For this Question: An electronics shop sells red and blue lamps. A red lamp costs X rupees and a blue lamp costs Y rupees. John is going to buy exactly N lamps from this shop. Find the minimum amount of money John needs to pay such that at least K of the lamps bought is red.The first line of input will contain a single integer T, denoting the number of test cases. Each test case consists of a single line containing four space- separated integers N, K, X, Y.

<b>Constraints</b>
1 &le; T &le; 10<sup>3</sup>
1 &le; N &le; 10<sup>8</sup>
0 &le; K &le; N
1 &le; X, Y &le; 10For each test case, output on a new line the minimum amount of money John needs to pay in order to buy N lamps such that at least K of the lamps bought are red.Sample Input :
4
2 2 5 1
4 1 3 1
3 0 4 7
5 2 3 4
Sample Output :
10
6
12
15
Explanation :
<ul><li>John buys 2 red lamps with 2 ⋅ 5 = 10 rupees. </li><li>John buys 1 red lamp and 3 blue lamps with 1. 3 + 3 . 1 = 6 rupees. </li><li>John buys 3 red lamps with 3 ⋅ 4 = 12 rupees., I have written this Solution Code: #include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

template <class key, class value, class cmp = std::less<key>>
using ordered_map = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;
// find_by_order(k)  returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;

#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif

#define int long long

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

inline int64_t random_long(int l = LLONG_MIN, int r = LLONG_MAX) {
    uniform_int_distribution<int64_t> generator(l, r);
    return generator(rng);
}

int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    auto start = std::chrono::high_resolution_clock::now();

    int tt = 1;
    cin >> tt;
    while (tt--) {
        int N, K, X, Y;
        cin >> N >> K >> X >> Y;
        int l = N - K;
        int sum = K * X;
        int sum1 = sum + (l * Y);
        int sum2 = sum + (l * X);
        if (sum1 < sum2) {
            cout << sum1;
        } else {
            cout << sum2;
        }
        cout << endl;
    }

    auto stop = std::chrono::high_resolution_clock::now();
    auto duration = std::chrono::duration_cast<std::chrono::nanoseconds>(stop - start);
    cerr << "Time taken : " << ((long double)duration.count()) / ((long double)1e9) << "s " << endl;
    return 0;
};
, In this Programming Language: C++, Now tell me if this Code is compilable or not?