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 INF 1001001001
typedef long long ll;
using namespace std;
template <typename T>
void show(vector<T> &a) {
	for (int i = 0; i < a.size(); i++) {
		cout << a[i];
		if (i == a.size() - 1) {
			cout << endl;
		} else {
			cout << " ";
		}
	}
}
template <typename T>
void show(T &a) {
	cout << a << endl;
}

int main() {
	int tt;
	cin >> tt;
	while (tt--) {
		int n, m;
		cin >> n >> m;
		vector<int> a(n);
		for (int i = 0; i < n; i++) {
			cin >> a[i];
		}
		vector<int> b(n);
		for (int i = 0; i < n; i++) {
			cin >> b[i];
		}

		vector<int> v1, v2;
		for (int i = 0; i < n; i++) {
			if (b[i] == 1) {
				v1.push_back(a[i]);
			} else {
				v2.push_back(a[i]);
			}
		}
		sort(v1.begin(), v1.end(), greater<int>());
		sort(v2.begin(), v2.end(), greater<int>());
		int nv1 = v1.size(), nv2 = v2.size();
		vector<ll> sv1(nv1 + 1, 0), sv2(nv2 + 1, 0);
		for (int i = 0; i < nv1; i++) {
			sv1[i + 1] += sv1[i] + v1[i];
		}
		for (int i = 0; i < nv2; i++) {
			sv2[i + 1] += sv2[i] + v2[i];
		}

		// show(sv1);
		// show(sv2);

		int ans = INF;
		for (int i = 0; i <= nv1; i++) {
			ll tot = sv1[i];
			ll left = m - tot;
			if (left > sv2[nv2]) continue;
			int ok = nv2, ng = -1, mid;
			while (abs(ok - ng) > 1) {
				mid = (ok + ng) / 2;
				if (left <= sv2[mid]) {
					ok = mid;
				} else {
					ng = mid;
				}
			}
			// show(i);
			// show(ok);
			ans = min(ans, i + ok * 2);
		}
		if (ans == INF) ans = -1;
		cout << ans << endl;
	}
	return 0;
}