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.
// first second push_back unordered return continue break vector visited check flag bool while iterator begin end lower_bound upper_bound temp true false ll_MAX ll_MIN insert erase clear pop push compare ll64_MAX ll64_MIN  reverse replace stringstream string::npos length substr front priority_queue

#ifndef ONLINE_JUDGE
#include "debug.h"
#else
#include <bits/stdc++.h>
using namespace std;
#define debug(...) 42
#endif

#define endl '\n'
#define ll long long
#define minpq priority_queue <int, vector<int>, greater<int> >
using db = long double;
using pll = pair < ll, ll >;
#define all(a)      (a).begin(),(a).end()
#define unique_sort(x) sort(all(x)), x.resize(distance(x.begin(), unique(all(x))))
#define hell 1000000007
#define infl LLONG_MAX
#define countBits(n)	__builtin_popcount(n)
#define cout_p(x, p) cout << fixed << setprecision((p)) << x << endl     //print with precision

#define tc(tt) \
    ll tt;    \
    cin >> tt; \
    for (ll _tt = 0; _tt < tt; _tt++) // testcase 

int32_t main() {
	ios::sync_with_stdio(0); 		cin.tie(0); cout.tie(0);

	ll t, i, x, j, y, z, k, n;
	tc(tt) {
		cin >> n >> k;
		vector<pll> a(n);
		ll org = 0, ans = infl;
		for (auto &zx : a)	cin >> zx.first;
		for (auto &zx : a)	cin >> zx.second;

		std::vector<ll> v1, v2;
		for (auto &[zx, y] : a) {
			if (y == 1)	v1.push_back(zx);
			else v2.push_back(zx);
			org += y;
		}

		if (v2.size() > 0) {
			sort(all(v2));		reverse(all(v2));
		}
		if (v1.size() > 0) {
			sort(all(v1));		reverse(all(v1));
		}

		ll pre = 0;
		for (ll i = 0; i < v1.size(); i++)	pre += v1[i], v1[i] = pre;
		pre = 0;
		for (ll i = 0; i < v2.size(); i++)	pre += v2[i], v2[i] = pre;
		debug(k, v1, v2);

		for (ll i = 0; i < v1.size(); i++) {
			x = v1[i];
			if (x >= k) {
				ans = min(ans, i + 1);
			} else {
				y = i + 1;
				j = lower_bound(all(v2), k - x) - v2.begin();
				y += 2 * (j + 1);
				if (j != v2.size())	ans = min(ans, y);
			}
		}

		for (ll i = 0; i < v2.size(); i++) {
			x = v2[i];
			if (x >= k) {
				ans = min(ans, 2 * i + 2);
			} else {
				y = 2 * i + 2;
				j = lower_bound(all(v1), k - x) - v1.begin();
				y += (j + 1);
				if (j != v1.size())	ans = min(ans, y);
			}
		}

		if (v2.size() > 0 and v2.back() >= k)	ans = min(ans, (ll)(2 * int(v2.size())));
		if (v1.size() > 0 and v1.back() >= k)	ans = min(ans, (ll)(int(v1.size())));

		if (ans == infl)	ans = -1;
		cout << ans << "\n";
		debug(ans);
	}
}

