For this Question: There are n rectangles of the same size: w in width and h in length. It is required to find a square of the smallest size into which these rectangles can be packed. Rectangles cannot be rotated.The first line of the input contains three space-separated integers w, h and n.

<b>Constraints</b>
1 ≤ w, h, n ≤ 10<sup>9</sup>Output the minimum length of a side of a square, into which the given rectangles can be packed.Sample Input :
2 3 10
Sample Output:
9

<b>Explanation:</b>
9 is the minimum size of the square in which all these rectangles can fit.
<img src = "https://d3dyfaf3iutrxo.cloudfront.net/general/upload/7aef090772524a368cf0cbe754a484a9.png"></img>, 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 w, h, n;
    cin >> w >> h >> n;
    int l = 1, r = (int)1e18;
    while (l <= r) {
        int mid = l + (r - l) / 2;
        int x = mid / w;
        int y = mid / h;
        if ((y > 0 && x >= ((n - 1) / y + 1)) || (x > 0 && y >= ((n - 1) / x + 1))) {
            r = mid - 1;
        } else {
            l = mid + 1;
        }
    }
    cout << l << "\n";

    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?