#include <bits/stdc++.h>
long long mod = 1000000007;
using namespace std;
long long ceiling(long long a, long long b) {
  if (a % b == 0)
    return a / b;
  else
    return a / b + 1;
}
long long power(long long a, long long b) {
  if (b == 0)
    return 1;
  else if (b % 2 == 1) {
    long long k = power(a, b / 2) % mod;
    return (((a * k) % mod) * k) % mod;
  } else {
    long long k = power(a, b / 2) % mod;
    return (k * k) % mod;
  }
}
long long gcd(long long a, long long b) {
  if (a > b) swap(a, b);
  if (a == 0) return b;
  return gcd(b % a, a);
}
bool comparePairs(vector<long long> &a, vector<long long> &b) {
  long long cnt = 0;
  for (int i = 0; i < 5; i++) {
    if (a[i] < b[i]) cnt++;
  }
  return cnt >= 3;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long q;
  cin >> q;
  while (q--) {
    long long n, k, i, j;
    cin >> n >> k;
    vector<long long> a(n);
    for (i = 0; i < n; i++) {
      cin >> a[i];
    }
    vector<vector<long long>> dp(n + 1, vector<long long>(n + 1, 0));
    long long ans = -1;
    for (i = 1; i < n + 1; i++) {
      for (j = 1; j < i + 1; j++) {
        long long k = (a[i - 1] == j) ? 1 : 0;
        dp[i][j] = max(k + dp[i - 1][j - 1], dp[i - 1][j]);
      }
    }
    for (i = 0; i <= n; i++) {
      if (dp[n][i] >= k) ans = max(ans, i);
    }
    if (ans != -1)
      cout << n - ans << endl;
    else
      cout << -1 << endl;
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
long long mod = 1000000007;
using namespace std;
long long ceiling(long long a, long long b) {
  if (a % b == 0)
    return a / b;
  else
    return a / b + 1;
}
long long power(long long a, long long b) {
  if (b == 0)
    return 1;
  else if (b % 2 == 1) {
    long long k = power(a, b / 2) % mod;
    return (((a * k) % mod) * k) % mod;
  } else {
    long long k = power(a, b / 2) % mod;
    return (k * k) % mod;
  }
}
long long gcd(long long a, long long b) {
  if (a > b) swap(a, b);
  if (a == 0) return b;
  return gcd(b % a, a);
}
bool comparePairs(vector<long long> &a, vector<long long> &b) {
  long long cnt = 0;
  for (int i = 0; i < 5; i++) {
    if (a[i] < b[i]) cnt++;
  }
  return cnt >= 3;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long q;
  cin >> q;
  while (q--) {
    long long n, k, i, j;
    cin >> n >> k;
    vector<long long> a(n);
    for (i = 0; i < n; i++) {
      cin >> a[i];
    }
    vector<vector<long long>> dp(n + 1, vector<long long>(n + 1, 0));
    long long ans = -1;
    for (i = 1; i < n + 1; i++) {
      for (j = 1; j < i + 1; j++) {
        long long k = (a[i - 1] == j) ? 1 : 0;
        dp[i][j] = max(k + dp[i - 1][j - 1], dp[i - 1][j]);
      }
    }
    for (i = 0; i <= n; i++) {
      if (dp[n][i] >= k) ans = max(ans, i);
    }
    if (ans != -1)
      cout << n - ans << endl;
    else
      cout << -1 << endl;
  }
}

```