The followers of Psycho-Helmet religion follow a peculiar calendar – a normal year contains N days. Every K-th year is a “MOB” year. For example, if K = 4, then years $4, 8, 12, 16 \ldots$ are “MOB” years. A “MOB” year contains M additional days i.e. it contains N+M days.

Given X, determine whether the day after (X-1) days from Day 1 of Year 1 falls in a “MOB” year.

------ Input Format ------

• The first line of input will contain a single integer T, denoting the number of test cases.
• Each test case consists of a single line containing N, M, K and X — the number of days in a normal year, the number of days in a “MOB” year, the number of years between two “MOB” years and the integer X.

------ Output Format ------

For each test case, output YES on a new line if the day after (X-1) days from Year 1, Day 1 falls in a “MOB” year, and NO otherwise.

You may print each character of YES and NO in uppercase or lowercase (for example, yes, yEs, Yes will be considered identical).

------ Constraints ------

$1 ≤ T ≤ 10^{5}$
$1 ≤ N, M, K ≤ 10^{4}$
$1 ≤ X ≤ 10^{9}$

----- Sample Input 1 ------
3
2 1 4 17
3 7 5 50
3 7 5 61
----- Sample Output 1 ------
YES
NO
YES
----- explanation 1 ------
Test case 1: Since $N = 2$, $M = 1$ and $K = 4$, years $4, 8, 12, 16 \ldots$ are “MOB” years and:
- Year $1$ has $N = 2$ days.
- Year $2$ has $N = 2$ days.
- Year $3$ has $N = 2$ days.
- Year $4$ is a “MOB” year and has $N + M = 3$ days.
- Year $5$ has $N = 2$ days.
- Year $6$ has $N = 2$ days.
- Year $7$ has $N = 2$ days.
- Year $8$ is a “MOB” year and has $N + M = 3$ days.

The $X = 17$-th day falls on the second day of Year $8$, which is a “MOB” year. Hence the answer is YES.

Test case 2: The $X = 50$-th day falls on the third day of Year $12$, which is not a “MOB” year. Hence the answer is NO.

Test case 3: The $X = 61$-st day falls on the fifth day of Year $15$, which is a “MOB” year. Hence the answer is YES.

见题面

见题面