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: Time Limit Exceeded)

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

int compare_long(const void *p, const void *q){
    return *(long*)p - *(long*)q;
}

int main(){
    int  i,N;
    int  flg;
    int  count=0;
    long A,B,C,D;
    long AB[1000000] = {};
    long noon,night;
    long *ptr;
    
    scanf("%d", &N);
    for(i=0;i<N;i++){
        scanf("%ld %ld %ld %ld", &A,&B,&C,&D);
        night = A;
        count = 0;
        memset(AB,0,sizeof(AB));
        
        // 同じA-Bか出現するかA-Bがマイナスになるまでループ
        while(1){
            noon = night-B;
            // マイナスになったらNo終了
            if(noon < 0) {
               flg = 1;
               break;
            }
            
            // 過去出たリストに同値があるか走査
            ptr = (long*)bsearch(&noon,AB,count,sizeof(long),compare_long);
            // 過去に出てない？
            if( ptr == NULL ){
                // 過去出たリストに追加してソート
                AB[count++] = noon;
                qsort(AB, count, sizeof(long), compare_long);
                
                // C以下ならD本補充
                if(noon<=C){
                    night = noon + D;
                }
                else{
                    night = noon;
                }
            }
            // 過去と同じA-Bが出現したらYes終了
            else{
                flg=0;
                break;
            }
        }

        if(flg==0){
            printf("Yes\n");
        }
        else{
            printf("No\n");
        }
    }

    return 0;
}