Solve the programming task below in a Python markdown code block.
Chef wants to make a purchase. For this, he needs X gold coins, but he has none at the moment.

Chef has N fixed deposits, the i^{th} of which is worth A_{i} coins. He wants to open the minimum number of these deposits so that he has at least X coins.

You have to tell Chef the minimum number of fixed deposits he must open in order to have X coins, or say that this is impossible.

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

- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains two space-separated integers — N and X, as described in the statement.
- The second line of each test case contains N space-separated integers — the i^{th} of which is A_{i}.

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

For each test case, output one line containing the answer — the minimum number of FDs Chef must open to have at least X coins. If it is not possible for him to open FDs such that he has at least X coins, output -1. 

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

$1 ≤ T, A_{i}, N ≤ 100$
$1 ≤ X ≤ 10^{4}$

------ subtasks ------ 

Subtask #1 (100 points): Original constraints

----- Sample Input 1 ------ 
4
4 6
4 3 5 1
3 15
1 5 3
2 5
10 3
4 7
1 2 3 4
----- Sample Output 1 ------ 
2
-1
1
2
----- explanation 1 ------ 
Test case $1$: Chef can open the first and second FDs to get $4 + 3 = 7$ coins, which is more than the $6$ coins he needs. There are other combinations of two FDs giving at least $6$ coins — for example, the first and third (for $4 + 5 = 9$) and third and fourth (for $5 + 1 = 6$), but there is no way to open only one FD and get $6$ or more coins.

Test case $2$: No matter which FDs Chef opens, he cannot have $≥ 15 $ coins, so the answer is $-1$ — even if he opens all of them, he will have only $1 + 5 + 3 = 9$ coins.

Test case $3$: Chef can open the first FD to obtain $10$ coins, which is larger than the $5$ coins he needs.

Test case $4$: Chef can open the third and fourth FDs to get $3 + 4 = 7$ coins, which is enough because he needs $7$ coins.