Consider a sequence of integers a_1, a_2, …, a_n. In one move, you can select any element of the sequence and delete it. After an element is deleted, all elements to the right are shifted to the left by 1 position, so there are no empty spaces in the sequence. So after you make a move, the sequence's length decreases by 1. The indices of the elements after the move are recalculated.

E. g. let the sequence be a=[3, 2, 2, 1, 5]. Let's select the element a_3=2 in a move. Then after the move the sequence will be equal to a=[3, 2, 1, 5], so the 3-rd element of the new sequence will be a_3=1 and the 4-th element will be a_4=5.

You are given a sequence a_1, a_2, …, a_n and a number k. You need to find the minimum number of moves you have to make so that in the resulting sequence there will be at least k elements that are equal to their indices, i. e. the resulting sequence b_1, b_2, …, b_m will contain at least k indices i such that b_i = i.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.

Each test case consists of two consecutive lines. The first line contains two integers n and k (1 ≤ k ≤ n ≤ 2000). The second line contains a sequence of integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n). The numbers in the sequence are not necessarily different.

It is guaranteed that the sum of n over all test cases doesn't exceed 2000.

Output

For each test case output in a single line:

  * -1 if there's no desired move sequence; 
  * otherwise, the integer x (0 ≤ x ≤ n) — the minimum number of the moves to be made so that the resulting sequence will contain at least k elements that are equal to their indices. 

Example

Input


4
7 6
1 1 2 3 4 5 6
5 2
5 1 3 2 3
5 2
5 5 5 5 4
8 4
1 2 3 3 2 2 5 5


Output


1
2
-1
2

Note

In the first test case the sequence doesn't satisfy the desired condition, but it can be provided by deleting the first element, hence the sequence will be [1, 2, 3, 4, 5, 6] and 6 elements will be equal to their indices.

In the second test case there are two ways to get the desired result in 2 moves: the first one is to delete the 1-st and the 3-rd elements so that the sequence will be [1, 2, 3] and have 3 elements equal to their indices; the second way is to delete the 2-nd and the 3-rd elements to get the sequence [5, 2, 3] with 2 desired elements.
import java.io.*;
import java.util.*;

import static java.lang.Math.*;

public class Round {
    FastScanner in;
    PrintWriter out;

    private void solve() throws IOException {
        for (int t = in.nextInt(); t-- > 0; )
            solveE();
    }

    private void solveE() throws IOException {
        int n = in.nextInt(), k = in.nextInt();

        int[] needRemove = new int[n];
        for (int pos = 0; pos < n; pos++)
            needRemove[pos] = pos - (in.nextInt() - 1);

        out.println(Arrays.toString(needRemove));

        long[][] dp = new long[n][n + 1]; // dp[prefLen - 1][deleted] = max right placed

        if (needRemove[0] == 0)
            dp[0][0]++;
        // out.println(0 + ": " + Arrays.toString(dp[0]));

        for (int i = 1; i < n; i++) {
            System.arraycopy(dp[i - 1], 0, dp[i], 0, n);
            for (int j = 0; j <= min(i - 1, needRemove[i]); j++)
                dp[i][needRemove[i]] = max(dp[i][needRemove[i]], dp[i - j - 1][needRemove[i] - j] + 1);
            // out.println(i + ": " + Arrays.toString(dp[i]));
        }

        int ans = 0;
        while (ans < n && dp[n - 1][ans] < k)
            ans++;
        out.println(ans == n ? -1 : ans);
    }

    class FastScanner {
        StringTokenizer st;
        BufferedReader br;

        FastScanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }
    }

    private void run() throws IOException {
        in = new FastScanner(System.in);
        out = new PrintWriter(System.out);

        solve();

        out.flush();
        out.close();
    }

    public static void main(String[] args) throws IOException {
        new Round().run();
    }
}
