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>

#define fo(i,n) for(int i=0;i<n;i++)
#define dsort(arr,n) sort(arr, arr + n, greater<int>())
#define vsortd(a) sort(a.begin(), a.end(), greater<double>())
#define vsort(a) sort(a.begin(), a.end())
#define dbg(x) cout<<#x<<'='<<(x)<<endl;
#define dbg2(x,y) cout<<#x<<'='<<(x)<<endl<<#y<<'='<<(y)<<endl;
#define dbg3(x,y,z) cout<<#x<<'='<<(x)<<endl<<#y<<'='<<(y)<<endl<<#z<<"="<<(z)<<endl;
#define dbg4(x,y,z,w) cout<<#x<<'='<<(x)<<endl<<#y<<'='<<(y)<<endl<<#z<<"="<<(z)<<endl<<#w<<"="<<(w)<<endl;
#define minheap priority_queue <int, vector<int>, greater<int> >
#define maxheap priority_queue <int>
#define test int tc;cin>>tc;while(tc--)
#define ll long long int
#define pb push_back
#define ff first
#define ss second
#define mpr make_pair
#define mkv vector<int>
#define allzero(dp) memset(dp,0,sizeof(dp))
#define endl "\n"
#define print(x) cout<<x<<endl;
#define full(v) v.begin(),v.end()
#define inf INT_MAX

const int mod=1e9+7;
using namespace std;

void fastio()
{
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    #ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #endif
}

bool compare(pair<ll,ll> a,pair<ll,ll> b){
    if(a.first>b.first)return true;
    if(a.first==b.first)return a.second<b.second;
    return false;
}

ll power(ll x, ll y){
    ll res = 1;x=x%mod;
    while(y>0){
        if(y&1)res=(res*x)%mod;
        y=y>>1;x=(x*x)%mod;}
    return res;
}

void solve()
{
    ll n,m;cin>>n>>m;
    ll a[n],b[n];
    fo(i,n)cin>>a[i];
    fo(i,n)cin>>b[i];
    vector<ll>hi,lo;
    fo(i,n){
        if(b[i]==2)
            hi.pb(a[i]);
        else
            lo.pb(a[i]);
    }
    vsortd(hi);
    vsortd(lo);
    int i=0,j=0;
    ll tmp=0;
    while(i<hi.size()&&j<lo.size()){
        if(hi[i]>lo[j]){
            tmp+=hi[i];
            i+=1;
        }
        else{
            tmp+=lo[j];
            j+=1;
        }
        if(tmp>=m)
            break;
    }
    if(tmp<m){
        while(i<hi.size()){
            tmp+=hi[i];
            i+=1;
            if(tmp>=m)
                break;
        }
        while(j<lo.size()){
            tmp+=lo[j];
            j+=1;
            if(tmp>=m)
                break;
        }
    }
    if(tmp<m){
        print(-1)
        return;
    }
    if(i==0){
        print(j)
        return;
    }
    if(j==lo.size()){
        int ans=i*2+j;
        print(ans)
        return;
    }
    i-=1;
    while(i>=0){
        if(j>=lo.size()){
            break;
        }
        if(tmp-hi[i]+lo[j]>=m){
            tmp=tmp-hi[i]+lo[j];
            i-=1;
            j+=1;
            continue;
        }
        if(j>=lo.size()-1){
            break;
        }
        if(tmp+lo[j]+lo[j+1]-hi[i]>=m){
            tmp=tmp-hi[i]+lo[j]+lo[j+1];
            j+=2;
            i-=1;
        }
        else{
            break;
        }
    }
    int ans=(i+1)*2+j;
    print(ans)
}

int main()
{
    //cout<<fixed<<setprecision(20);

  //  fastio();
    test
    solve();
}
