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:
/**
 * @author Finn Lidbetter
 */
import java.util.*;
import java.io.*;
import java.awt.geom.*;

public class Main {
  public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringBuilder sb = new StringBuilder();
    
    int nTests = Integer.parseInt(br.readLine());
    for (int t=0; t<nTests; t++) {
      String[] s = br.readLine().split(" ");
      long a = Long.parseLong(s[0]);
      long b = Long.parseLong(s[1]);
      long c = Long.parseLong(s[2]);
      long d = Long.parseLong(s[3]);
      if (a<b) {
        sb.append("No\n");
      } else {
        if (d<b) {
          sb.append("No\n");
        } else {
          d = d%b;
          if (d==0) {
            if (a%b>c) {
              sb.append("No\n");
            } else {
              sb.append("Yes\n");
            }
          } else {
            if (c+1>b-1) {
              sb.append("Yes\n");
            } else {
              d += b;
              long div = gcd(b,d);
              long badLo = (c+1);
              long badHi = b-1;
              if (badHi-badLo+1>=div) {
                sb.append("No\n");
              } else {
                long mBadLo = badLo%div;
                long mBadHi = badHi%div;
                if (mBadLo>mBadHi) {
                  mBadHi += div;
                }
                long mA = a%div;
                if (mA>=mBadLo && mA<=mBadHi) {
                  sb.append("No\n");
                } else {
                  sb.append("Yes\n");
                }
              }
            }
          }
        }
      }
    }
    System.out.print(sb);
  }
  static long gcd(long a, long b) {
    return (b==0 ? a : gcd(b, a%b));
  }
}