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 <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
#include <functional>
using namespace std;
 
typedef long long ll;
 
const int MAXN = 200005;
const int EPS = 1e-8;
const int INF = 1e9+7;
 
ll a[MAXN], b[MAXN];
ll s[MAXN];
 
bool cmp(ll a,ll b){
	return a > b;
}
 
int main(){
	int t;
	scanf("%d", &t);
	while(t--){
		ll n, m;
		scanf("%lld %lld", &n, &m);
		ll ans = 0;
		for (int i = 0; i < n;i++){
			scanf("%lld", &s[i]);
			ans += s[i];
		}
		if(ans<m){
			for (int i = 0; i < n;i++){
				int k;
				scanf("%d", &k);
			}
			printf("-1\n");
			continue;
		}
		int sa = 0, sb = 0;
		for (int i = 0; i < n;i++){
			int k;
			scanf("%d", &k);
			if(k==1) a[sa++] = s[i];
			else b[sb++] = s[i];
		}
		sort(a, a + sa, cmp);
		sort(b, b + sb, cmp);
		ans = 0;
		int sum = 0;
		int smin = INF;
		int i, j;
		for (i = 0; i < sa;i++){
			if(ans>=m){
				smin = min(smin, sum);
				break;
			}
			ans += a[i];
			sum++;
		}
		i--;
		if(ans>=m) smin = min(smin, sum);
		for (j = 0; j < sb;j++){
			ans += b[j];
			sum += 2;
			if(ans>=m&&i>=0){
				while(ans>=m&&i>=0){
					ans -= a[i];
					sum--;
					i--;
				}
				if(ans<m){
					i++;
					ans += a[i];
					sum++;
				}
			}
			if(ans>=m) smin = min(smin, sum);
		}
		printf("%d\n", smin);
	}
	return 0;
}