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)

#include <iostream>
#include <cstdio>
#include<list>
#include<iomanip>
#include<cmath>
#include <random>
#include<queue>
#include <functional>
#include<stdio.h>
#include<assert.h>
#include<stack>
#include<sstream>
#include <cstdlib>
#include<map>
#include<algorithm>
#include<iostream>
#include<set>
#include<utility>
#include<memory.h>
#include<string>
#include<vector>
#include <unordered_map>
#include<numeric>
using namespace std;
#define ios std::ios_base::sync_with_stdio(false);
#define ll  long long
#define pb push_back
#define fi(ss) freopen (ss,"r",stdin)
#define fo(ss) freopen (ss,"w",stdout)
#define sz(v)               ((int)((v).size()))
#define all(x)          (x).begin(),(x).end()
#define REP(i, v)       for(int i=0;i<sz(v);i++)
#define ld long double
#define lp(i,n) for(int i = 0 ; i < n ; ++i)
typedef pair<double,double> pii;
typedef pair<pii,int> tri;
using namespace std;

int main() {
	ios
	int t;
	cin >> t;
	ll a,b,c,d;
	while (t--) {
		cin >> a >> b >> c >> d;
		int ops = 0;
		set<ll> seen;
		bool bad = 0;
		ll old_e = 1ll<<62, old_o = 1ll<<62;
		while (a > 0) {
			seen.insert(a);
			ll cur = a;
			if (a < b) break;
			ll r = max(0LL,(a - c)) / b;

			if (max(0LL,(a - c)) % b != 0) {
				++r;
			}
			if (r == 0) ++r;
			a -= 1ll*r*b;

			if (a < 0) break;
			a += 1ll*d;

			if (a % b) {
				if (a >= old_o) {
					bad = 1; break;
				}
				old_o = a;
			} else {
				if (a >= old_e) {
					bad = 1; break;

				}
				old_e = a;
			}
			//seen.insert(a);
		//	++ops;

		}
		if (!bad) cout << "No\n";
		else cout << "Yes\n";

	}

}