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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

function myin(){return require("fs").readFileSync("/dev/stdin", "utf8").trim();}
function myout(t){console.log(t);}//standard output
//[no]param
//0:何もしない  1:数値へ変換  2:半角SPで分割  3:改行で分割  4:半角SPで分割し、数値配列へ
//5:改行で分割し、数値配列へ  6:1文字に分割  7:1文字に分割し、数値配列へ
function myconv(i,no){switch(no){case 0:return i;case 1:return parseInt(i);case 2:return i.split(" ");case 3:return i.split("\n");case 4:return i.split(" ").map((a)=>Number(a));case 5:return i.split("\n").map((a)=>Number(a));case 6:return i.split("");case 7:return i.split("").map((a)=>Number(a));}}

function Main(input) {
  input = myconv(input,3);
  input.shift();
  var output = [];
  for(var i = 0; i < input.length; i++){
    //a:最初の在庫 b:すぬけ君が買う個数 c:閾値 d:在庫が閾値以外の時に次の日のaに追加する個数
    var a = myconv(input[i],4)[0];
    var initA = myconv(input[i],4)[0];
    var b = myconv(input[i],4)[1];
    var c = myconv(input[i],4)[2];
    var d = myconv(input[i],4)[3];
	
    //a % b:すぬけ君がbだけ買うことができなくなる個数
    var noFlag = false;
    var count = 0;
    while(true){
      count++;
      if(a < b){
        noFlag = true;
        	break;
   		 }
      a = a % b;
      if(a <= c){
           a += d;
      }
      if(a == initA || count >= 1000000){
           break;
      }
    }
    if(noFlag){
       output.push("No");
    }else{
      output.push("Yes");
    }
  }
  myout(output.join("\n"));
}

Main(myin());