Solve the programming task below in a Python markdown code block.
A sorting algorithm A is said to have more time complexity than a sorting algorithm B if it uses more number of comparisons for sorting the same array than algorithm B.

Given that algorithm A uses X comparisons to sort an array and algorithm B uses Y comparisons to sort the same array, find whether algorithm A has more time complexity.

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

- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of two space-separated integers X and Y — the number of comparisons used by algorithms A and B to sort the array respectively.

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

For each test case, output on a new line, YES, if the algorithm A has more time complexity than B and NO otherwise.

You may print each character of the string in uppercase or lowercase (for example, the strings YES, yEs, yes, and yeS will all be treated as identical).

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

$1 ≤ T ≤ 100$
$1 ≤ X, Y ≤ 100$

----- Sample Input 1 ------ 
4
9 9
15 7
10 19
21 20

----- Sample Output 1 ------ 
NO
YES
NO
YES
----- explanation 1 ------ 
Test case $1$: The number of comparisons used by algorithm $A$ is $9$ and that used by $B$ is also $9$. Since the number of comparisons used by $A$ is not more than that of $B$, $A$ does not have more time complexity than $B$.

Test case $2$: The number of comparisons used by algorithm $A$ is $15$ and that used by $B$ is $7$. Since the number of comparisons used by $A$ is more than that of $B$, $A$ does have more time complexity than $B$.

Test case $3$: The number of comparisons used by algorithm $A$ is $10$ and that used by $B$ is $19$. Since the number of comparisons used by $A$ is not more than that of $B$, $A$ does not have more time complexity than $B$.

Test case $4$: The number of comparisons used by algorithm $A$ is $21$ and that used by $B$ is $20$. Since the number of comparisons used by $A$ is more than that of $B$, $A$ does have more time complexity than $B$.