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 forn(i, x, n) for (int i = int(x); i <= int(n); ++i)
#define for1(i, n, x) for (int i = int(n); i >= int(x); --i)
#define x first
#define y second
#define pb push_back

typedef long long ll;
typedef pair <int, int> pii;
typedef long double ld;
typedef vector <ll> vi;

const int N = 2e6 + 1;
const int INF = 1e9 + 9;
const int B = 1e9 + 7;

int main() {
    #ifdef black
    freopen("in", "r", stdin);
    #endif // black
    ios_base :: sync_with_stdio(0);
    cin.tie(0);

    ll t;
    cin >> t;
    forn(it, 1, t) {
    //cerr << "test " << it << "\n";
        ll a, b, c, d;
        cin >> a >> b >> c >> d;
        if (d < b || a < b) {
        //cerr << "1\n";
            cout << "No\n";
            continue;
        }
        if (a > c) {
            a -= ((a / b) - (c / b)) * b;
            if (a <= c) a += b;
        }
        if (a < b) {
            //cerr << "2 " << a << "\n";
            cout << "No\n";
            continue;
        }
        if (c + 1 <= b - 1) {
            ll g = __gcd(b, d);
            ++c, --b;
            if (b / g > 1 + c / g || (c % g <= b % g ? c % g <= a % g && a % g <= b % g : c % g <= a % g || a % g <= b % g)) {
                //cerr << "3 " << g << "\n";
                cout << "No\n";
                continue;
            }
        }
        cout << "Yes\n";
    }

    return 0;
}