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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashSet;
import java.util.Set;

public class Main {
    BufferedReader rd;

    Main() throws IOException {
        rd = new BufferedReader(new InputStreamReader(System.in));
        compute();
    }

    private void compute() throws IOException {
        int t = pint();
        for(int i=0;i<t;i++) {
            out(solve());
        }
    }

    private String solve() throws IOException {
        long[] q = longarr();
        long a = q[0];
        long b = q[1];
        long c = q[2];
        long d = q[3];

        boolean res = solve(a,b,c,d);

        return res?"Yes":"No";
    }

    private boolean solve(long a, long b, long c, long d) {
        boolean res = false;
        if(b <= a) {
            if(c >= a-b) {
                res = p(a-b,b,c,d);
            } else {
                long diff = a-c;
                long e = diff % b;
                if(e == 0) {
                    res = p(c,b,c,d);
                } else {
                    long s = c - (b-e);
                    if(s >= 0) {
                        res = p(s,b,c,d);
                    }
                }
            }
        }
        return res;
    }

    private boolean p(long s, long b, long c, long d) {
        long e = d % b;
        if(d >= b && e == 0) {
            return true;
        }
        if(d > b) {
            Set<Long> visited = new HashSet<>();
            long cur = s;
            while(cur >= 0 && visited.add(cur)) {
                if (cur + e <= c) {
                    long f = (c - cur) % e;
                    cur = c + e - f - b;
                } else {
                    cur = cur + e - b;
                }
            }
            return cur >= 0;
        }
        return false;
    }

    private int pint() throws IOException {
        return pint(rd.readLine());
    }

    private int pint(String s) {
        return Integer.parseInt(s);
    }

    private long[] longarr() throws IOException {
        return longarr(rd.readLine());
    }

    private long[] longarr(String s) {
        String[] q = split(s);
        int n = q.length;
        long[] a = new long[n];
        for(int i=0;i<n;i++) {
            a[i] = Long.parseLong(q[i]);
        }
        return a;
    }

    private String[] split(String s) {
        if(s == null) {
            return new String[0];
        }
        int n = s.length();
        int start = -1;
        int end = 0;
        int sp = 0;
        boolean lastWhitespace = true;
        for(int i=0;i<n;i++) {
            char c = s.charAt(i);
            if(isWhitespace(c)) {
                lastWhitespace = true;
            } else {
                if(lastWhitespace) {
                    sp++;
                }
                if(start == -1) {
                    start = i;
                }
                end = i;
                lastWhitespace = false;
            }
        }
        if(start == -1) {
            return new String[0];
        }
        String[] res = new String[sp];
        int last = start;
        int x = 0;
        lastWhitespace = true;
        for(int i=start;i<=end;i++) {
            char c = s.charAt(i);
            boolean w = isWhitespace(c);
            if(w && !lastWhitespace) {
                res[x++] = s.substring(last,i);
            } else if(!w && lastWhitespace) {
                last = i;
            }
            lastWhitespace = w;
        }
        res[x] = s.substring(last,end+1);
        return res;
    }

    private boolean isWhitespace(char c) {
        return c==' ' || c=='\t';
    }

    private static void out(Object x) {
        System.out.println(x);
    }

    public static void main(String[] args) throws IOException {
        new Main();
    }
}