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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
typedef long long ll;

int main()
{
    int t;
    ll a,b,c,d;
    cin>>t;
    for(int i=0;i<t;i++)
    {
        cin>>a>>b>>c>>d;
        if(a<b)
        {
            cout<<"No"<<endl;
            continue;
        }
        if(a-b<=c)
        {
            if(d<b) {
                cout<<"No"<<endl;
                continue;
            }
            if(a-b+d-b<0)
            {
                cout<<"No"<<endl;
                continue;
            }
            ll lo=0,hi=1000000000000000000,mid,ans=-1;
            while(lo<=hi)
            {
                mid=(lo+hi)/2;
                ll pp=1LL*mid*(d-b);
                if(a-b+pp<=c)
                {
                    lo=mid+1;
                }
                else if(a-b+pp>=b)
                {
                    hi=mid-1;
                }
                else if(a-b+pp>c && a-b+pp<b)
                {
                    ans=mid;
                    break;
                }
            }
            if(ans==-1)
            {
                cout<<"Yes"<<endl;
            }
            else
            {
                cout<<"No"<<endl;
                continue;
            }
        }
        else
        {

            ll lo=0,hi=100000000000000000,mid,ans=-1;
            while(lo<=hi)
            {
                mid=(lo+hi)/2;
                ll pp=1LL*mid*b;
                if(a-pp>c)
                {
                    ans=mid;
                    lo=mid+1;
                }
                else if(a-pp<=c)
                {
                    hi=mid-1;
                }
            }
            if(ans==-1)
            {
                cout<<"Yes"<<endl;
                continue;
            }
            a=a-1LL*ans*b;
            if(d<b)
            {
                cout<<"No"<<endl;
                continue;
            }
            if(a>=b )
            {
                cout<<"Yes"<<endl;

            }
            else cout<<"No"<<endl;
        }
    }
    return 0;
}