#include <bits/stdc++.h>
using namespace std;
const int N = 2005;
int dp[N][N], mark[N][N], id, a[N];
int solve(int idx, int rem) {
  if (idx == -1) {
    return (rem != 0) * -1e9;
  }
  if (idx + 1 == rem) return 0;
  if (mark[idx][rem] == id) return dp[idx][rem];
  mark[idx][rem] = id;
  int cur = idx + 1 - rem;
  int op1 = (cur == a[idx]) + solve(idx - 1, rem);
  int op2 = -1e9;
  if (rem > 0) op2 = solve(idx - 1, rem - 1);
  return dp[idx][rem] = max(op1, op2);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) {
    id++;
    int n, k;
    cin >> n >> k;
    for (int i = 0; i < n; i++) cin >> a[i];
    int ans = -1;
    for (int i = 0; i < n; i++) {
      if (solve(n - 1, i) >= k) {
        ans = i;
        break;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2005;
int dp[N][N], mark[N][N], id, a[N];
int solve(int idx, int rem) {
  if (idx == -1) {
    return (rem != 0) * -1e9;
  }
  if (idx + 1 == rem) return 0;
  if (mark[idx][rem] == id) return dp[idx][rem];
  mark[idx][rem] = id;
  int cur = idx + 1 - rem;
  int op1 = (cur == a[idx]) + solve(idx - 1, rem);
  int op2 = -1e9;
  if (rem > 0) op2 = solve(idx - 1, rem - 1);
  return dp[idx][rem] = max(op1, op2);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) {
    id++;
    int n, k;
    cin >> n >> k;
    for (int i = 0; i < n; i++) cin >> a[i];
    int ans = -1;
    for (int i = 0; i < n; i++) {
      if (solve(n - 1, i) >= k) {
        ans = i;
        break;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

```