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 int int64_t

using namespace std;

const int N = 1e6+1;
const int mod = 1000000007;
int fact[N];
int invFact[N];


int gcd(int a,int b){
	if(a==0)
		return b;
	
	return gcd(b%a, a);
}

int fast_pow(int a, int p) {
    int res = 1;
    while (p) {
        if (p % 2 == 0) {
            a = (a * a) % mod;
            p /= 2;
        } else {
            res = (res * a) % mod;
            p--;
        }
    }
    return res;
}

int C(int n, int k) {
    if (k > n) {
        return 0;
    }
    return fact[n] * invFact[k] % mod * invFact[n - k] % mod;
}

bool comp(pair<int,int> p1, pair<int,int> p2){
	if(p1.first < p2.first)
		return true;
	else
		if( p1.second > p2.second)
			return true;
		
	return false;
}



void solve(){
	
}





signed main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	
	/*fact[0] = invFact[0] = 1;
    for (int i = 1; i < N; i++) {
        fact[i] = (fact[i - 1] * i) % mod;
        invFact[i] = fast_pow(fact[i], mod - 2);
    }*/
 
	int t = 1;
	cin>>t;
	
	while(t--){
		int n,m;
		cin>>n>>m;
		
		vector<int> a(n);
		vector<int> b(n);
		
		for(int i =0 ;i < n; i++)
			cin>>a[i];
			
		for(int i = 0; i < n ; i++)
			cin>>b[i];
			
		vector<int> ones,twos;
		
		for(int i = 0;i < n;i++){
			if(b[i] == 2) twos.push_back(a[i]);
			else	ones.push_back(a[i]);
		}

		sort(twos.begin(), twos.end(), greater<int>());
		sort(ones.begin(), ones.end(), greater<int>());
		
		
		vector<int> prf1(ones.size()+1);
		vector<int> prf2(twos.size()+1);
	
		
		prf1[0] = 0;
		
		for(int i = 1;i <= ones.size(); i++){
			prf1[i] = prf1[i-1] + ones[i-1];
		}
		
		prf2[0] = 0;
		
		for(int i = 1;i <= twos.size(); i++){
			prf2[i] = prf2[i-1] + twos[i-1];
		}
		int ans = INT_MAX;
		for(int i = 0;i <=  ones.size(); i++){
			
			int x,y;
		
			int toFind = m;
			if(i)
				toFind -= prf1[i];
			 
			x = i;
			y = lower_bound(prf2.begin(), prf2.end(), toFind) - prf2.begin();
			
		//	cout<<y<<"\n";
			if(y > twos.size())
				continue;
			
		//	cout<<x<<" "<<y<<"\n";
			ans = min(ans, x+2*y);
		}
		
		if(ans == INT_MAX) cout<<"-1\n";
		else cout<<ans<<"\n";
		
	}
}