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)

//#pragma GCC optimize("O3")
#include <bits/stdc++.h>
#define enl printf("\n")
#define case(t) printf("Case #%d: ", (t))
#define ni(n) scanf("%d", &(n))
#define nl(n) scanf("%lld", &(n))
#define nai(a, n) for (int i = 0; i < (n); i++) ni(a[i])
#define nal(a, n) for (int i = 0; i < (n); i++) nl(a[i])
#define pri(n) printf("%d\n", (n))
#define prl(n) printf("%lld\n", (n))
#define pii pair<int, int>
#define pll pair<long long, long long>
#define vii vector<pii>
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
typedef long long ll;
const double pi = acos(-1);
const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;
const int MAXN = 2e6 + 5;
const double eps = 1e-9;
using namespace std;

void solve()
{
    ll a, b, c, d;
    nl(a), nl(b), nl(c), nl(d);
    if (d < b || a < b)
    {
        printf("No\n");
        return;
    }
    ll delta = d - b;
    if (c >= a && delta != 0)
    {
        ll l = (c + 1 - a + delta - 1) / delta;
        a += l * delta;
    }
    if (b == d)
    {
        if (c >= a)
        {
            printf("Yes\n");
            return;
        }
        ll nx = (a - c) % b + c;
        if ((a - c) % b == 0)
            nx += b;
        if (nx - b < 0)
            printf("No\n");
        else
            printf("Yes\n");
        return;
    }
    ll nx = (a - c) % b + c;
    if ((a - c) % b == 0)
        nx += b;
    if (nx - b < 0)
    {
        printf("No\n");
        return;
    }
    if (b <= c + 1)
    {
        printf("Yes\n");
        return;
    }
    if (b - 1 >= nx && nx >= c + 1)
    {
        printf("No\n");
        return;
    }
    ll cur = a;
    if (cur % b >= c + 1)
    {
        printf("No\n");
        return;
    }
    a %= b, delta %= b;
    if (__gcd(b, delta) == 1)
    {
        printf("No\n");
        return;
    }
    //cur = 0;
    //ll tmp = (c + 1 - a + b) % b;
    ll tmp = c + 1;
    for (int i = 0; i <= tmp; i++)
    {
        cur = (cur + delta) % b;
        if (cur >= tmp)
        {
            printf("No\n");
            return;
        }
    }
    printf("Yes\n");
}

int main()
{
    int t;
    ni(t);
    while (t--)
        solve();
    return 0;
}