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;

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define REP(i, n) for(int i=0; i<(n); ++i)
#define FOR(i, a, b) for(int i=(a); i<(b); ++i)
#define FORR(i, a, b) for(int i=(b)-1; i>=(a); --i)

#define DEBUG(x) cout<<#x<<": "<<x<<'\n'
#define DEBUG_VEC(v) cout<<#v<<":";REP(i, v.size())cout<<' '<<v[i];cout<<'\n'
#define ALL(a) (a).begin(), (a).end()

#define CHMIN(a, b) a=min((a), (b))
#define CHMAX(a, b) a=max((a), (b))

const ll MOD=1000000007ll;
// const ll MOD=998244353ll;
#define FIX(a) ((a)%MOD+MOD)%MOD

const double EPS=1e-11;
#define EQ0(x) (abs((x))<EPS)
#define EQ(a, b) (abs((a)-(b))<EPS)

ll gcd(ll a, ll b){
	return b==0 ? a : gcd(b, a%b);
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	int t;
	ll a, b, c, d;
	cin>>t;
	bool can;
	REP(i, t){
		can=false;
		cin>>a>>b>>c>>d;
		if(a>=b && b<=d){
			if(c>=b-1){
				can=true;
			}
			else{
				ll g=gcd(b, d);
				if(c>=b-g+a%g){
					can=true;
				}
			}
		}
		cout<<(can ? "Yes" : "No")<<'\n';
	}
	return 0;
}