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.
#####################################   
import atexit, io, sys, collections, math, heapq, fractions
buffer = io.BytesIO() 
sys.stdout = buffer
@atexit.register 
def write(): sys.__stdout__.write(buffer.getvalue())   
#####################################


def g(m,ais, bis):
	
	t =  [(u,v) for u,v in zip(ais,bis)]
	if sum([e[0] for e in t]) < m: return -1
	
	onepoints = [u for u,v in zip(ais,bis) if v == 1]
	onepoints.sort(key = lambda x:-x)
	twopoints = [u for u,v in zip(ais,bis) if v == 2]
	twopoints.sort(key = lambda x:-x)


	if len(onepoints) == 0 or len(twopoints) == 0:
		if len(twopoints):
			cumula = 0
			for i in range(len(twopoints)):
				cumula += twopoints[i]
				if cumula >= m:
					return (i+1) * 2
		if len(onepoints):
			cumula = 0
			for i in range(len(onepoints)):
				cumula += onepoints[i]
				if cumula >= m:
					return i+1

	counttwo = 0
	cumul = 0
	ans = +float('inf')

	c = 0
	countone = 0
	
	for j in range(len(onepoints)):
		if j: onepoints[j] += onepoints[j-1]
	if onepoints and onepoints[-1] >= m:
		ans = min(ans, len(onepoints))


	lo,hi = 0, len(onepoints)-1
	while(lo < hi):
		med= (lo + hi) //2
		if cumul + onepoints[med] >= m:
			hi = med
		else:
			lo = med +1
	if cumul + onepoints[lo] >= m:
		ans = min(ans, (2 * counttwo) + lo +1)
	
	for e in twopoints:
		counttwo += 1
		cumul += e
		if cumul >= m:
			ans = min(ans, 2 * counttwo)
		else:
			lo,hi = 0, len(onepoints)-1
			while(lo < hi):
				med= (lo + hi) //2
				if cumul + onepoints[med] >= m:
					hi = med
				else:
					lo = med +1
			if cumul + onepoints[lo] >= m:
				ans = min(ans, (2 * counttwo) + lo +1)

	return ans


nt = int(raw_input())
for u in range(nt):
	n,m = map(int,raw_input().split())
	ais = map(int, raw_input().split())
	bis = map(int, raw_input().split())
	print g(m,ais,bis)
	