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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

/*
for (int i = 0; i < N; ++i){}

// 各桁の和を計算する関数
int findSumOfDigits(int n) {
  int sum = 0;
  while (n > 0) { // n が 0 になるまで
    sum += n % 10;
    n /= 10;
  }
  return sum;
}

sort(a, a + N, greater<int>());  // a[0:N] を大きい順にソート

int num[110] = {0};  // バケット
for (int i = 0; i < N; ++i) {
    num[d[i]]++;  // d[i] が 1 個増える
}

map<string, int> mp; // 連想配列 map<キー型, 値型> オブジェクト名
for (int i = 0; i < N; ++i) {
    auto itr = mp.find(s[i]);        // s[i] が設定されているか？
    if(itr != mp.end() ) {
         mp[s[i]] += 1;
    }
    else {
         mp[s[i]] += 1 ;
    }
}
stack<int> s;	//intをデータとするスタックを用意
s.push(1);		//{} -> {1}
printf("%d\n", s.top());	// 3
s.pop();

queue<int> que;	//intをデータとするキューを用意
que.push(1);		//{} -> {1}
printf("%d\n", que.front());	// 1
que.pop();
*/
#include <iostream>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <bitset>

using namespace std;

// #define for(i,a,b) for (int i=(a);i<(b);++i)
typedef long long ll;
typedef pair<int,int> point;
#define rep(i,n) for(i,0,n)
 
int main() {
	// 入力
	int T;
	cin >> T;
	ll A[T], B[T], C[T], D[T];
	for (int i = 0; i < T; ++i) cin >> A[i] >> B[i] >> C[i] >> D[i];
 
	// 解法
	for (int i = 0; i < T; ++i){
		bool ans;
		if (A[i]<B[i]) ans = false; // 初回でアウト
		else if (B[i]>D[i]) ans = false; // 消費が常に勝る
		// else if (B[i]<C[i]) ans = true; // stock=Bになる前に必ず補充する
		else{
			ll threshold = B[i] - C[i];
			ll K = __gcd(B[i], D[i]);
			ll AmodK = A[i]%K;
			ll CmodK = C[i]%K;
			if (CmodK<AmodK){
				ll tmp = C[i] - CmodK+AmodK;
				if (tmp<B[i]) ans = false;
				else ans = true;
			}
			else if (CmodK==AmodK){
				ll tmp = C[i] + K;
				if (tmp<B[i]) ans = false;
				else ans = true;
			}
			else{
				int tmp = C[i] - CmodK+AmodK + K;
				if (tmp<B[i]) ans = false;
				else ans = true;
				
			}
			// cout <<"K " << K <<" AmodK " << AmodK <<" CmodK " << CmodK << " ";
			
		}
		if (ans) printf("Yes\n");
		else printf("No\n");
	}
}