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 mp make_pair 
#define pb push_back 
#define MAX(a, b) a = max(a, b) 
#define MIN(a, b) a = min(a, b) 
#define SQR(x) ((LL)(x) * (x)) 
#define RESET(a, b) memset(a, b, sizeof(a))
#define ALL(v) v.begin(), v.end() 
#define SORT(v) sort(ALL(v)) 
#define PERMUTE next_permutation
#define fi first 
#define se second

#define MOD 1000000009 //1E9+9

using namespace std;

typedef int64_t i64;

typedef pair<i64,i64> p64;
typedef vector<i64> v64;
typedef vector<p64>  v6464;
typedef vector < v64 > vv64;

typedef double dle;
typedef long double ldle;

typedef pair<int,int> pii; 
typedef vector<int> vi;
typedef vector< pii > vii;
typedef vector< vi > vvi;


typedef pair<string,string> pss;
typedef vector<string> vs;
typedef vector<pss> vss;

typedef pair<int,string> pis;
typedef vector<pis> vis;
typedef pair<string,int> psi;
typedef vector<psi> vsi;

typedef map<string,string> mss;
typedef map<int,string> mis;
typedef map<string,int> msi;

int main(){
    ios_base::sync_with_stdio(0);cin.tie(0);
    int t,n,m,bi;
    cin >> t;
    while(t--){
        cin >> n >> m ;
        vi a1,a(n);
        v64 a2; 
        int ans = INT_MAX;
        for(int i = 0 ; i < n ; i ++)
            cin >> a[i];
        for(int i = 0 ; i < n ; i ++){
            cin >> bi;
            if(bi==2)
                a2.pb(a[i]);
            else
                a1.pb(a[i]);
        }
        sort(a1.begin(),a1.end()),a1.pb(0),reverse(a1.begin(),a1.end());
        sort(a2.begin(),a2.end()),reverse(a2.begin(),a2.end());
        int s1=a1.size(),s2=a2.size();
        for(int i = 1 ; i < s2 ; i ++)
            a2[i]+=a2[i-1];
        i64 acum = 0 ; 
        for(int i = 0 ; i < s1; i ++){
            acum += a1[i];
            ///pure
            if(acum>=m){
                MIN(ans,i);
                break;
            }else if(s2){
                int l = 0, r = s2-1;
                while(l+1<r){
                    int mid = (l+r)/2;
                    if(acum+a2[mid]<m)
                        l = mid + 1;
                    else
                        r = mid;
                }
                ///answer is in r
                int ind = acum+a2[l]>=m ? l : r;
                if(acum+a2[ind]>=m)
                    MIN(ans,i+(ind+1)*2);
            }
        }
        cout << (ans==INT_MAX?-1:ans) << "\n";
    }
    return 0;
}
 	 	  		 					  			 					  		 	