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 sys
import math
import itertools
import functools
import collections
import operator
import fileinput
import copy
from collections import *
 
ORDA = 97  # a
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return [int(i) for i in input().split()]
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def revn(n): return str(n)[::-1]
def dd(): return collections.defaultdict(int)
def ddl(): return collections.defaultdict(list)
def sieve(n):
    if n < 2: return list()
    prime = [True for _ in range(n + 1)]
    p = 3
    while p * p <= n:
        if prime[p]:
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 2
    r = [2]
    for p in range(3, n + 1, 2):
        if prime[p]:
            r.append(p)
    return r
def divs(n, start=2):
    r = []
    for i in range(start, int(math.sqrt(n) + 1)):
        if (n % i == 0):
            if (n / i == i):
                r.append(i)
            else:
                r.extend([i, n // i])
    return r
def divn(n, primes):
    divs_number = 1
    for i in primes:
        if n == 1:
            return divs_number
        t = 1
        while n % i == 0:
            t += 1
            n //= i
        divs_number *= t
def prime(n):
    if n == 2: return True
    if n % 2 == 0 or n <= 1: return False
    sqr = int(math.sqrt(n)) + 1
    for d in range(3, sqr, 2):
        if n % d == 0: return False
    return True
def convn(number, base):
    new_number = 0
    while number > 0:
        new_number += number % base
        number //= base
    return new_number
def cdiv(n, k): return n // k + (n % k != 0)
def ispal(s): # Palindrome
    for i in range(len(s) // 2 + 1):
        if s[i] != s[-i - 1]:
            return False
    return True
   
# a = [1,2,3,4,5] -----> print(*a) ----> list print krne ka new way

def function(arr, size,dp):
	if size == 1 or size == 0:
		return 1
	key = ''.join(map(str, arr[:size]))
	if dp[key]:
		return dp[key]
	output = function(arr, size-1, dp)
	key = ''.join(map(str, arr[:size-1]))
	dp[key] = output
	if (arr[size-2]*10 + arr[size-1]) <= 26:
		output += function(arr, size-2, dp)
		key = ''.join(map(str, arr[:size-2]))
		dp[key] = output
	return output

def function(arr,s,n,dp):
	if s == 0:
		return 0
	if n == 0 and s!= 0:
		return s
	if dp[n][s] != 0:
		return dp[n][s]
	if arr[n-1] <= s:
		dp[n][s] = max(arr[n-1], function(arr,s-arr[n-1],n-1,dp), function(arr,s,n-1,dp))
		return dp[n][s]
	else:
		dp[n][s] = function(arr,s,n-1,dp)
		return dp[n][s]
	
def main():
	for _ in range(ii()):
		n, m= mi()
		arr = li()
		brr = li()
		a, b = [],[]
		s = 0
		as1 = 0
		cost = 0
		for i in range(n):
			s += arr[i]
			if brr[i] == 1:
				a.append(arr[i])
				as1 += arr[i]
			else:
				b.append(arr[i])
		if s < m:
			print(-1)
		else:
			a = sorted(a, reverse = True)
			b = sorted(b, reverse = True)
			if as1 >= m:
				cost = len(a)
				i = cost -1
				while as1> m:
					as1 -= a[i]
					i -= 1
					cost -= 1
				if as1 < m:
					cost += 1
					as1 += a[i+1]
			else:
				i = 0
				cost = len(a)
				while as1 < m:
					as1 += b[i]
					i += 1
				cost += i*2
				i = len(a) -1
				#print(as1)
				#print(cost)
				while as1 >m and i >= 0:
					as1 -= a[i]
					i -= 1
					cost -= 1
				if as1 < m:
					cost += 1
			print(cost)
					
				
				
				
				
main()