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 void main(String args[]) {
    Scanner sc = new Scanner(System.in);

    int n = sc.nextInt();
    long[] a = new long[n];
    long[] b = new long[n];
    long[] c = new long[n];
    long[] d = new long[n];

    for (int i = 0; i < n; i++) {
      a[i] = sc.nextLong();
      b[i] = sc.nextLong();
      c[i] = sc.nextLong();
      d[i] = sc.nextLong();
    }

    for (int i=0; i < n; i++) {
      int ans = 1; // true

      if (a[i] < b[i]) {
        ans = -1; // false
      } else if (d[i] < b[i]) {
        ans = -1; // false
      } else if ((a[i]%b[i]) > c[i]) {
        ans = -1; // false
      } else if (b[i] > c[i] + (long)1) {
        long tmp = d[i] % b[i];
        long init = a[i] % b[i];
        long diff = b[i] - c[i];
        if (diff > tmp && tmp != (long)0) {
          ans = -1; // false
        } else if (tmp != (long)0) {
          if (gcd(b[i], tmp) < diff) {
            ans = -1; // false
          }
        }
      }

      if (ans == 1) {
        System.out.println("Yes");
      } else {
        System.out.println("No");
      }
    }
  }

  private static long gcd(long m, long n) {
    if(m < n) return gcd(n, m);
    if(n == 0) return m;
    return gcd(n, m % n);
  }
}