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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

def judge(morning, buy, thresh, adder)
  # 朝より購入のほうが多い -> 即NG
  return false if morning < buy

  # 購入より補充のほうが少ない -> 即NG
  return false if adder < buy

  # 購入よりしきい値のほうが多い -> 即OK
  # 購入-1 <= しきい値           -> 即OK
  return true if buy-1 <= thresh

  # 初期処理
  # morningをthreshぎりぎりまで減らす
  if thresh < morning
    d = (morning - thresh) / buy
    morning -= buy * d
  end
  # adderもthreshぎりぎりまで減らす
  if thresh < adder
    d = (adder - thresh) / buy
    adder -= buy * d
  end

  s = morning
  s_ary = []
  inc = adder - buy

  while true do
    if s <= thresh
      # ここの回数を減らしたい！
      # s += adder
      # 単純に加算するのではなく、(adder-buy)ずつ増やしていって、threshを超えさせよう

      # thresh < s + (adder - buy) * XXX
      return true if inc == 0

      d = (thresh - s) / inc + 1
      s += inc * d
    else
      # 補充なしターン
      if s < buy
        return false
      end
      if s_ary.index(s)
        return true
      end
      # thresh上に出たところだけ記憶しておく
      s_ary << s
      s -= buy
    end
  end
end

def main
  t = ARGF.gets.to_i

  t.times do
    morning, buy, thresh, adder = ARGF.gets.split.map(&:to_i)
    if judge(morning, buy, thresh, adder)
      puts 'Yes'
    else
      puts 'No'
    end
  end
end

if __FILE__ == $0
  main
end