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 <iostream>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <utility>
#include <algorithm>
#include <cstdio>
#include <iomanip>
#include <queue>
#include <deque>
#include <stack>
#include <fstream>
#include <cmath>
#include <random>
#include <complex>
#include <functional>

#define ll int64_t
#define Rep(i, n) for (ll i = 0; i < n; i++)
using namespace std;
typedef vector<ll> vec;
typedef vector<vec> mat;

const ll inf = 1LL << 60;

template<class T> inline void chmin(T& a, T b) {
    if (a > b) {
        a = b;
    }
}
template<class T> inline void chmax(T& a, T b) {
    if (a < b) {
        a = b;
    }
}

ll GCD (ll a, ll b) { return b ? GCD(b, a%b) : a;}

ll LCM (ll a, ll b) {
    ll g = GCD (a, b);
    return a * b / g;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll T;
    cin >> T;
    while (T--) {
        ll A, B, C, D;
        cin >> A >> B >> C >> D;
        bool ans;
        if (A < B) ans = false;
        else if (B > D) ans = false;
        else if (C >= B) ans = true;
        else if (B - GCD(B, D) + A % GCD(B, D) > C) ans = false;
        else ans = true;

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