Polycarp often uses his smartphone. He has already installed n applications on it. Application with number i takes up a_i units of memory.

Polycarp wants to free at least m units of memory (by removing some applications).

Of course, some applications are more important to Polycarp than others. He came up with the following scoring system — he assigned an integer b_i to each application: 

  * b_i = 1 — regular application; 
  * b_i = 2 — important application. 



According to this rating system, his phone has b_1 + b_2 + … + b_n convenience points.

Polycarp believes that if he removes applications with numbers i_1, i_2, …, i_k, then he will free a_{i_1} + a_{i_2} + … + a_{i_k} units of memory and lose b_{i_1} + b_{i_2} + … + b_{i_k} convenience points.

For example, if n=5, m=7, a=[5, 3, 2, 1, 4], b=[2, 1, 1, 2, 1], then Polycarp can uninstall the following application sets (not all options are listed below): 

  * applications with numbers 1, 4 and 5. In this case, it will free a_1+a_4+a_5=10 units of memory and lose b_1+b_4+b_5=5 convenience points; 
  * applications with numbers 1 and 3. In this case, it will free a_1+a_3=7 units of memory and lose b_1+b_3=3 convenience points. 
  * applications with numbers 2 and 5. In this case, it will free a_2+a_5=7 memory units and lose b_2+b_5=2 convenience points. 



Help Polycarp, choose a set of applications, such that if removing them will free at least m units of memory and lose the minimum number of convenience points, or indicate that such a set does not exist.

Input

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

The first line of each test case contains two integers n and m (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ m ≤ 10^9) — the number of applications on Polycarp's phone and the number of memory units to be freed.

The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the number of memory units used by applications.

The third line of each test case contains n integers b_1, b_2, …, b_n (1 ≤ b_i ≤ 2) — the convenience points of each application.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case, output on a separate line: 

  * -1, if there is no set of applications, removing which will free at least m units of memory; 
  * the minimum number of convenience points that Polycarp will lose if such a set exists. 

Example

Input


5
5 7
5 3 2 1 4
2 1 1 2 1
1 3
2
1
5 10
2 3 2 3 2
1 2 1 2 1
4 10
5 1 3 4
1 2 1 2
4 5
3 2 1 2
2 1 2 1


Output


2
-1
6
4
3

Note

In the first test case, it is optimal to remove applications with numbers 2 and 5, freeing 7 units of memory. b_2+b_5=2.

In the second test case, by removing the only application, Polycarp will be able to clear only 2 of memory units out of the 3 needed.

In the third test case, it is optimal to remove applications with numbers 1, 2, 3 and 4, freeing 10 units of memory. b_1+b_2+b_3+b_4=6.

In the fourth test case, it is optimal to remove applications with numbers 1, 3 and 4, freeing 12 units of memory. b_1+b_3+b_4=4.

In the fifth test case, it is optimal to remove applications with numbers 1 and 2, freeing 5 units of memory. b_1+b_2=3.
#from itertools import product
#from itertools import combinations
#from collections import defaultdict
#from collections import Counter
#from sys import setrecursionlimit
from sys import stderr

def dbp(*args, **kwargs): # hacky debug print, maybe make ('varname:', value) version?
    pass#print(*args, file=stderr, **kwargs)

def get_int_list():
	return [int(i) for i in input().strip().split()]


def do_thing():
	N, freeme = get_int_list()
	mlist = get_int_list()
	clist = get_int_list()

	if sum(mlist) < freeme:
		return -1

	# too general, use the fact that convenience values are only 1 or 2
	c1list = []
	c2list = []

	rsum = 0
	for i in range(N):
		if clist[i] == 1:
			c1list.append(mlist[i])
		else:
			t = mlist[i]
			c2list.append(t)
			rsum += t

	c1list.sort(reverse=True)
	c2list.sort(reverse=True)

	dbp('target:', freeme)
	dbp(c1list)
	dbp(c2list)

	lsum = 0
	r = len(c2list)
	# rsum = sum(c2list)
	bcc = float('inf') # best convenience cost

	for l in range(len(c1list)+1): # can include last element
		dbp('before:', l, lsum, r, rsum, bcc)
		while r > 0 and lsum + rsum - c2list[r-1] >= freeme:
			r -= 1
			rsum -= c2list[r]
		if lsum + rsum >= freeme:
			cost = 2 * r + l
			dbp(l, lsum, r, rsum, 'goodcost:', cost)
			bcc = min(cost, bcc)
		if l < len(c1list): # but can't add past last element, also choose better names
			lsum += c1list[l] # l itself moved by loop
		dbp('after:', l, lsum, r, rsum, bcc)

	return bcc if bcc != float('inf') else -1

# prep goes here

maxcc = int(input().strip())
for cc in range(maxcc):
	print(do_thing())
