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)

// finish date: 2018/07/14
#include <bits/stdc++.h>

using namespace std;

#define FOR(i, a, b) for(int (i)=a;(i)<(b);(i)++)
#define rep(i, n)  FOR(i,0,n)
typedef long long ll;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<vector<vector<int>>>
        vvvi;
typedef vector<ll> vl;
typedef vector<vector<ll>> vvl;
typedef vector<double> vd;
typedef vector<vector<double>> vvd;
typedef vector<vector<vector<double>>>
        vvvd;
typedef vector<bool> vb;
typedef vector<vector<bool>> vvb;
typedef vector<string> vs;
typedef vector<char> vc;
typedef vector<vector<char>> vvc;
typedef pair<int, int> pii;
typedef pair<ll, int> pli;
typedef pair<ll, pair<int, int>> plii;
const int bigmod = 1000000007;
const int INF = 1050000000;
const long long INFll = 100000000000000000;


//最大公約数
ll gcd(ll a, ll b) {
    if (a < 0) a = -a;
    if (b < 0) b = -b;
    if (a < b) {
        swap(a, b);
    }
    ll r = 1; //0以外であれば何でも
    while (r != 0) {
        r = a % b;
        a = b;
        b = r;
    }
    return a;
}


int main() {
    int T;
    cin >> T;
    rep(i, T) {
        ll A, B, C, D;
        cin >> A >> B >> C >> D;
        if (A < B || A % B > C || B > D) {
            cout << "No" << endl;
            continue;
        }
        if (B == D) {
            cout << "Yes" << endl;
            continue;
        }
        if (B - C <= 1 || D % B == 0) {
            cout << "Yes" << endl;
            continue;
        }

        ll G = gcd(D % B, B);
        if (G == 1) {
            cout << "No" << endl;
            continue;
        } else {
            if (A > C) {
                A = A % B;
            }
            ll n = (ceil(((double)C+1 - A) / G));
            ll check = A + n * G;
            if (check < B) {
                cout << "No" << endl;
            } else {
                cout << "Yes" << endl;
            }
        }
    }

    return 0;
}