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.
#ifndef _GLIBCXX_NO_ASSERT
#include <cassert>
#endif
#include <cctype>
#include <cerrno>
#include <cfloat>
#include <ciso646>
#include <climits>
#include <clocale>
#include <cmath>
#include <csetjmp>
#include <csignal>
#include <cstdarg>
#include <cstddef>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>

#if __cplusplus >= 201103L
#include <ccomplex>
#include <cfenv>
#include <cinttypes>
#include <cstdbool>
#include <cstdint>
#include <ctgmath>
#include <cwchar>
#include <cwctype>
#endif

// C++
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>

#if __cplusplus >= 201103L
#include <array>
#include <atomic>
#include <chrono>
#include <condition_variable>
#include <forward_list>
#include <future>
#include <initializer_list>
#include <mutex>
#include <random>
#include <ratio>
#include <regex>
#include <scoped_allocator>
#include <system_error>
#include <thread>
#include <tuple>
#include <typeindex>
#include <type_traits>
#include <unordered_map>
#include <unordered_set>
#endif

using namespace std;
#define tenp3 1001
#define tenp5 100001
#define tenp6 1000001
#define tenp9 1000000001
#define infinity 100000000000000000
#define mod 1000000007

typedef long long ll;
typedef vector<long long int> vll;
typedef pair<long long int, long long int> pll;
typedef vector<pair<long long int, long long int> > vpll;
// #define mat

#define pushb push_back
#define popb pop_back
#define F first
#define S second
#define mp make_pair

#define repe(i, a, b) for (long long int i = (long long int)(a); i < (long long int)(b); i++)
#define repv(x, y) for (auto x : y)
#define repvref(x, y) for (auto &x : y)
#define repit(it, x) for (auto it = x.begin(); it != x.end(); it++)
#define justSort(x) sort(x.begin(), x.end())
#define justSortR(x) sort(x.rbegin(), x.rend())
#define vec(typeOfVec, nameOfVec, sizeOfVec)      \
    vector<typeOfVec> nameOfVec(sizeOfVec);       \
    for (long long int i = 0; i < sizeOfVec; i++) \
    {                                             \
        cin >> nameOfVec[i];                      \
    }
#define print(x) cout << (x) << "\n";
#define rev(x) reverse(x.begin(), x.end())
#define copy(x, y) copy(x.begin(), x.end(), back_inserter(y));
#define mat(nameOfMat, typeOFMat, rows, cols, intitialValue) vector<vector<typeOFMat>> nameOfMat(rows, vector<typeOFMat>(cols, intitialValue));
#define gmat(nameOfMat, typeOFMat, rows, cols)                             \
    vector<vector<typeOFMat>> nameOfMat(rows, vector<typeOFMat>(cols, 0)); \
    for (long long int i = 0; i < rows; i++)                               \
    {                                                                      \
        for (long long int j = 0; j < cols; j++)                           \
        {                                                                  \
            cin >> nameOfMat[i][j];                                        \
        }                                                                  \
    }
#define pmat(nameOfMat)                                         \
    for (long long int i = 0; i < nameOfMat.size(); i++)        \
    {                                                           \
        for (long long int j = 0; j < nameOfMat[0].size(); j++) \
        {                                                       \
            cout << nameOfMat[i][j] << " ";                     \
        }                                                       \
        cout << "\n";                                           \
    }                                                           \
    cout << "\n";
#define preSum(fromArray, newArray, dataType)               \
    vector<dataType> newArray(fromArray.size() + 1);        \
    newArray[0] = 0;                                        \
    for (long long int qw = 0; qw < fromArray.size(); qw++) \
    {                                                       \
        newArray[qw + 1] = newArray[qw] + fromArray[qw];    \
    }
void printx(vll &x)
{
    for (auto i = x.begin(); i != x.end(); i++)
    {
        cout << (*i) << " ";
    }
    cout << "\n";
}
void printy(vll &x)
{
    for (auto i = x.begin(); i != x.end(); i++)
    {
        cout << (*i) << "\n";
    }
}

ll answer, n, k, m, p, current, best, median,tp;
ll gcd(ll a, ll b){
    if (b == 0)
       return a;
    return gcd(b, a % b); 
}

int solve(int tn)
{
    cin>>n>>m;
    vll a(n);
    vll b1,b2;
    repe(i,0,n){
        cin>>a[i];
    }
    b1.push_back(0);
    b2.push_back(0);
    repe(i,0,n){
        cin>>tp;
        if(tp==1){
            b1.push_back(a[i]);
        }
        else if(tp==2){
            b2.push_back(a[i]);
        }
    }
    // justSort(b1);
    sort(b1.rbegin(),b1.rend()-1);
    sort(b2.rbegin(),b2.rend()-1);
    repe(i,1,b1.size()){
        b1[i]+=b1[i-1];
    }
    repe(i,1,b2.size()){
        b2[i]+=b2[i-1];
    }
    if(b2.back()+b1.back()<m){
        cout<<-1<<"\n";
        return 0;
    }
    // repe()
    // printx(b1);
    // printx(b2);
    // justSort(b2);
    ll best=infinity;
    vll::iterator foundIt;
    repe(i,0,b1.size()){
        foundIt=lower_bound(b2.begin(),b2.end(),m-b1[i]);
        if(foundIt!=b2.end()){
            best=min(best,(i)+2*( foundIt-b2.begin()));
        }
    }
    cout<<best<<"\n";

    return 0;
}

int main()
{
    ll t = 1;
    cin >> t;
    repe(w, 1, t + 1)
    {
        solve(w);
    }
    return 0;
}
