For this Question: Shreya has two integers A and B (A &le; B). Shreya can choose any non- negative integer X and add them to both A and B. Find whether it is possible to make A a divisor of B.The first line of input will contain a single integer T, denoting the number of test cases. Each test case consists of two integers A and B.

<b>Constraints</b>
1 &le; T &le; 10<sup>5</sup>
1 &le; A &le; B &le; 10<sup>9</sup>For each test case, output YES if it is possible to make A a factor of B, NO otherwise.Sample Input :
 3
3 6
4 14
9 10

Sample Output :
YES
YES
NO

Explanation :
<ul><li>We can choose X = 0 and add them to 3 and 6. Thus, 3 is a factor of 6. </li><li>We can choose X = 1 and add them to 4 and 14. Thus, 4 + 1 = 5 is a factor of 14 + 1 = 15. </li><li>There is no possible value of X to add such that A becomes a factor of B. </li>, 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(long long l = LLONG_MIN, long long 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;
    cin >> tt;
    while (tt--) {
        int x, y;
        cin >> x >> y;

        if (y % x == 0)
            cout << "YES\n";
        else if ((y - x) >= x)
            cout << "YES\n";
        else
            cout << "NO\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\n";

    return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?