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 collections
import string
import math
import copy
import os
import sys
from io import BytesIO, IOBase

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)


def input(): return sys.stdin.readline().rstrip("\r\n")


# n = 0
# m = 0

# n = int(input())

# li = [int(i) for i in input().split()]

# s = sorted(li)

mo = 998244353


def exgcd(a, b):
    if not b:
        return 1, 0
    y, x = exgcd(b, a % b)
    y -= a//b * x
    return x, y


def getinv(a, m):
    x, y = exgcd(a, m)
    return -(-1) if x == 1 else x % m


def comb(n, b):
    res = 1
    b = min(b, n-b)
    for i in range(b):
        res = res*(n-i)*getinv(i+1, mo) % mo
        # res %= mo
    return res % mo


def quickpower(a, n):
    res = 1
    while n:
        if n & 1:
            res = res * a % mo
        n >>= 1
        a = a*a % mo
    return res


def dis(a, b):
    return abs(a[0]-b[0]) + abs(a[1]-b[1])


def getpref(x):
    if x > 1:
        return (x)*(x-1) >> 1
    else:
        return 0


t = int(input())

for ti in range(t):
    n, m = map(int, input().split())
    l1 = [int(i) for i in input().split()]
    l2 = [int(i) for i in input().split()]
    s1 = []
    s2 = []
    for i, j in zip(l1, l2):
        if j == 2:
            s2.append(i)
        else:
            s1.append(i)
    s1.sort()
    s2.sort()
    # if len(s1) + 2*len(s2) < m:
    #     print(-1)
    #     continue
    cur = 0
    loss = 0
    ans = 1145141919810
    idx = 0
    for ind, i in enumerate(s1[::-1]):
        # if cur >= m:
        #     idx = len(s1) - ind - 1
        #     break
        cur += i
        loss += 1
    while idx < len(s1) and cur-s1[idx] >= m:
        cur -= s1[idx]
        idx += 1
        loss -= 1
    if cur >= m:
        ans = min(ans, loss)
    while s2:
        cur += s2.pop()
        loss += 2
        while idx < len(s1) and cur-s1[idx] >= m:
            cur -= s1[idx]
            idx += 1
            loss -= 1
        if cur >= m:
            ans = min(ans, loss)

    if ans == 1145141919810:
        print(-1)
    else:
        print(ans)
