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: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AGC026_B
{
    class Program
    {
        static void Main(string[] args)
        {
            int t = int.Parse(Console.ReadLine());

            bool[] flags = new bool[t];

            for (int i = 0; i < t; ++i)
            {
                string[] strs = Console.ReadLine().Split(' ');
                long a = long.Parse(strs[0]);
                long b = long.Parse(strs[1]);
                long c = long.Parse(strs[2]);
                long d = long.Parse(strs[3]);

                flags[i] = Check(a, b, c, d);

            }

            for (int i = 0; i < t; ++i)
            {
                Console.WriteLine("{0}", flags[i] ? "Yes" : "No");
            }
        }

        static bool Check(long a, long b, long c, long d)
        {
            if (a < b) return false;

            long xMod = a % b;
            long x = a / b;

            long yMod = (a - c) % b;
            long y = (a - c) / b;

            if (x == y && yMod != 0) return false;

            long a2 = xMod + d;

            if (xMod + d < b) return false;


            long xMod2 = a2 % b;


            x = a2 / b;
            yMod = (a2 - c) % b;
            y = (a2 - c) / b;

            if (x == y && yMod != 0) return false;

            //if (xMod == xMod2) return true;



            if (d == b) return true;
            if (d < b) return false;


            long a3 = xMod2 + d;
            long xMod3 = a3 % b;

            if (xMod2 == xMod3) return true;

            long gcd = Gcd(b, d);

            long e = a3 - c;

            long f = (e / gcd) * gcd;
            long fMod = e % gcd;
            if (fMod == 0) f -= gcd;

            if (f + b > a3) return false;

            ////if (c + gcd >= b && d >= b) return true;
            //if ((c + gcd) - b + d >= b) return true;

            if (d > b) return true;


            return false;
        }

        public static long Gcd(long a, long b)
        {
            if (a < b)
            {
                // 引数を入替えて自分を呼び出す
                return Gcd(b, a);
            }

            while (b != 0)
            {
                var remainder = a % b;
                a = b;
                b = remainder;
            }
            return a;
        }
    }
}