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 ll long long
#define vi vector<int>
#define pb push_back
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()

#define pii pair<int, int>
#define mp make_pair
#define f first
#define s second

#define mii map<int, int>
#define umii unordered_map<int, int>
#define li list<int>
#define MOD 1000000007

#define loop(i,s,n) for(int i=s; i<n; i++)

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(0);
	
	#ifdef nkTy
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	#endif

	/* ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Code ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ */
	int t; cin >> t;
	int n, m, x, cnt1, cnt2, ans;
	vi arr1(2e5), arr2(2e5);
	ll sum1, sum2;

	while (t--) {
		cin >> n >> m;
		cnt1 = 0; cnt2 = 0;

		loop(i,0,n) cin >> arr1[i];
		loop(i,0,n) {
			cin >> x;
			if (x == 2) {
				arr2[cnt2] = arr1[i];
				cnt2 += 1;
			} else {
				arr1[cnt1] = arr1[i];
				cnt1 += 1;
			}
		}
		sort(arr1.begin(), arr1.begin()+cnt1);
		sort(arr2.begin(), arr2.begin()+cnt2);

		sum1 = 0; sum2 = 0;
		loop(i,0,cnt2) sum2 += arr2[i];
		ans = INT_MAX;

		/*
		5 3 2 1 4	--->	2 3 4 ; 1 5
		2 1 1 2 1	--->	1 1 1 ; 2 2
		*/

		int j=cnt1, k=0;
		while (1) {
			while (k<cnt2 && sum1+sum2-arr2[k] >= m) {
				sum2 -= arr2[k];
				k++;
			}

			if (sum1+sum2 >= m) ans = min(ans, 2*(cnt2-k)+(cnt1-j));

			if (j<=0) break;
			else {
				j--;
				sum1 += arr1[j];
			}
		}

		if (ans == INT_MAX) ans = -1;
		cout << ans << endl;
	}

	return 0;
}

/*
	2
   -1
	6
	4
	3
	3
*/
