A. Payment Without Change（Codeforces Round #598 (Div. 3)）

A. Payment Without Change（Codeforces Round #598 (Div. 3)）

time limit per test：1 second
memory limit per test：256 megabytes
input：standard input
output：standard output

Description

You have a coins of value $n$ and $b$ coins of value $1$. You always pay in exact change, so you want to know if there exist such $x$ and $y$ that if you take $x (0≤x≤a)$ coins of value $n$ and $y (0≤y≤b)$ coins of value $1$, then the total value of taken coins will be $S$.

You have to answer $q$ independent test cases.

Input

The first line of the input contains one integer $q (1≤q≤10^4)$ — the number of test cases. Then $q$ test cases follow.

The only line of the test case contains four integers $a, b, n$ and $S (1≤a,b,n,S≤10^9)$ — the number of coins of value $n$, the number of coins of value $1$, the value $n$ and the required total value.

Output

For the $i$-th test case print the answer on it — YES(without quotes) if there exist such $x$ and $y$ that if you take $x$ coins of value $n$ and $y$ coins of value $1$, then the total value of taken coins will be $S$, and NOotherwise.

You may print every letter in any case you want (so, for example, the strings yEs, yes, Yesand YESwill all be recognized as positive answer).

Example

input

4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18

output

YES
NO
NO
YES

题解

列出等式 $x*n+y*1=S$，找到最大的不大于 $S$ 的 $x$ ，再判断如果 $xn+b$不小于 $S$ 则有答案，否则无答案
时间复杂度： $O(n log n)$

代码

#include <iostream>
#include <algorithm>
#include <vector>
#define maxn 300005
#define _for(i, a) for(LL i = 0; i < (a); i++)
using namespace std;
typedef long long LL;

int main() {
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	//freopen("in.txt", "r", stdin);

	LL n;
	cin >> n;
	_for(q, n) {
		LL a, b, n, S;
		cin >> a >> b >> n >> S;
		LL ans = min(S / n, a);
		if (S - ans * n <= b) cout << "YES\n";
		else cout << "NO\n";
	}
	return 0;
}