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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <iostream>

using namespace std;
#define ll long long
int main(){
    int ans=0;
    int t=0;
    cin>>t;
    for(int i=0;i<t;i++){
        ll a,b,c,d;
        cin >>a>>b>>c>>d;
        //zaiko<b && c<zaikoとなるときがあったらアウト
        ll zaiko=a;
        bool flag=true;
        if (a<b)flag=false;
        if (d<b)flag=false;
        ll n=(d%b==0)?0:c/(d%b)-1;
        ll m=(d%b==0)?0:b/(d%b)+1;
        ll l=0;
//        cout <<n <<" "<<b<<" "<<c<<" "<<d<<endl;
        if (c < a%b && b > a%b)flag=false;
        else if(d%b!=0 && d%b<b-c)flag=false;
        ll lo=n;
        ll hi=m;
        l=(lo+hi)/2;
        ll dif=d%b;
        ll err=a%b;
        if(dif!=0 && flag ){
            if(dif<b-c)flag=false;
            do{
                if (c < err%dif+l*dif && b > err%dif+l*dif){
                    flag=false;
                    break;
                }else{
                    bool ba=false;
                    if (c >= l*(d%b)){
                        lo=l+1;
                        ba=true;
                    }
                    if(b <= l*(d%b)){
                        hi=l-((ba)?0:1);
                    }
                        l=(lo+hi)/2;
                }
            }while(lo<hi && d%b!=0);
        }

/*        for(int i=0;i<c;i++){
            if((cinit+d)%b>c)flag=false;
            if (cinit+d<b)flag=false;
            cinit=(cinit+d)%b;
        }*/
        if(flag){
            cout <<"Yes"<<endl;
        }else{
            cout <<"No"<<endl;
        }

    }
//    cout <<ans<<endl;
    return 0;
}