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;

#define _for(i, a, b) for (int i = (a); i < (b); ++i)
#define _rep(i, a, b) for (int i = (a); i <= (b); ++i)
#define FOR(i, a, b) for (int i = (a); i >= (b); --i)

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

const int N = 200000 + 10, INF = 0x3f3f3f3f;
int a[N];
ll suma[N], sumb[N];

int main()
{
#ifdef LOCAL
	freopen("data.in", "r", stdin);
#endif
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	int T;
	cin >> T;
	while (T--)
	{
		memset(suma, 0, sizeof suma);
		memset(sumb, 0, sizeof sumb);
		int n, m;
		ll sum = 0;
		cin >> n >> m;
		_for(i, 0, n) cin >> a[i], sum += a[i];
		vi cnta, cntb;
		_for(i, 0, n)
		{
			int x; cin >> x;
			if (x == 1) cnta.push_back(a[i]);
			else cntb.push_back(a[i]);
		}

		if (sum < m)
		{
			cout << -1 << endl;
			continue;
		}

		sort(cnta.begin(), cnta.end(), greater<int>());
		sort(cntb.begin(), cntb.end(), greater<int>());
		int sza = cnta.size(), szb = cntb.size();
		_for(i, 0, sza) suma[i + 1] = suma[i] + cnta[i];
		_for(i, 0, szb) sumb[i + 1] = sumb[i] + cntb[i];

		int ans = INF;
		_rep(i, 0, sza)
		{
			if (suma[i] >= m)
			{
				ans = min(ans, i);
				break;
			}
			
			int l = 1, r = szb;
			while (l < r)
			{
				int mid = l + r >> 1;
				if (sumb[mid] >= m - suma[i]) r = mid;
				else l = mid + 1;
			}
			if (l == r && sumb[l] >= m - suma[i]) ans = min(ans, i + l * 2);
		}
		cout << ans<< endl;
	}
	
	return 0;
}