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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)



import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Scanner;
import java.util.List;

public class Main {
    public static void main(String[] args){
        PrintWriter out = new PrintWriter(System.out);
        Scanner sc = new Scanner(System.in);
        Task task = new Task();
        task.solve(sc, out);
        out.flush();
        sc.close();
    }
    static class Task{
        public void solve(Scanner sc, PrintWriter out){
            int T = nint(sc);
            List<long[]> inputLine = new ArrayList<>();
            for (int i= 0 ; i < T; i++){
                long[] line = longLine(sc, 4);
                inputLine.add(line);
            }
            for (int i=0; i<T; i++){
                long A = inputLine.get(i)[0];
                long B = inputLine.get(i)[1];
                long C = inputLine.get(i)[2];
                long D = inputLine.get(i)[3];

                if (A < B){
                    // 初日アウト
                    out.println("No");
                    continue;
                }
                if(B > D){
                    // 必ずいつか尽きる
                    out.println("No");
                    continue;
                }

                if (C >= B ){
                    // 危ない状況にならない
                    out.println("Yes");
                    continue;
                }

                long remain = A % B;
                long div = D / B;
                long margin = D - B;
                long result = remain;
                if (div >=1){
                    if (margin !=0){
                        for (long j = 1; result <C+1; j++){
                            result +=margin;
                        }
                    }else{
                        if(remain <= C){
                            out.println("Yes");
                            continue;
                        }else {
                            out.println("No");
                            continue;
                        }
                    }
                    if (result >= B){
                        out.println("Yes");
                    }else {

                        out.println("No");
                    }
                }else {
                    out.println("Yes");
                }
            }
        }
    }

    static int nint(Scanner sc){
        return Integer.parseInt(sc.next());
    }
    static long nlong(Scanner sc){
        return Long.parseLong(sc.next());
    }
    static double ndouble(Scanner sc){
        return Double.parseDouble(sc.next());
    }
    static float nfloat(Scanner sc){
        return Float.parseFloat(sc.next());
    }
    static String nstr(Scanner sc){
        return  String.valueOf(sc.next());
    }
    static long[] longLine(Scanner sc, int size){
        long[] lLine = new long[size];
        for (int i = 0; i < size; i++) {
            lLine[i] = nlong(sc);
        }
        return lLine;
    }
    static int[] intLine(Scanner sc, int size){
        int[] iLine = new int[size];
        for (int i = 0; i < size; i++) {
            iLine[i] = nint(sc);
        }
        return iLine;
    }
    static String[] strLine(Scanner sc, int size){
        String[] strLine = new String[size];
        for (int i = 0; i < size; i++) {
            strLine[i] = nstr(sc);
        }
        return strLine;
    }
}