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.
#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long int ull;
typedef long long int ll;
#define int long long
#define fi(i,n) for(int i=0;i<n;i++)
#define f(i, a, b) for(int i=a;i<b;i++)
#define vi vector<int>
#define pb push_back
#define MOD 1000000007
#define pii pair<int, int>
#define ff first
#define ss second
#define setzero(a) memset(a,0,sizeof(a))
#define prDouble(x) cout<<fixed<<setprecision(10)<<x;
#define SetBits(x) __builtin_popcount(x)
#define in_range(x, y, r, c) (x >= 0 && x < r && y >= 0 && y < c)
ll digits(ll n)
{
    return floor(log10(double(n))) + 1;
}
void faster()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
}
ll modularExponentiation(ll x,ll n,ll M)
{
    ll result = 1;
    while(n>0)
    {
        if(n % 2 ==1)
        {
            result = ((result%M)*(x%M))%M;
        }
        x  = ((x%M)*(x%M))%M;
        n = n/2;
    }
    return result%M;
}
bool ispower(int n)
{
    if(n%2==1)
        return false;
    while(n>0)
    {
        if(n%2 == 1)
            return false;
        n = n/2;
    }
    return true;
}

// {1,{2,3}} {2,3}
//4 4

int32_t main()
{
    faster();
    int t;
    cin>>t;
    while(t--)
    {
        int n,m;
        cin>>n>>m;
        int points[n];
        fi(i,n)cin>>points[i];
        vector<int>one_penalty,two_penalty;
        vector<pair<int,pair<int,int>>>dp(2*n + 1);//dp[i] will store max ans till i and penalty
        fi(i,n)
        {
            int x;
            cin>>x;
            if(x==1)
                one_penalty.push_back(points[i]);
            else
                two_penalty.push_back(points[i]);
        }
        int si1 = one_penalty.size();
        int si2 = two_penalty.size();
        sort(one_penalty.begin(),one_penalty.end());
        sort(two_penalty.begin(),two_penalty.end());
        if(si1 - 1>=0)
        {
            dp[1].first = one_penalty[si1 - 1];
            dp[1].second.first = si1 - 2;
            dp[1].second.second = si2 -1;
        }
        else
        {
            dp[1].first = -1;
            dp[1].second.first = -1;
            dp[1].second.second = -1;
        }
        dp[0].first = 0;
        dp[0].second.first = si1 - 1;
        dp[0].second.second = si2 - 1;
        //cout<<dp[1].first<<" ";
        int ans = -1;
        if(dp[1].first>=m)
        {
            ans = 1;
            cout<<ans<<endl;
            continue;
        }
        for(int i = 2; i<=2*n; i++)
        {
            int ma1 = -1;
            int ma2 = -1;
            int pre1 = dp[i - 1].first;
            int pre1first = dp[i - 1].second.first;
            int pre1second = dp[i - 1].second.second;
            //cout<<pre1<<" "<<pre1first<<" "<<pre1second<<endl;
            if(pre1first>=0 && pre1 != -1)
            {
                ma1 = pre1 +one_penalty[pre1first];
            }
            int pre2 = dp[i - 2].first;
            int pre2first = dp[i - 2].second.first;
            int pre2second = dp[i - 2].second.second;
            if(pre2second>=0 && pre2 != -1)
            {
                ma2 = pre2 + two_penalty[pre2second];
            }
            if(ma1 == -1 && ma2 == -1)
            {
                dp[i].first = -1;
                dp[i].second.first = -1;
                dp[i].second.first = -1;
            }
            else if(ma1>ma2)
            {
                dp[i].first = ma1;
                dp[i].second.first = dp[i - 1].second.first - 1;
                dp[i].second.second = dp[i -1].second.second;
            }
            else
            {
                dp[i].first =ma2;
                dp[i].second.first = dp[i - 2].second.first;
                dp[i].second.second = dp[i - 2].second.second - 1;
            }
            //cout<<dp[i].first<<" ";
            if(dp[i].first >= m)
            {
                ans = i;
                break;
            }
            //cout<<dp[i].first<<" ";

        }
        cout<<ans<<endl;

    }
}
