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 string str;
typedef long long ll;
typedef double db;
typedef pair<int, int> pii;
typedef map<int, int> mii;
typedef vector<int> vi;
typedef vector<bool> vb;

#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define fi first
#define se second
#define pb push_back

#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)
#define iter(m) for(auto it = m.begin(); it != m.end(); it++)
#define iterr(m) for(auto it = m.rbegin(); it != m.rend(); it++)

#define speed ios_base :: sync_with_stdio(0);cin.tie(0);cout.tie(0);

const int MID = 1e5 + 3;
const int INF = 1e9 + 5;
const int MOD = 1e9 + 7;
#define PI 3.14159265358979323846264338327950L

int main(){
	speed;
	int q;
	cin >> q;
	while(q--){
		int n, m;
		cin >> n >> m;
		vector<ll> v(n), a, b;
		ll s = 0;
		for(int i = 0; i < n; ++i){
			cin >> v[i];
			s += v[i];
		}
		for(int i = 0; i < n; ++i){
			int x;
			cin >> x;
			if(x == 1){
				a.pb(v[i]);
			}
			else{
				b.pb(v[i]);
			}
		}
		if(m > s){
			cout << "-1\n";
		}
		else{
			sort(all(a));
			sort(all(b));
			a.pb(0);
			b.pb(0);
			reverse(all(a));
			reverse(all(b));
			for(int i = 1; i < a.size(); ++i){
				a[i] += a[i - 1];
			}
			for(int i = 1; i < b.size(); ++i){
				b[i] += b[i - 1];
			}
			int mn = INF;
			for(int i = 0; i < a.size(); ++i){
				int k = lower_bound(all(b), m - a[i]) - b.begin();
				if(k == b.size()){
					continue;
				}
				// cout << k << " ";
				mn = min(mn, i + 2 * k);
			}
			cout << mn << "\n";
		}
	}
}