Solve the programming task below in a Python markdown code block.
A problem setter is called an *expert* if at least 50 \% of their problems are approved by Chef.

Munchy submitted X problems for approval. If Y problems out of those were approved, find whether Munchy is an *expert* or not.

------ 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 two space-separated integers X and Y - the number of problems submitted and the number of problems that were approved by Chef.

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

For each test case, output on a new line YES, if Munchy is an *expert*. Otherwise, print NO.

The output is case-insensitive. Thus, the strings YES, yes, yeS, and Yes are all considered the same.

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

$1 ≤ T ≤ 1000$
$1 ≤ Y ≤X ≤ 10^{6}$

----- Sample Input 1 ------ 
4
5 3
1 1
4 1
2 1

----- Sample Output 1 ------ 
YES
YES
NO
YES
----- explanation 1 ------ 
Test case $1$: We are given that $3$ out of $5$ problems were approved. Thus, $60 \%$ of the problems were approved. Since at least $50 \%$ of the problems were approved, Munchy is an *expert*. 

Test case $2$: We are given that $1$ out of $1$ problems were approved. Thus, $100 \%$ of the problems were approved. Since at least $50 \%$ of the problems were approved, Munchy is an *expert*. 

Test case $3$: We are given that $1$ out of $4$ problems were approved. Thus, $25 \%$ of the problems were approved. Since at least $50 \%$ of the problems were not approved, Munchy is not an *expert*. 

Test case $4$: We are given that $1$ out of $2$ problems were approved. Thus, $50 \%$ of the problems were approved. Since at least $50 \%$ of the problems were approved, Munchy is an *expert*.