Problem: 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.

Examples

Input

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


Output

No
Yes
No
Yes
Yes
No
No
Yes
No
Yes
Yes
No
No
Yes


Input

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


Output

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 JAVA):

Solution:
import java.io.PrintWriter;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int T = sc.nextInt();
        for (int i = 0; i < T; i++) {
            new Main().solve(sc, pw);
        }
        pw.flush();
        pw.close();
        sc.close();
    }

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

    void solve(Scanner sc, PrintWriter pw) {
        long A = sc.nextLong();
        long B = sc.nextLong();
        long C = sc.nextLong();
        long D = sc.nextLong();
        pw.write(solve(A, B, C, D) ? "Yes" : "No");
        pw.write("\n");
    }

    boolean solve(long A, long B, long C, long D) {
        if (A < B) {
            return false;
        }
        long g = gcd(B, D);
        long remainderA = A % B;
        if (remainderA > C) {
            return false;
        }
        long delta = remainderA % g;
        if (g > 1) {
             if (delta + D < B) {
                 return false;
             }
        } else {
            if (D < B) {
                return false;
            }
        }
        long cPlusOne = C - delta;
        long bMinusOne = B - delta;
//        return bMinusOne / g <= cPlusOne / g;
        long candidate =  (cPlusOne + g - 1) / g;
        long candidate2 =  (cPlusOne) / g + 1;
        return !(valide(candidate, g, cPlusOne, bMinusOne) || valide(candidate2, g, cPlusOne, bMinusOne));
    }

    private boolean valide(long candidate2, long g, long cPlusOne, long bMinusOne) {
        long value = candidate2 * g;
        return value > cPlusOne && value < bMinusOne;
    }
}