Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

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

Sample Output 1

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

In the first query, the number of cans of juice in stock changes as follows: (D represents daytime and N represents night.)

9
→D 2
→N 11
→D 4
→N 13
→D 6
→N 6
→D x

In the second query, the number of cans of juice in stock changes as follows:

9
→D 2
→N 11
→D 4
→N 13
→D 6
→N 15
→D 8
→N 8
→D 1
→N 10
→D 3
→N 12
→D 5
→N 14
→D 7
→N 7
→D 0
→N 9
→D 2
→N 11
→D …

and so on, thus Snuke can buy juice indefinitely.

Sample Input 2

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

Sample Output 2

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 C (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
#include<stdlib.h>

long long gcd(long long a, long long b);

int main(void){

  int T;
  scanf("%d\n", &T);

  long long *A, *B, *C, *D;
  A = (long long*)malloc(T*sizeof(long long));
  B = (long long*)malloc(T*sizeof(long long));
  C = (long long*)malloc(T*sizeof(long long));
  D = (long long*)malloc(T*sizeof(long long));
  for(int i=0; i<T; i++){
    scanf("%lld %lld %lld %lld", &A[i], &B[i], &C[i], &D[i]);
  }

  long long R=0;
  long long *rist;
  long long a,b,c,d;
  int flag=0, g=0, jMax=0;

  for(int i=0; i<T; i++){
    a = A[i];
    b = B[i];
    c = C[i];
    d = D[i];

    if(a<b){
      printf("No\n");
      flag++;
    }

    R = a % b;
    g = gcd(b,d);
    jMax = b/g;

    for(int j=0; j<jMax && !flag; j++){
      if(R>c || R+d<b){
        printf("No\n");
        flag++;
      }else{
        R = (R+d)%b;
      }
    }

    if(!flag) printf("Yes\n");
    flag = 0;
  }

  return 0;
}


long long gcd(long long a, long long b){

  long long r, tmp;
  /* 自然数 a > b を確認・入替 */
  if(a<b){
    tmp = a;
    a = b;
    b = tmp;
  }
 
  /* ユークリッドの互除法 */
  r = a % b;
  while(r!=0){
    a = b;
    b = r;
    r = a % b;
  }

  return b;

}