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 <random>
#include <stdio.h>
#include <bits/stdc++.h>
#include <unordered_set>
#include <unordered_map>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> pll;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;
typedef pair<int, int> pii;

#define ls rt<<1
#define rs ls|1
#define fst first
#define sed second
#define psb push_back
#define lowbit(x) ((x) & -(x))
#define SZ(x) ((int)(x).size())
#define sqr(x) (1LL * (x) * (x))
#define ALL(x) (x).begin(), (x).end()
#define MEM(x, y) memset((x), (y), sizeof(x))
#ifdef LOCAL
#define debug(x) cout << "[" __FUNCTION__ ": " #x " = " << (x) << "]\n"
#define TIME cout << "RuningTime: " << clock() << "ms\n", 0
#else
#define TIME 0
#endif
#define hash_ 1000000009
#define gc p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2) ? EOF : *p1++;

template<class T> inline bool IsPri(T x) { if (x < 2) return false; for (T i = 2; i * i <= x; ++i) if (x % i == 0) return false; return true; }
template<class T> T gcd(T a, T b) { return b == 0 ? a : gcd(b, a % b); };
template<class T, class S, class D> T qpow(T a, S b, D mod) { T res = 1; for (; b > 0; b >>= 1) { if (b & 1) res = res * a % mod; a = a * a % mod; } return res; }
template<class T>double dpow(double a, T b) { double res = 1.0; for (; b > 0; b >>= 1) { if (b & 1)res = res * a; a = a * a; }return res; }
inline int read() { static char buf[1000000], * p1 = buf, * p2 = buf; int x = false; char ch = gc; bool sgn = false; while (ch != '-' && (ch < '0' || ch > '9')) ch = gc; if (ch == '-') sgn = true, ch = gc; while (ch >= '0' && ch <= '9') x = (x << 1) + (x << 3) + (ch ^ 48), ch = gc; return sgn ? -x : x; }

const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const int N = 2e5 + 10;
const double PI = acos(-1);
const double E = exp(1.0);

ll a[N], b[N];
ll k1[N], k2[N];
int main() {
#ifdef LOCAL
	freopen("D:/input.txt", "r", stdin);
	//freopen("D:/output.txt", "w", stdout);
#endif
	int t;
	cin >> t;
	while (t--) {
		int n, m;
		cin >> n >> m;
		ll sum = 0;
		for (int i = 1; i <= n; i++)
			cin >> a[i], sum += a[i];
		for (int i = 1; i <= n; i++)
			cin >> b[i];
		int tot1 = 0, tot2 = 0;
		for (int i = 1; i <= n; i++)
			if (b[i] == 1)
				k1[++tot1] = a[i];
			else
				k2[++tot2] = a[i];
		sort(k1 + 1, k1 + 1 + tot1), sort(k2 + 1, k2 + 1 + tot2);
		if (sum < m) {
			puts("-1");
			continue;
		}
		sum = 0;
		for (int i = 1; i <= tot1; i++)
			sum += k1[i];
		int j = tot2, i = 0;
		ll ans = LINF;
		while (i <= tot1) {
			while (j >= 1 && sum < m) {
				sum += k2[j--];
			}
			if (sum >= m) {
				ans = min(ll(tot1 - i + 2 * (tot2 - j)), ans);
			}
			sum -= k1[++i];
		}
		cout << ans << endl;
	}
	return TIME;
}
