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.
for _ in range(int(input())):
    n,k = map(int, input().split())
    x = k
    a = list(map(int,input().split()))
    b = list(map(int,input().split()))

    dp = []
    for i in range(n+1):
        dp.append([0]*(k+1))

    infy = float("inf")
    for i in range(1,k+1):
        dp[0][i] = infy

    for i in range(1,n+1):
        for j in range(1,k+1):
            # print(a[i-1])
            if(a[i-1]<=j):
                dp[i][j] = min(b[i-1]+dp[i-1][j-a[i-1]] , dp[i-1][j])
            else:
                dp[i][j] = dp[i-1][j]

    # for i in dp:
    #     print(i)
    # if(dp[n][k] == infy):
    #     print(-1)
    # else:
    #     print(dp[n][k])










    o = []
    t = []
    for i in range(n):
        if(b[i]==1):
            o.append(a[i])
        else:
            t.append(a[i])
    o.sort()
    t.sort()
    i = len(o)-1
    j = len(t)-1
    c = 0
    while(k>0 and i>0 and j>=0):
        if(o[i]+o[i-1] >= t[j]):
            if(o[i]>=k):
                k -= o[i]
                i -= 1
                c += 1
            else:
                k -= o[i]+o[i-1]
                i -= 2
                c += 2
        else:
            k -= t[j]
            j -= 1
            c += 2
    # print(k,c,i,j)
    # print(o[:i+1])
    # print(t[:j+1])
    if(k>0):
        if(i>=0 and j>=0):
            while(k>0 and i==0 and j>=0):
                if(o[i] >= t[j] or (k<o[i] and k<t[j])):
                    k -= o[i]
                    i -= 1
                    c += 1
                else:
                    k -= t[j]
                    c += 2
                    j -= 1

        if(i>=0 and j<0):
            while(k>0 and i>=0):
                k -= o[i]
                i -= 1
                c += 1
        if(j>=0 and i<0):
            while(k>0 and j>=0):
                k -= t[j]
                j -= 1
                c += 2


    # print("k",k,c)
    if(k>0):
        print(min(-1,dp[n][x]))
    else:
        if(dp[n][x]==infy):
            print(c)
        else:
            print(min(c,dp[n][x]))
