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 (GCC 5.4.1) (judge verdict: Wrong Answer)

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

int main(){
  int t,i,j;
  scanf("%d",&t);
  long int a[t],b[t],c[t],d[t];
  int s[t];
  
  for(i=0;i<t;i++){
    scanf("%ld %ld %ld %ld",&a[i],&b[i],&c[i],&d[i]);
    s[i]=0;
  }
  
  for(i=0;i<t;i++){
    
    if(d[i]>b[i] && a[i]%b[i] !=0){
      
      if( (b[i]-a[i]%b[i])%(d[i]-b[i])!=0  )
        j=(b[i]-a[i]%b[i])/(d[i]-b[i]);
      
      if((b[i]-a[i]%b[i])/(d[i]-b[i])==0)
        j=(b[i]-a[i]%b[i])/(d[i]-b[i])-1;
      
      if(a[i]%b[i]+j*(d[i]-b[i])>c[i])
        s[i]=1;
    }
  }
        
  
  
  for(i=0;i<t;i++){
    if(a[i]<b[i])
      printf("No\n");
    else if(d[i]<b[i])
      printf("No\n");
    else if(s[i]==1)
      printf("No\n");
    
    else if(b[i]==d[i]){
      if(a[i]%b[i]<c[i])
        printf("Yes\n");
      else printf("No\n");
    }
    
    else if(d[i]-b[i]>=b[i]-c[i])
      printf("Yes\n");
    
    else printf("No\n");
  }
  
}