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>
#define mod 1000000007
#define pb push_back
#define lb lower_bound
#define fi first
#define se second
#define int long long
using namespace std;
typedef long long LL;
typedef pair<LL, LL> pll;
LL tc, n, m, a[200200], b[200200], c[200200], d[200200];
int32_t main()
{
    scanf("%lld", &tc);
    while(tc--)
    {
        scanf("%lld %lld", &n, &m);
        for(int i=0; i<n; i++) scanf("%lld", &a[i]);
        for(int i=0; i<n; i++) scanf("%lld", &b[i]);
        LL cnt1=0, cnt2=0;
        for(int i=0; i<n; i++)
        {
            if(b[i]==1) c[cnt1++]=a[i];
            else d[cnt2++]=a[i];
        }
        sort(c, c+cnt1);
        sort(d, d+cnt2);
        LL k1=0, k2=0, cur=0, ans=0;
        while(k1<cnt1 || k2<cnt2)
        {
            if(cur>=m) break;
            if(k1<cnt1 && (2*c[cnt1-1-k1]>=d[cnt2-1-k2] || k2>=cnt2 || cur+c[cnt1-1-k1]>=m))
                cur+=c[cnt1-1-k1], ans+=1, k1++;
            else cur+=d[cnt2-1-k2], ans+=2, k2++;
        }
        k1--;
        LL ans1=ans;
        while(k2<=cnt2)
        {
            while(cur-c[cnt1-1-k1]>=m && k1>=0) cur-=c[cnt1-1-k1], ans1-=1, k1--;
            ans=min(ans, ans1);
            cur+=d[cnt2-1-k2], ans1+=2, k2++;
        }
        if(cur<m) printf("-1\n");
        else printf("%lld\n", ans);
    }
}
