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.Collections.Generic;
using System.Text;

namespace AtTest.AGC026
{
    class B
    {
        static void Main(string[] args)
        {
            Method(args);
            Console.ReadLine();
        }

        static void Method(string[] args)
        {
            int t = int.Parse(Console.ReadLine());
            long[,] query = new long[t, 4];
            for (int i = 0; i < t; i++)
            {
                string[] input = Console.ReadLine().Split(' ');
                query[i, 0] = long.Parse(input[0]);
                query[i, 1] = long.Parse(input[1]);
                query[i, 2] = long.Parse(input[2]);
                query[i, 3] = long.Parse(input[3]);
            }

            for (int i = 0; i < t; i++)
            {
                if (Eternal(query[i, 0], query[i, 1], query[i, 2], query[i, 3]))
                {
                    Console.WriteLine("Yes");
                }
                else
                {
                    Console.WriteLine("No");
                }
            }
        }

        static bool Eternal(long a, long b, long c, long d)
        {/*
            if (b > d || b > a || a % b > c) return false;
            if (b == d) return true;
            long margin = a % b;//ここが違う
            while (margin + b < c)
            {
                margin += b;
            }
            long delta = d - b;
            long shou = (c-margin) / delta;
            if (b <= delta * (shou + 1) + margin) return true;
            return false;
            */
            if (b > d || b > a) return false;//絶対無理のパターン
            
            long firstStock = 0;
            if (a <= c)
            {
                firstStock = a - b;
            }
            else
            {
                long mar = a % b;
                if (mar > c) return false;

                long ite = (c - mar) / b;
                firstStock = mar + ite * b;
            }
            if (firstStock < 0) return false;
            if (b == d) return true;//絶対okのパターン

            long delta = d - b;//入荷後に購入した場合の在庫変化
            long cnt = (c - firstStock) / delta + 1;
            if (b <= firstStock + delta * cnt) return true;

            return false;
        }
    }
}