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>

long euclid( long a, long b )
{
    long temp;

    if( a < b ) { temp = a; a = b; b = temp; }
    if( b < 1 ) return -1;

    if( a % b == 0 ) return b;
    return euclid( b, a % b );
}

int main()
{
    int T;
    scanf("%d", &T);
    long a[T], b[T], c[T], d[T];
    int i;
    for(i=0;i<T;i++) scanf("%ld %ld %ld %ld", &a[i], &b[i], &c[i], &d[i]);
    
    for(i=0;i<T;i++){
        if(a[i]<b[i]) printf("No\n");
        else if(b[i]>d[i]) printf("No\n");
        else{
            if(b[i]==d[i]){
                if(a[i]%b[i]>c[i]) printf("No\n");
                else printf("Yes\n");
            }else{
                long sa = d[i]-b[i];
                long amari = a[i]%b[i];
                long left, right;
                left = (c[i]+1-amari)/sa;
                right = (b[i]-1-amari)/sa;
                if(c[i]+1==b[i]-1){
                    if((c[i]+1-amari)%sa==0) printf("No\n");
                    else printf("Yes\n");
                }else if(c[i]+1<b[i]-1){
                   if(amari+sa*left==c[i]+1) printf("No\n");
                   else if(amari+sa*(left+1)<=b[i]-1) printf("No\n");
                   else{
                       if(euclid(sa, b[i])==1) printf("No\n");
                       else printf("Yes\n");
                   }
                }else printf("Yes\n");
            }
        }
    }
    

    return 0;
}