find the minimum number of coins needed to defeat the monster by dealing at least $z$ damage.

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

The only line of each test case contains 4 integers $x$, $y$, $z$, and $k$ ($1\leq x, y, z, k\leq 10^8$) — the first operation's cost, the second operation's cost, the monster's health, and the limitation on the first operation.

Output: For each test case, output the minimum number of coins needed to defeat the monster.

Examples:
input:
4
2 3 5 5
10 20 40 5
1 60 100 10
60 1 100 10
output:
12
190
280
160

Note: In the first test case, $x = 2$, $y = 3$, $z = 5$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $12$ coins:

  * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.

You deal a total of $3 + 3 = 6$ damage, defeating the monster who has $5$ health. The total number of coins you use is $2 + 2 + 2 + 3 + 3 = 12$ coins.

In the second test case, $x = 10$, $y = 20$, $z = 40$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $190$ coins:

  * Increase damage by $5$, costing $5\cdot x$ = $50$ coins.    * Attack the monster once, dealing $5$ damage, costing $20$ coins.    * Increase damage by $2$, costing $2\cdot x$ = $20$ coins.    * Attack the monster $5$ times, dealing $5\cdot 7 = 35$ damage, costing $5\cdot y$ = $100$ coins.

You deal a total of $5 + 35 = 40$ damage, defeating the monster who has exactly $40$ health. The total number of coins you use is $50 + 20 + 20 + 100 = 190$ coins.

Buggy Code:
def ceil_div(a, b):
    return (a + b - 1) // b

def calc(d):
    res = d * x
    take = d // k
    if d % k == 0:
        take -= 1
    res += take * y
    term = z - k * (take * (take + 1) // 2)
    res += ceil_div(max(term, 0), d) * y
    return res

def main():
    global x, y, z, k

    t = int(input().strip())

    for _ in range(t):
        l = 1
        x, y, z, k = map(int, input().strip().split())
        r = z
        res = 10**20

        while l <= r:
            mid1 = l + (r - l) // 3
            mid2 = r - (r - l) // 3
            res1 = calc(mid1)
            res2 = calc(mid2)

            res = min(res, res1, res2)

            if res1 > res2:
                l = mid1 + 1
            elif res1 < res2:
                r = mid2 - 1
            else:
                l = mid1 + 1
                r = mid2 - 1

        print(res)

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
```plaintext
1
1 100000000 100000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Monster
time_limit_ms: 2000
memory_limit_mb: 256
Description: Man, this Genshin boss is so hard. Good thing they have a top-up of $6$ coins for only $ \$4.99$. I should be careful and spend no more than I need to, lest my mom catches me...

You are fighting a monster with $z$ health using a weapon with $d$ damage. Initially, $d=0$. You can perform the following operations.

  * Increase $d$ — the damage of your weapon by $1$, costing $x$ coins.    * Attack the monster, dealing $d$ damage and costing $y$ coins.

You cannot perform the first operation for more than $k$ times in a row.

Find the minimum number of coins needed to defeat the monster by dealing at least $z$ damage.

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

The only line of each test case contains 4 integers $x$, $y$, $z$, and $k$ ($1\leq x, y, z, k\leq 10^8$) — the first operation's cost, the second operation's cost, the monster's health, and the limitation on the first operation.

Output: For each test case, output the minimum number of coins needed to defeat the monster.

Examples:
input:
4
2 3 5 5
10 20 40 5
1 60 100 10
60 1 100 10
output:
12
190
280
160

Note: In the first test case, $x = 2$, $y = 3$, $z = 5$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $12$ coins:

  * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.

You deal a total of $3 + 3 = 6$ damage, defeating the monster who has $5$ health. The total number of coins you use is $2 + 2 + 2 + 3 + 3 = 12$ coins.

In the second test case, $x = 10$, $y = 20$, $z = 40$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $190$ coins:

  * Increase damage by $5$, costing $5\cdot x$ = $50$ coins.    * Attack the monster once, dealing $5$ damage, costing $20$ coins.    * Increase damage by $2$, costing $2\cdot x$ = $20$ coins.    * Attack the monster $5$ times, dealing $5\cdot 7 = 35$ damage, costing $5\cdot y$ = $100$ coins.

You deal a total of $5 + 35 = 40$ damage, defeating the monster who has exactly $40$ health. The total number of coins you use is $50 + 20 + 20 + 100 = 190$ coins.

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

typedef long long ll;

// Función para calcular el mínimo costo para un caso de prueba
ll findMinCost(ll x, ll y, ll z, ll k) {
    // Calcular el valor óptimo de d basado en la fórmula matemática
    double dOptDouble = sqrt((double)y * (double)z / (double)x);
    ll dOpt = floor(dOptDouble);

    // Definir el rango de iteración alrededor de dOpt
    ll startD = max(1LL, dOpt > 100 ? dOpt - 100 : 1LL);
    ll endD = min(z, dOpt + 100);

    ll minCost = LLONG_MAX;

    for(ll d = startD; d <= endD; d++) {
        // Calcular el número de bloques de incrementos
        ll attacksBuild = d / k;

        // Calcular el daño total durante la construcción utilizando __int128 para evitar desbordamientos
        ll sumBuild;
        if(attacksBuild > 0){
            __int128 temp = (__int128)k * attacksBuild * (attacksBuild + 1);
            temp /= 2;
            if(temp >= z){
                sumBuild = z;
            }
            else{
                sumBuild = (ll)temp;
            }
        }
        else{
            sumBuild = 0;
        }

        // Calcular los ataques finales necesarios
        ll attacksFinal;
        if(sumBuild >= z){
            attacksFinal = 0;
        }
        else{
            attacksFinal = (z - sumBuild + d - 1) / d;
        }

        // Calcular el costo total
        ll totalCost = d * x + (attacksBuild + attacksFinal) * y;

        // Actualizar el costo mínimo
        if(totalCost < minCost){
            minCost = totalCost;
        }
    }

    // Considerar d = z si no está dentro del rango anterior
    if(z < startD || z > endD){
        ll d = z;
        ll attacksBuild = d / k;
        ll sumBuild;
        if(attacksBuild > 0){
            __int128 temp = (__int128)k * attacksBuild * (attacksBuild + 1);
            temp /= 2;
            if(temp >= z){
                sumBuild = z;
            }
            else{
                sumBuild = (ll)temp;
            }
        }
        else{
            sumBuild = 0;
        }

        ll attacksFinal;
        if(sumBuild >= z){
            attacksFinal = 0;
        }
        else{
            attacksFinal = (z - sumBuild + d - 1) / d;
        }

        ll totalCost = d * x + (attacksBuild + attacksFinal) * y;

        if(totalCost < minCost){
            minCost = totalCost;
        }
    }

    return minCost;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while(t--){
        ll x, y, z, k;
        cin >> x >> y >> z >> k;
        ll resultado = findMinCost(x, y, z, k);
        cout << resultado << "\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
1 1 100000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Monster
time_limit_ms: 2000
memory_limit_mb: 256
Description: Man, this Genshin boss is so hard. Good thing they have a top-up of $6$ coins for only $ \$4.99$. I should be careful and spend no more than I need to, lest my mom catches me...

You are fighting a monster with $z$ health using a weapon with $d$ damage. Initially, $d=0$. You can perform the following operations.

  * Increase $d$ — the damage of your weapon by $1$, costing $x$ coins.    * Attack the monster, dealing $d$ damage and costing $y$ coins.

You cannot perform the first operation for more than $k$ times in a row.

Find the minimum number of coins needed to defeat the monster by dealing at least $z$ damage.

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

The only line of each test case contains 4 integers $x$, $y$, $z$, and $k$ ($1\leq x, y, z, k\leq 10^8$) — the first operation's cost, the second operation's cost, the monster's health, and the limitation on the first operation.

Output: For each test case, output the minimum number of coins needed to defeat the monster.

Examples:
input:
4
2 3 5 5
10 20 40 5
1 60 100 10
60 1 100 10
output:
12
190
280
160

Note: In the first test case, $x = 2$, $y = 3$, $z = 5$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $12$ coins:

  * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.

You deal a total of $3 + 3 = 6$ damage, defeating the monster who has $5$ health. The total number of coins you use is $2 + 2 + 2 + 3 + 3 = 12$ coins.

In the second test case, $x = 10$, $y = 20$, $z = 40$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $190$ coins:

  * Increase damage by $5$, costing $5\cdot x$ = $50$ coins.    * Attack the monster once, dealing $5$ damage, costing $20$ coins.    * Increase damage by $2$, costing $2\cdot x$ = $20$ coins.    * Attack the monster $5$ times, dealing $5\cdot 7 = 35$ damage, costing $5\cdot y$ = $100$ coins.

You deal a total of $5 + 35 = 40$ damage, defeating the monster who has exactly $40$ health. The total number of coins you use is $50 + 20 + 20 + 100 = 190$ coins.

Buggy Code:
#include <bits/stdc++.h>
#define ll long long
#define mp make_pair
#define F first
#define S second
#define pb push_back
#define pii pair < int, int >
#define FIO ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)

using namespace std;

ll x, y, z, k;

void upis( void ) {

	cin >> x >> y >> z >> k;

	return;
}

void solve( void ) {

	ll res = 1e18;
	ll sol = 1e18;
	ll tempres;
	int cur;
	for(ll i = 1; i < z + 1; i++) {
		res = i / k * y + i * x + (max((ll)0, (z - (i / k + 1) * (i / k) / 2 * k)) + i - 1) / i * y;
		sol = min(sol, res);
	}
	cout << sol << "\n";

	return;
}

int main( void ) {

	FIO;
	int tt;
	cin >> tt;
	while(tt--) {
		upis();
		solve();
	}

	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
1 100 100000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Monster
time_limit_ms: 2000
memory_limit_mb: 256
Description: Man, this Genshin boss is so hard. Good thing they have a top-up of $6$ coins for only $ \$4.99$. I should be careful and spend no more than I need to, lest my mom catches me...

You are fighting a monster with $z$ health using a weapon with $d$ damage. Initially, $d=0$. You can perform the following operations.

  * Increase $d$ — the damage of your weapon by $1$, costing $x$ coins.    * Attack the monster, dealing $d$ damage and costing $y$ coins.

You cannot perform the first operation for more than $k$ times in a row.

Find the minimum number of coins needed to defeat the monster by dealing at least $z$ damage.

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

The only line of each test case contains 4 integers $x$, $y$, $z$, and $k$ ($1\leq x, y, z, k\leq 10^8$) — the first operation's cost, the second operation's cost, the monster's health, and the limitation on the first operation.

Output: For each test case, output the minimum number of coins needed to defeat the monster.

Examples:
input:
4
2 3 5 5
10 20 40 5
1 60 100 10
60 1 100 10
output:
12
190
280
160

Note: In the first test case, $x = 2$, $y = 3$, $z = 5$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $12$ coins:

  * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.

You deal a total of $3 + 3 = 6$ damage, defeating the monster who has $5$ health. The total number of coins you use is $2 + 2 + 2 + 3 + 3 = 12$ coins.

In the second test case, $x = 10$, $y = 20$, $z = 40$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $190$ coins:

  * Increase damage by $5$, costing $5\cdot x$ = $50$ coins.    * Attack the monster once, dealing $5$ damage, costing $20$ coins.    * Increase damage by $2$, costing $2\cdot x$ = $20$ coins.    * Attack the monster $5$ times, dealing $5\cdot 7 = 35$ damage, costing $5\cdot y$ = $100$ coins.

You deal a total of $5 + 35 = 40$ damage, defeating the monster who has exactly $40$ health. The total number of coins you use is $50 + 20 + 20 + 100 = 190$ coins.

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

public class Main {
    static PrintWriter pw = new PrintWriter(System.out);
    static Scanner sc = new Scanner(System.in);

    public static long sumMultiplesOfK(long k, long n) {
        long m = n / k;
        return k * m * (m + 1) / 2;
    }

    static long ans;

//    public static long ternarySearch(long low, long high, long x, long y, long z, long k) {
//        long mid1 = low + (high - low) / 3;
//        long mid2 = high - (high - low) / 3;
//
//        long cnt1 = mid1 * x;
//        long ans1 = z - sumMultiplesOfK(k, mid1);
//        cnt1 += (mid1 / k) * y;
//        cnt1 += ((ans1 + mid1 - 1) / mid1) * y;
//
//
//        long cnt2 = mid2 * x;
//        long ans2 = z - sumMultiplesOfK(k, mid2);
//        cnt2 += (mid2 / k) * y;
//        cnt2 += ((ans2 + mid2 - 1) / mid2) * y;
//
//        if(cnt1 < cnt2) {
//            high = mid2 - 1;
//            ans = Math.min(ans, cnt1);
//            return ternarySearch(low, mid1 - 1, x, y, z, k);
//        }
//        else{
//            low = mid1 + 1;
//            ans = Math.min(ans, cnt2);
//            return ternarySearch(mid2 + 1, high, x, y, z, k);
//        }
//
//    }

    static class Pair implements Comparable<Pair>{
        Long x ;
        Long y ;

        public 	Pair(Long i, Long j) {
            this.x = i;
            this.y = j;
        }
        public int compareTo(Pair p) {
            if (this.x.compareTo(p.x) != 0)
                return -this.x.compareTo(p.x);
            else
                return this.y.compareTo(p.y);
        }

        public String toString(){
            return "("+this.x + "," + this.y + ")";
        }
    }

    public static void solve() throws IOException {
        long x = sc.nextInt(), y = sc.nextInt(), z = sc.nextInt(), k = sc.nextInt();
        ans = Long.MAX_VALUE;
        if(z <= 1000){
            for(int i = 1; i <= z; i++){
                long cur = i * x;
                long rem = z - sumMultiplesOfK(k, i);
                cur += (i / k) * y;
                cur += ((rem + i - 1) / i) * y;
                ans = Math.min(ans, cur);
            }
        }
        long cur1 = -1, cur2 = -1;
        int idx1 = -1, idx2 = -1;
        for(int i = 1 ; i <= z ; i+= 10){
            long cur = i * x;
            long rem = z - sumMultiplesOfK(k, i);
            cur += (i / k) * y;
            cur += ((rem + i - 1) / i) * y;
            if(cur1 == -1){
                cur1 = cur;
                idx1 = i;
            }
            else if(cur2 == -1){
                cur2 = cur;
                idx2 = i;
            }
            else{
                if(cur <= cur2){
                    cur1 = cur2;
                    idx1 = idx2;
                    cur2 = cur;
                    idx2 = i;
                }
                else{
                    break;
                }
            }
        }
        for(int i = Math.max(1, idx1 - 5 * 100); i <= Math.min(z, idx2 + 5 * 100); i++){
            long cur = i * x;
            long rem = z - sumMultiplesOfK(k, i);
            cur += (i / k) * y;
            cur += ((rem + i - 1) / i) * y;
            ans = Math.min(ans, cur);
        }
        pw.println(ans);
    }

    public static void main(String[] args) throws IOException {
        int t = sc.nextInt();
        while(t -- > 0)
            solve();
        pw.flush();
    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(String s) throws FileNotFoundException {
            br = new BufferedReader(new FileReader(s));
        }

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

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

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

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

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec)
                        f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

        public long[] nextlongArray(int n) throws IOException {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public Long[] nextLongArray(int n) throws IOException {
            Long[] a = new Long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public int[] nextIntArray(int n) throws IOException {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public Integer[] nextIntegerArray(int n) throws IOException {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public boolean ready() throws IOException {
            return br.ready();
        }
    }
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. The code has a few potential pitfalls, such as the way it handles the sum of multiples of `k` and the ternary search logic (which is commented out). Let's consider the following scenarios:

1. **Edge Case with Small Values:**
   - When `z` is very small, the code might not handle it correctly if it relies on large steps.
   - Example: `x = 1`, `y = 1`, `z = 1`, `k = 1`

2. **Edge Case with Large Values:**
   - When `z` is very large, the code might struggle with large values and the sum calculation.
   - Example: `x = 100000000`, `y = 100000000`, `z = 100000000`, `k = 100000000`

3. **Edge Case with Large `k` and Small `x`:**
   - When `k` is very large and `x` is small, the code might not handle the large steps correctly.
   - Example: `x = 1`, `y = 100000000`, `z = 100000000`, `k = 100000000`

4. **Edge Case with Small `k` and Large `x`:**
   - When `k` is very small and `x` is large, the code might not handle the small steps correctly.
   - Example: `x = 100000000`, `y = 1`, `z = 100000000`, `k = 1`

5. **Edge Case with Balanced Values:**
   - When `x`, `y`, `z`, and `k` are balanced, the code might still fail if it doesn't handle the balance correctly.
   - Example: `x = 10000`, `y = 10000`, `z = 100000`, `k = 10000`

Let's choose the most challenging and compact input that can expose the bug:

```plaintext
5
1 1 1 1
100000000 100000000 100000000 100000000
1 100000000 100000000 100000000
100000000 1 100000000 1
10000 10000 100000 10000
```

This input includes a variety of edge cases and balanced values to thoroughly test the code's ability to handle different scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Monster
time_limit_ms: 2000
memory_limit_mb: 256
Description: Man, this Genshin boss is so hard. Good thing they have a top-up of $6$ coins for only $ \$4.99$. I should be careful and spend no more than I need to, lest my mom catches me...

You are fighting a monster with $z$ health using a weapon with $d$ damage. Initially, $d=0$. You can perform the following operations.

  * Increase $d$ — the damage of your weapon by $1$, costing $x$ coins.    * Attack the monster, dealing $d$ damage and costing $y$ coins.

You cannot perform the first operation for more than $k$ times in a row.

Find the minimum number of coins needed to defeat the monster by dealing at least $z$ damage.

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

The only line of each test case contains 4 integers $x$, $y$, $z$, and $k$ ($1\leq x, y, z, k\leq 10^8$) — the first operation's cost, the second operation's cost, the monster's health, and the limitation on the first operation.

Output: For each test case, output the minimum number of coins needed to defeat the monster.

Examples:
input:
4
2 3 5 5
10 20 40 5
1 60 100 10
60 1 100 10
output:
12
190
280
160

Note: In the first test case, $x = 2$, $y = 3$, $z = 5$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $12$ coins:

  * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.

You deal a total of $3 + 3 = 6$ damage, defeating the monster who has $5$ health. The total number of coins you use is $2 + 2 + 2 + 3 + 3 = 12$ coins.

In the second test case, $x = 10$, $y = 20$, $z = 40$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $190$ coins:

  * Increase damage by $5$, costing $5\cdot x$ = $50$ coins.    * Attack the monster once, dealing $5$ damage, costing $20$ coins.    * Increase damage by $2$, costing $2\cdot x$ = $20$ coins.    * Attack the monster $5$ times, dealing $5\cdot 7 = 35$ damage, costing $5\cdot y$ = $100$ coins.

You deal a total of $5 + 35 = 40$ damage, defeating the monster who has exactly $40$ health. The total number of coins you use is $50 + 20 + 20 + 100 = 190$ coins.

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

public class Main {
    static PrintWriter pw = new PrintWriter(System.out);
    static Scanner sc = new Scanner(System.in);

    public static long sumMultiplesOfK(long k, long n) {
        long m = n / k;
        return k * m * (m + 1) / 2;
    }
    static long ans;

    public static void solve() throws IOException {
        long x = sc.nextInt(), y = sc.nextInt(), z = sc.nextInt(), k = sc.nextInt();
        ans = Long.MAX_VALUE;
        if(z <= 1000){
            for(int i = 1; i <= z; i++){
                long cur = i * x;
                long rem = z - sumMultiplesOfK(k, i);
                cur += (i / k) * y;
                cur += ((rem + i - 1) / i) * y;
                ans = Math.min(ans, cur);
            }
            pw.println(ans);
            return;
        }
        long cur1 = -1, cur2 = -1;
        int idx1 = -1, idx2 = -1;
        for(int i = 1 ; i <= z ; i+= 100){
            long cur = i * x;
            long rem = z - sumMultiplesOfK(k, i);
            cur += (i / k) * y;
            cur += ((rem + i - 1) / i) * y;
            if(cur1 == -1){
                cur1 = cur;
                idx1 = i;
            }
            else if(cur2 == -1){
                cur2 = cur;
                idx2 = i;
            }
            else{
                if(cur <= cur2){
                    cur1 = cur2;
                    idx1 = idx2;
                    cur2 = cur;
                    idx2 = i;
                }
            }
        }
        for(int i = Math.max(1, idx1 - 100000); i <= Math.min(z, idx2 + 100000); i++){
            long cur = i * x;
            long rem = z - sumMultiplesOfK(k, i);
            cur += (i / k) * y;
            cur += ((rem + i - 1) / i) * y;
            ans = Math.min(ans, cur);
        }
        pw.println(ans);
    }

    public static void main(String[] args) throws IOException {
        int t = sc.nextInt();
        while(t -- > 0)
            solve();
        pw.flush();
    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(String s) throws FileNotFoundException {
            br = new BufferedReader(new FileReader(s));
        }

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

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

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

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

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec)
                        f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

        public long[] nextlongArray(int n) throws IOException {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public Long[] nextLongArray(int n) throws IOException {
            Long[] a = new Long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public int[] nextIntArray(int n) throws IOException {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public Integer[] nextIntegerArray(int n) throws IOException {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public boolean ready() throws IOException {
            return br.ready();
        }
    }
}

**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
1 1 100000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Monster
time_limit_ms: 2000
memory_limit_mb: 256
Description: Man, this Genshin boss is so hard. Good thing they have a top-up of $6$ coins for only $ \$4.99$. I should be careful and spend no more than I need to, lest my mom catches me...

You are fighting a monster with $z$ health using a weapon with $d$ damage. Initially, $d=0$. You can perform the following operations.

  * Increase $d$ — the damage of your weapon by $1$, costing $x$ coins.    * Attack the monster, dealing $d$ damage and costing $y$ coins.

You cannot perform the first operation for more than $k$ times in a row.

Find the minimum number of coins needed to defeat the monster by dealing at least $z$ damage.

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

The only line of each test case contains 4 integers $x$, $y$, $z$, and $k$ ($1\leq x, y, z, k\leq 10^8$) — the first operation's cost, the second operation's cost, the monster's health, and the limitation on the first operation.

Output: For each test case, output the minimum number of coins needed to defeat the monster.

Examples:
input:
4
2 3 5 5
10 20 40 5
1 60 100 10
60 1 100 10
output:
12
190
280
160

Note: In the first test case, $x = 2$, $y = 3$, $z = 5$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $12$ coins:

  * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.

You deal a total of $3 + 3 = 6$ damage, defeating the monster who has $5$ health. The total number of coins you use is $2 + 2 + 2 + 3 + 3 = 12$ coins.

In the second test case, $x = 10$, $y = 20$, $z = 40$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $190$ coins:

  * Increase damage by $5$, costing $5\cdot x$ = $50$ coins.    * Attack the monster once, dealing $5$ damage, costing $20$ coins.    * Increase damage by $2$, costing $2\cdot x$ = $20$ coins.    * Attack the monster $5$ times, dealing $5\cdot 7 = 35$ damage, costing $5\cdot y$ = $100$ coins.

You deal a total of $5 + 35 = 40$ damage, defeating the monster who has exactly $40$ health. The total number of coins you use is $50 + 20 + 20 + 100 = 190$ coins.

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

using namespace std;

typedef long long ll;

#define pb push_back
#define sz(x) (int)x.size()
#define all(x) begin(x),end(x)
#define lb(x,y) lower_bound(all(x),y)-begin(x)

mt19937 rng;

__int128 x, y, z, k;

__int128 f(__int128 g, __int128 r) {
    return (z - g * (g + 1) / 2 * k + g * k + r - 1) / (g * k + r) * y + g * (k * x + y) + r * x;
}

void print(__int128 n) {
    if (n < 0) { cout << "-"; print(-n); return; }
    if (n > 9) print(n / 10);
    cout << (int)(n % 10);
}

__int128 part(int g) {
    __int128 top = z - g * (g + 1) / 2 * k;
    __int128 res = (__int128)LLONG_MAX * LLONG_MAX;
    if (top <= 0) return res;
    for (int i = (g == 0 ? 1 : 0); (top + g * k + i - 1) / (g * k + i) >= (g * k + i) && i <= k; i++) {
        res = min(res, f(g, i));
    }
    for (int x = 1; x * x <= top; x++) {
        __int128 r = (top + x - 1) / x - g * k;
        if ((g == 0 ? 1 : 0) <= r && r <= k) res = min(res, f(g, r));
    }
    return res;
}

void solve() {
    ll tX, tY, tZ, tK; cin >> tX >> tY >> tZ >> tK;
    x = tX; y = tY; z = tZ; k = tK;
    __int128 mxG = 0;
    while ((mxG) * (mxG + 1) / 2 * k <= z) mxG++;
    int l = 0, r = mxG;
    while (r - l >= 3) {
        int m1 = l + (r - l) / 3, m2 = r - (r - l) / 3;
        if (part(m1) < part(m2)) r = m2;
        else l = m1;
    }
    __int128 res = (__int128)LLONG_MAX * LLONG_MAX;
    for (int g = max(0, l - 5); g <= min((int)mxG, r + 5); g++) {
        res = min(res, part(g));
    }
    print(res);
    cout << "\n";
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    rng = mt19937(chrono::steady_clock::now().time_since_epoch().count());

    int T; cin >> T;
    for (int t = 0; t < T; t++) {
        solve();
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
1 100000000 100000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Monster
time_limit_ms: 2000
memory_limit_mb: 256
Description: Man, this Genshin boss is so hard. Good thing they have a top-up of $6$ coins for only $ \$4.99$. I should be careful and spend no more than I need to, lest my mom catches me...

You are fighting a monster with $z$ health using a weapon with $d$ damage. Initially, $d=0$. You can perform the following operations.

  * Increase $d$ — the damage of your weapon by $1$, costing $x$ coins.    * Attack the monster, dealing $d$ damage and costing $y$ coins.

You cannot perform the first operation for more than $k$ times in a row.

Find the minimum number of coins needed to defeat the monster by dealing at least $z$ damage.

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

The only line of each test case contains 4 integers $x$, $y$, $z$, and $k$ ($1\leq x, y, z, k\leq 10^8$) — the first operation's cost, the second operation's cost, the monster's health, and the limitation on the first operation.

Output: For each test case, output the minimum number of coins needed to defeat the monster.

Examples:
input:
4
2 3 5 5
10 20 40 5
1 60 100 10
60 1 100 10
output:
12
190
280
160

Note: In the first test case, $x = 2$, $y = 3$, $z = 5$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $12$ coins:

  * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Increase damage by $1$, costing $2$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.    * Attack the monster, dealing $3$ damage, costing $3$ coins.

You deal a total of $3 + 3 = 6$ damage, defeating the monster who has $5$ health. The total number of coins you use is $2 + 2 + 2 + 3 + 3 = 12$ coins.

In the second test case, $x = 10$, $y = 20$, $z = 40$, and $k = 5$. Here's a strategy that achieves the lowest possible cost of $190$ coins:

  * Increase damage by $5$, costing $5\cdot x$ = $50$ coins.    * Attack the monster once, dealing $5$ damage, costing $20$ coins.    * Increase damage by $2$, costing $2\cdot x$ = $20$ coins.    * Attack the monster $5$ times, dealing $5\cdot 7 = 35$ damage, costing $5\cdot y$ = $100$ coins.

You deal a total of $5 + 35 = 40$ damage, defeating the monster who has exactly $40$ health. The total number of coins you use is $50 + 20 + 20 + 100 = 190$ coins.

Buggy Code:
#include <bits/stdc++.h>
#define forr(i, a, b) for (int i = (a); i < (b); i++)
#define forn(i, n) forr(i, 0, n)
#define dforn(i, n) for (int i = (n) - 1; i >= 0; i--)
#define forall(it, v) for (auto it = v.begin(); it != v.end(); it++)
#define sz(c) ((int)c.size())
#define rsz resize
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define fst first
#define snd second

#ifdef ANARAP
// local
#else
// judge
#endif

using namespace std;

typedef long long ll;
typedef pair<int, int> ii;

const int MAGIC = 1e5;

ll x,y,z,k;

ll f(ll t0) {
	ll baseops = t0/k;
    ll basedam = min(__int128(z), baseops * (baseops+1) / 2 * __int128(k));
    return baseops*y + t0*x + (z-basedam+t0-1)/t0 * y;
}

int main() {
// agregar g++ -DANARAP en compilacion
#ifdef ANARAP
  freopen("input.in", "r", stdin);
  // freopen("output.out","w", stdout);
#endif
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  forn(T,t) {
    cin >> x >> y >> z >> k;
    ll ans = z*y + x;
    int L = 1, R = int(z);
    while(R - L > MAGIC) {
		int M1 = L + (R-L)/3;
		int M2 = L + 2*(R-L)/3;
		ll f1 = f(M1), f2 = f(M2);
		if(f1 > f2) L = M1+1;
		else R = M2;
	}
	forr(i,L,R+1) ans = min(ans, f(i));
	cout << ans << '\n';
  }
  return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
1 1 100000000 100000000
```