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.
''' ===============================
    -- @uthor : Kaleab Asfaw
    -- Handle : kaleabasfaw2010
    -- Bio    : High-School Student
    ==============================='''

# Fast IO
import sys
import os
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); input = lambda: sys.stdin.readline().rstrip("\r\n")
# Others
# from math import floor, ceil, gcd
# from decimal import Decimal as d
mod = 10**9+7
def lcm(x, y): return (x * y) / (gcd(x, y))
def fact(x, mod=mod):
    ans = 1
    for i in range(1, x+1): ans = (ans * i) % mod
    return ans
def arr2D(n, m, default=0): return [[default for j in range(m)] for i in range(n)]
def arr3D(n, m, r, default=0): return [[[default for k in range(r)] for j in range(m)] for i in range(n)]
def sortDictV(x): return {k: v for k, v in sorted(x.items(), key = lambda item : item[1])}
class DSU:
    def __init__(self, length): self.length = length; self.parent = [-1] * self.length  # O(log(n))
    def getParent(self, node, start):  # O(log(n))
        if node >= self.length: return False
        if self.parent[node] < 0:
            if start != node: self.parent[start] = node
            return node
        return self.getParent(self.parent[node], start)
    def union(self, node1, node2):  # O(log(n))
        parent1 = self.getParent(node1, node1); parent2 = self.getParent(node2, node2)
        if parent1 == parent2: return False
        elif self.parent[parent1] <= self.parent[parent2]: self.parent[parent1] += self.parent[parent2]; self.parent[parent2] = parent1
        else: self.parent[parent2] += self.parent[parent1]; self.parent[parent1] = parent2
        return True
    def getCount(self, node):  return -self.parent[self.getParent(node, node)] # O(log(n))

def exact(num):
    if abs(num - round(num)) <= 10**(-9):return round(num)
    return num

def solve(n, m, a, b):
    if sum(a) < m: return -1

    ones = []
    twos = []
    
    for i in range(n):
        if b[i] == 1:
            ones.append(a[i])
        else:
            twos.append(a[i])

    ones = sorted(ones, reverse=True)
    twos = sorted(twos, reverse=True)

    # L = 0
    # R = 2 * len(twos)

    # ans = -1
    # while L <= R:
    #     mid = L + (R - L) // 2
    #     can = False




    # preO = []
    # preT = []

    # for i in ones:
    #     if  len(preO) == 0: preO.append(i)
    #     else:
    #         preO.append(preO[-1] + i)
    
    # for i in twos:
    #     if  len(preT) == 0: preT.append(i)
    #     else:
    #         preT.append(preT[-1] + i)

    
    i, j = len(twos)-1, 0
    k = 0
    minn = float("inf")
    summ = sum(twos)

    # print(ones)
    # print(twos)
    # print(summ)
    while True:
        # print("O", summ, i, j, k, (i+1)*2 + j - k)
        if summ >= m:
            minn = min(minn, (i+1)*2 + j - k)
            if i == -1:
                summ -= ones[k]
                k += 1
            else:
                summ -= twos[i] 
                i -=1
        else:
            if j == len(ones) :
                break
            summ += ones[j]
            j += 1
        if summ >= m:
            minn = min(minn, (i+1)*2 + j - k)

    # Infinte loop
    
    return minn
    


for _ in range(int(input())): # Multicase
    n, m = list(map(int, input().split()))
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))
    print(solve(n, m, a, b))