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
input = sys.stdin.readline
from functools import cmp_to_key;

def pi():
    return(int(input()))
def pl():
    return(int(input(), 16))
def ti():
    return(list(map(int,input().split())))
def ts():
    s = input()
    return(list(s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))
f = [];
def factMod(n,m):
    global f;
    f = [1 for i in range(n+1)];
    f[0] = 1;
    for i in range(1,n+1):
        f[i] = (f[i-1]*i)%m;

def fact(n):
    global f;
    f = [1 for i in range(n)];
    for i in range(1,n):
        f[i] = f[i-1]*i;

def fast_mod_exp(a,b,m):
    res = 1;
    while b > 0:
        if b & 1:
            res = (res*a)%m;
        a = (a*a)%m;
        b = b >> 1;
    return res;

def inverseMod(n,m):
    return fast_mod_exp(n,m-2,m);

def ncrMod(n,r,m):
    if r == 0: return 1;
    if n < r: return 0;
    return ((f[n]*inverseMod(f[n-r],m))%m*inverseMod(f[r],m))%m;

def ncr(n,r):
    if r == 0: return 1;
    if n < r: return 0;
    return (f[n])/(f[r]*f[n-r]);
def nc2(n):
    return (n*(n-1))/2;

def main():
    D();

def cmp(a,b):
    if a[0] > b[0]: return 1;
    elif a[0] < b[0]: return -1;
    else: return 0;

def D():
    t = pi();
    z = 0;
    try:
        while z < t:
            z += 1;
            [n,m] = ti();
            a,b = ti(),ti();
            if z == 37:
                sa = '';
                sb = '';
                for i in range(n):
                    sb += (str(b[i])+'_');
                    sa += (str(b[i])+'_');
                print(str(n)+'_'+str(m));
                print(sa);
                print(sb);
            one,two = [],[];
            for i in range(n):
                if b[i] == 1: one.append(a[i]);
                else: two.append(a[i]);
            one = sorted(one,reverse=True);
            two = sorted(two,reverse=True);
            for i in range(1,len(one)):
                one[i] += one[i-1];

            for i in range(1,len(two)):
                two[i] += two[i-1]
            x = -1;
            for i in range(len(one)):
                if one[i] >= m:
                    x = i;
                    break;
            ans = 99999999999999999;
            if x != -1:
                ans = x+1;
                y = 0;
                while x >= 0 and y < len(two):
                    while x >= 0 and one[x]+two[y] >= m:
                        ans = min(ans, x+1+2*(y+1));
                        x -= 1;
                    y += 1;
                print(ans);
            else:
                x = len(one)-1;
                y = 0;
                f = 0;
                if x == -1:
                    x = 0;
                    one = [0];
                    f = 1;
                while y < len(two) and one[x]+two[y] < m:
                    y += 1;
                if y >= len(two):
                    print(-1);
                    continue;
                ans = (x+1 if f == 0 else 0)+2*(y+1);
                p = x;
                while y < len(two) and x >= 0:
                    while x >= 0 and one[x]+two[y] >= m:
                        ans = min(ans, x+1+2*(y+1));
                        x -= 1;
                    y += 1;
                print(ans);
    except: print(sys.exc_info()[0]);

main();