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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Linq;
using System.Collections.Generic;
using static System.Math;

class P
{

    static void Main()
    {
        checked
        {
            int t = int.Parse(Console.ReadLine());
            bool[] res = new bool[t];
            for (int i = 0; i < t; i++)
            {
                long[] abcd = Console.ReadLine().Split().Select(long.Parse).ToArray();
                long a = abcd[0];
                long b = abcd[1];
                long c = abcd[2];
                long d = abcd[3];
                //開始がb以上
                //補充ポイントがc以下にならない
                //増加と補充のgcd
                //増加と補充
                //一番下を求めたくて
                //bの減少とd-bの増加を繰り返す
                //b+c以下だったら増加
                //b以下にならないか

                //上限と下限が欲しい
                //上からb*nおりたところが以下だったら上昇
                //c - (d % c)が次に降りる変わる幅
                //これの累積がcを超えたら+cがおきる
                //これがいい感じでループしてくれたら嬉しい
                //20 -> -5がはいるところがいいかんじだったらむっちゃうれしいけど
                //最初の奴からいい感じで落ちたい
                if (a < b) goto no;
                //必ず小さくなければだめなのはそれはそう
                if (d < b) goto no;
                if (c < d && c >= b) goto yes;
                //1サイクル回した時に下がる幅
                long haba = c - (d % c != 0 ? d % c : c);
                //取りうる値の粒度
                long gcd = Gcd(haba, d);
                //最初に補給が起きるターンの買う前
                //a - (c + 1):落ちてもいい値 これをドリンクで割って
                long firstPoint = (a + d) - (((a + d) - (c + 1)) / b) * b;
                //補給が起きる前の買う前から粒度gcdで落ちていく 一番下
                //最初期位置から落ちてく 最初から落ちられる幅 
                if ((firstPoint - (((firstPoint - (c + 1)) / gcd) * gcd) < b)) goto no;
                yes:;
                res[i] = true;
                continue;
                no:;
                res[i] = false;
            }
            Console.WriteLine(string.Join("\n", res.Select(x => x ? "Yes" : "No")));
        }
    }

    //O(80)
    public static long Gcd(long a, long b)
    {
        if (a < b) return Gcd(b, a);
        while (b != 0)
        {
            long remainder = a % b;
            a = b;
            b = remainder;
        }
        return a;
    }
}