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: Wrong Answer)

/*
ЗАПУСКАЕМ 
░ГУСЯ░▄▀▀▀▄░РАБОТЯГУ░░
▄███▀░◐░░░▌░░░░░░░
░░░░▌░░░░░▐░░░░░░░
░░░░▐░░░░░▐░░░░░░░
░░░░▐░░░░░▐░░░░░░░
░░░░▐░░░░░▐░░░░░░░
░░░░▐░░░░░▐░░░░░░░
░░░░▐░░░░░▐░░░░░░░
░░░░▐░░░░░▐░░░░░░░
░░░░▐░░░░░▐░░░░░░░
░░░░▐░░░░░▐░░░░░░░
░░░░▐░░░░░▐░░░░░░░
░░░░▐░░░░░▐░░░░░░░
░░░░▐░░░░░▐░░░░░░░
░░░░▐░░░░░▐░░░░░░░
░░░░▐░░░░░▐░░░░░░░
░░░░▐░░░░░▐░░░░░░░
░░░░▐░░░░░▐░░░░░░░
░░░░▌░░░░░▐▄▄░░░░░
░░░░▌░░░░▄▀▒▒▀▀▀▀▄
░░░▐░░░░▐▒▒▒▒▒▒▒▒▀▀▄
░░░▐░░░░▐▄▒▒▒▒▒▒▒▒▒▒▀▄
░░░░▀▄░░░░▀▄▒▒▒▒▒▒▒▒▒▒▀▄
░░░░░░▀▄▄▄▄▄█▄▄▄▄▄▄▄▄▄▄▄▀▄
░░░░░░░░░░░▌▌░▌▌░░░░░
░░░░░░░░░░░▌▌░▌▌░░░░░
░░░░░░░░░▄▄▌▌▄▌▌░░░░░ 
 */
#include <iostream>
#include <complex>
#include <vector>
#include <string>
#include <algorithm>
#include <cstdio>
#include <numeric>
#include <cstring>
#include <ctime>
#include <cstdlib>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <list>
#include <cmath>
#include <bitset>
#include <cassert>
#include <queue>
#include <stack>
#include <deque>
    
     
using namespace std;
template<typename T1, typename T2>inline void chkmin(T1 &x, T2 y) { if (x > y) x = y; }
template<typename T1, typename T2>inline void chkmax(T1 &x, T2 y) { if (x < y) x = y; } 
template<typename T, typename U> inline ostream &operator<< (ostream &_out, const pair<T, U> &_p) { _out << _p.first << ' ' << _p.second; return _out; }
template<typename T, typename U> inline istream &operator>> (istream &_in, pair<T, U> &_p) { _in >> _p.first >> _p.second; return _in; }
template<typename T> inline ostream &operator<< (ostream &_out, const vector<T> &_v) { if (_v.empty()) { return _out; } _out << _v.front(); for (auto _it = ++_v.begin(); _it != _v.end(); ++_it) { _out << ' ' << *_it; } return _out; }
template<typename T> inline istream &operator>> (istream &_in, vector<T> &_v) { for (auto &_i : _v) { _in >> _i; } return _in; }
template<typename T> inline ostream &operator<< (ostream &_out, const set<T> &_s) { if (_s.empty()) { return _out; } _out << *_s.begin(); for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) { _out << ' ' << *_it; } return _out; }
template<typename T> inline ostream &operator<< (ostream &_out, const multiset<T> &_s) { if (_s.empty()) { return _out; } _out << *_s.begin(); for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) { _out << ' ' << *_it; } return _out; }
template<typename T> inline ostream &operator<< (ostream &_out, const unordered_set<T> &_s) { if (_s.empty()) { return _out; } _out << *_s.begin(); for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) { _out << ' ' << *_it; } return _out; }
template<typename T> inline ostream &operator<< (ostream &_out, const unordered_multiset<T> &_s) { if (_s.empty()) { return _out; } _out << *_s.begin(); for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) { _out << ' ' << *_it; } return _out; }
template<typename T, typename U> inline ostream &operator<< (ostream &_out, const map<T, U> &_m) { if (_m.empty()) { return _out; } _out << '(' << _m.begin()->first << ": " << _m.begin()->second << ')'; for (auto _it = ++_m.begin(); _it != _m.end(); ++_it) { _out << ", (" << _it->first << ": " << _it->second << ')'; } return _out; }
template<typename T, typename U> inline ostream &operator<< (ostream &_out, const unordered_map<T, U> &_m) { if (_m.empty()) { return _out; } _out << '(' << _m.begin()->first << ": " << _m.begin()->second << ')'; for (auto _it = ++_m.begin(); _it != _m.end(); ++_it) { _out << ", (" << _it->first << ": " << _it->second << ')'; } return _out; }
#define sz(c) (int)(c).size()
#define all(c) (c).begin(), (c).end()
#define rall(c) (c).rbegin(), (c).rend()
#define left left228
#define right right228
#define next next228
#define rank rank228
#define prev prev228
#define y1 y1228                                                         
#define read(FILENAME) freopen((FILENAME + ".in").c_str(), "r", stdin)
#define write(FILENAME) freopen((FILENAME + ".out").c_str(), "w", stdout)
#define files(FILENAME) read(FILENAME), write(FILENAME)
#define pb push_back
const string FILENAME = "input";



long long gcd(long long a, long long b) {
	if (a == 0) {
		return b;
	}
	return gcd(b % a, a);
}


int t;
long long a, b, c, d;



bool can(long long st, long long jump, long long l, long long r, long long mod) {
	st %= mod;
	l -= st;
	r -= st;
	l %= mod;
	r %= mod;
	vector<pair<long long, long long> > add;
	if (l < 0 && r < 0) {	
		l += mod;
		r += mod;
		add.pb({l, r});
	} else {
		if (l < 0) {
			l += mod;
			add.pb({l, mod - 1});
			add.pb({0, r});
		} else {
			add.pb({l, r});
		}
	}
	jump %= mod;
	if (jump == 0) {
		for (auto x: add) {
			if (x.first <= 0 && 0 <= x.second) {
				return true;
			}
		}
		return false;
	}
	long long f = gcd(jump, mod);
	for (auto x: add) {
		x.first %= f;
		x.second %= f;
		if (x.first > x.second || x.first == 0 || x.second == 0) {
			return true;
		}
	}
	return false;
}


int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	//read(FILENAME);
	cin >> t;
	for (int i = 0; i < t; i++) {
		cin >> a >> b >> c >> d;
		if (a < b) {
			cout << "No\n";
			continue;
		}
		if (b > d) {
			cout << "No\n";
			continue;
		}
		long long t = (a - c - 1) / b;
		a -= (t + 1) * b;
		if (a + d < b) {
			cout << "No\n";
			continue;
		}
		if (c + 1 < b) {
			if (can(a, d, c + 1, b - 1, b)) {
				cout << "No\n";
				continue;
			}
		}

			//cout << "opa" << endl;
		if (b - d > 0 && b - d <= c) {
			if (can(a, d, 0, b - d - 1, b)) {
				cout << "No\n";
				continue;
			}
		}
		cout << "Yes\n";
		//a % b < b - d
		//
		//x < b - d
		//x += b


	}
}