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.math.BigDecimal;
import java.math.MathContext;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        long[] as = new long[t];
        long[] bs = new long[t];
        long[] cs = new long[t];
        long[] ds = new long[t];
        for (int i = 0;i < t;++i){
            as[i] = scanner.nextLong();
            bs[i] = scanner.nextLong();
            cs[i] = scanner.nextLong();
            ds[i] = scanner.nextLong();
        }
        for (int i = 0;i < t;++i){
            long a = as[i];
            long b = bs[i];
            long c = cs[i];
            long d = ds[i];
            if(d < b||a < b){
                System.out.println("No");
                continue;
            }
            if(c >= b){
                System.out.println("Yes");
                continue;
            }
            long k1 = a - ((a - (c + 1)) / b + 1) * b;
            if(k1 < 0){
                System.out.println("No");
                continue;
            }
            long k2 = g(b,d);
            long k3 = k1 % k2 - k2;
            if(k3 + b > c){
                System.out.println("No");
            }
            else System.out.println("Yes");
        }
    }

    public static long g(long a,long b) {
        while (a != b){
            if(a > b){
                long k = (a % b);
                a = k == 0 ? b : k;
            }
            else{
                long k = (b % a);
                b = k == 0 ? a : k;
            }
        }
        return a;
    }
}