Problem:
Score : 600 points

Problem Statement

Ringo Mart, a convenience store, sells apple juice.

On the opening day of Ringo Mart, there were A cans of juice in stock in the morning.
Snuke buys B cans of juice here every day in the daytime.
Then, the manager checks the number of cans of juice remaining in stock every night.
If there are C or less cans, D new cans will be added to the stock by the next morning.

Determine if Snuke can buy juice indefinitely, that is, there is always B or more cans of juice in stock when he attempts to buy them.
Nobody besides Snuke buy juice at this store.

Note that each test case in this problem consists of T queries.

Constraints

1 \leq T \leq 300

1 \leq A, B, C, D \leq 10^{18}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

T
A_1 B_1 C_1 D_1
A_2 B_2 C_2 D_2
:
A_T B_T C_T D_T

In the i-th query, A = A_i, B = B_i, C = C_i, D = D_i.

Output

Print T lines. The i-th line should contain Yes if Snuke can buy apple juice indefinitely in the i-th query; No otherwise.

Sample Input 1

14
9 7 5 9
9 7 6 9
14 10 7 12
14 10 8 12
14 10 9 12
14 10 7 11
14 10 8 11
14 10 9 11
9 10 5 10
10 10 5 10
11 10 5 10
16 10 5 10
1000000000000000000 17 14 999999999999999985
1000000000000000000 17 15 999999999999999985

Sample Output 1

No
Yes
No
Yes
Yes
No
No
Yes
No
Yes
Yes
No
No
Yes

In the first query, the number of cans of juice in stock changes as follows: (D represents daytime and N represents night.)

9
→D 2
→N 11
→D 4
→N 13
→D 6
→N 6
→D x

In the second query, the number of cans of juice in stock changes as follows:

9
→D 2
→N 11
→D 4
→N 13
→D 6
→N 15
→D 8
→N 8
→D 1
→N 10
→D 3
→N 12
→D 5
→N 14
→D 7
→N 7
→D 0
→N 9
→D 2
→N 11
→D …

and so on, thus Snuke can buy juice indefinitely.

Sample Input 2

24
1 2 3 4
1 2 4 3
1 3 2 4
1 3 4 2
1 4 2 3
1 4 3 2
2 1 3 4
2 1 4 3
2 3 1 4
2 3 4 1
2 4 1 3
2 4 3 1
3 1 2 4
3 1 4 2
3 2 1 4
3 2 4 1
3 4 1 2
3 4 2 1
4 1 2 3
4 1 3 2
4 2 1 3
4 2 3 1
4 3 1 2
4 3 2 1

Sample Output 2

No
No
No
No
No
No
Yes
Yes
No
No
No
No
Yes
Yes
Yes
No
No
No
Yes
Yes
Yes
No
No
No

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>


using namespace std;
#define rep(i,n) for (int i = 0; i < (int)(n); ++i)
#define reps(i,n) for (int i = 1; i <= (int)(n); ++i)
#define repd(i,n) for (int i = (int)(n - 1); i >= 0; --i)
#define repds(i,n) for (int i = (int)(n); i > 0; --i)
#define loop(i,x,n) for (int i = (int)(x); i < (n); ++i)
#define loops(i,x,n) for (int i = (int)(x); i <= (n); ++i)
#define loopd(i,x,n) for (int i = (int)(x); i > (n); --i)
#define loopds(i,x,n) for (int i = (int)(x); i >= (n); --i)
#define itrep(i,s) for (auto i = begin(s); i != end(s); ++i)
#define itrepd(i,s) for (auto i = rbegin(s); i != rend(s); ++i)
#define all(f,x,...) [&](decltype(x)& whole) { return (f)(begin(whole), end(whole), ##__VA_ARGS__); }(x)
#define rall(f,x,...) [&](decltype(x)& whole) { return (f)(rbegin(whole), rend(whole), ##__VA_ARGS__); }(x)
using ll = long long;
using ld = long double;
constexpr ll inf = static_cast<ll>(1e18);
constexpr int iinf = static_cast<int>(1e9);
constexpr double dinf = 1e10;
constexpr ld ldinf = 1e18;
ll mod = static_cast<ll>(1e9 + 7);
std::ostream& endn(std::ostream& os) { return os.put(os.widen('\n')); }
template<class T> constexpr int sz(const T& a) { return (int)a.ne(); }
template<class T> constexpr T pow2(const T a) { return a * a; }
template<class T, class ... Args> constexpr T mins(T& a, Args... args) { return a = min<T>({ a, args... }); }
template<class T, class... Args> constexpr T maxs(T& a, Args... args) { return a = max<T>({ a,args... }); }
template<class A, int N, class T>
void fillArray(A(&array)[N], const T& val) { std::fill((T*)array, (T*)(array + N), val); }

template<class T>
constexpr T gcd(T a, T b) {
    if (a < b) gcd(b, a);
    T r = 0;
    while ((r = a % b)) { a = b; b = r; }
    return b;
}
template<class T>
constexpr T lcm(T a, T b) {
    T g = gcd<T>(a, b);
    return a / g * b;
}


int n;
bool ans[305];
void solve() {
    cin >> n;
    rep(i, n) {
        ll a, b, c, d;
        cin >> a >> b >> c >> d;

        if (a<b || b>d) {
            ans[i] = false;
        }
        else if (b <= c) {
            ans[i] = true;
        }
        else {
            auto g = gcd(b, d);
            if (c < a % g + b - g) ans[i] = false;
            else ans[i] = true;
        }
    }

    rep(i, n) cout << (ans[i] ? "Yes" : "No") << endn;
}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    solve();
    cout << flush;
    return 0;
}