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)

// header {{{
#include <bits/stdc++.h>
using namespace std;

// {U}{INT,LONG,LLONG}_{MAX,MIN}
#define INF         INT_MAX/3
#define LLINF       LLONG_MAX/3
#define MOD         (1000000007LL)
#define MODA(a, b)  a=((a)+(b))%MOD
#define MODP(a, b)  a=((a)*(b))%MOD
#define inc(i, l, r)   for(int i=(l);i<(r);i++)
#define dec(i, l, r)   for(int i=(r)-1;i>=(l);i--)
#define pb          push_back
#define se          second
#define fi          first
#define mset(a, b)  memset(a, b, sizeof(a))

using LL  = long long;
using G   = vector<vector<int>>;

int di[] = {0, -1, 0, 1};
int dj[] = {1, 0, -1, 0};
// }}}

// 最大公約数
LL gcd(LL x, LL y) {
    return y ? gcd(y, x%y) : x;
}

bool solve(LL a, LL b, LL c, LL d) {
    if(a < b){
        return false;
    }else{
        if(d < b){
            return false;
        }else{
            if(c < b){ // C < B <= A
                if(gcd(b, d) == 1){
                    if(b-c>1){
                        return false;
                    }else{
                        return true;
                    }
                }else{
                    // C < B <= A, B <= D
                    a -= (a-b)/b*b;
                    LL diff = d-b;
                    if(b-c>1){
                        if((b-c) <= diff){
                            if(diff == 0){
                                if(a-d > c && a-d < b){
                                    return false;
                                }else{
                                    return true;
                                }
                            }else{
                                if(c%diff == a%diff || b%diff == a%diff){
                                    return true;
                                }else{
                                    //cout << "#";
                                    if((c-(a-d)+1)%diff >= (b-(a-d)-1)%diff){
                                        return true;
                                    }else{
                                        return false;
                                    }
                                }
                            }
                        }else{
                            if(diff == 0){
                                if(a-d > c && a-d < b){
                                    return false;
                                }else{
                                    return true;
                                }
                            }else{
                                return false;
                            }
                        }
                    }else{
                        return true;
                    }
                }
            }else{
                return true;
            }
        }
    }
}

int main() {
    cin.tie(0);ios::sync_with_stdio(false);
    int t;cin >> t;
    inc(i, 0, t){
        LL a, b, c, d;
        cin >> a >> b >> c >> d;
        cout << (solve(a, b, c, d) ? "Yes" : "No") << endl;
    }
    return 0;
}