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.util.*;

public class Main {
    public static long gcd(long x, long y) {
        if (y == 0) {
            return x;
        } else {
            return gcd(y, x - (x / y) * y);
        }
    }

    public static long quotient(long n, long k) {
        return n >= 0 ? n / k : n / k - 1;
    }

    public static void main(String[] args) {
        // read inputs
        Scanner in = new Scanner(System.in);
        final int N = in.nextInt();
        long[] as = new long[N], bs = new long[N], cs = new long[N], ds = new long[N];
        for (int i = 0; i < N; i++) {
            as[i] = in.nextLong();
            bs[i] = in.nextLong();
            cs[i] = in.nextLong();
            ds[i] = in.nextLong();
        }

        // solve
        for (int i = 0; i < N; i++) {
            final long a = as[i], b = bs[i], c = cs[i], d = ds[i];
            if (b > d) {
                System.out.printf("No\n");
            } else {
                if (Math.min(a, c) >= b) {
                    System.out.printf("Yes\n");
                } else if (a < b) {
                    System.out.printf("No\n");
                } else { // a >= b && c < b
                    final long gcdBD = gcd(b, d);
                    final long a0 = a % b;
                    final long num = a0 + (quotient(c - a0, gcdBD) + 1) * gcdBD;
                    if (num < b) {
                        System.out.printf("No\n");
                    } else {
                        System.out.printf("Yes\n");
                    }
                }
            }
        }
    }
}