#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int N = 2e5 + 1;
const long long MOD = 998244353;
long long POW(long long base, long long exponent) {
  long long result = 1;
  while (exponent > 0) {
    if (exponent % 2 == 1) result = (result * base) % mod;
    exponent = exponent >> 1;
    base = (base * base) % mod;
  }
  return result;
}
long long Inv(long long x) { return POW(x, mod - 2); }
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> q;
  while (q--) {
    int n, k;
    cin >> n >> k;
    vector<int> a(n + 1);
    for (int i = 1; i < n + 1; i++) cin >> a[i];
    vector<vector<int> > df(n + 1, vector<int>(n + 2, 0)),
        dp(n + 1, vector<int>(n + 1, 0)), mx(n + 1, vector<int>(n + 1, 0));
    for (int diff = 0; diff < n; diff++) {
      for (int i = n; i >= 0; i--) {
        if (i - a[i] == diff)
          df[diff][i] = df[diff][i + 1] + 1;
        else
          df[diff][i] = df[diff][i + 1];
      }
    }
    for (int i = 1; i < n + 1; i++) dp[0][i] = df[0][1], mx[0][i] = df[0][1];
    for (int i = 1; i < n + 1; i++)
      dp[1][i] = df[1][i + 1] + df[0][1] - df[0][i],
      mx[1][i] = max(mx[1][i - 1], dp[1][i]);
    for (int mov = 2; mov < n; mov++) {
      for (int i = 1; i < n + 1; i++) {
        dp[mov][i] = df[mov][i + 1] - df[mov - 1][i];
        dp[mov][i] += mx[mov - 1][i - 1];
        mx[mov][i] = dp[mov][i];
        mx[mov][i] = max(mx[mov][i], mx[mov][i - 1]);
      }
    }
    if (df[0][1] >= k) {
      cout << 0 << '\n';
      continue;
    }
    int ans = -1;
    for (int i = 1; i < n; i++) {
      int f = 0;
      for (int j = 1; j < n + 1; j++) {
        if (dp[i][j] >= k) {
          f = 1;
          break;
        }
      }
      if (f) {
        ans = i;
        break;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Your challenge is to write 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 long long mod = 1e9 + 7;
const int N = 2e5 + 1;
const long long MOD = 998244353;
long long POW(long long base, long long exponent) {
  long long result = 1;
  while (exponent > 0) {
    if (exponent % 2 == 1) result = (result * base) % mod;
    exponent = exponent >> 1;
    base = (base * base) % mod;
  }
  return result;
}
long long Inv(long long x) { return POW(x, mod - 2); }
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> q;
  while (q--) {
    int n, k;
    cin >> n >> k;
    vector<int> a(n + 1);
    for (int i = 1; i < n + 1; i++) cin >> a[i];
    vector<vector<int> > df(n + 1, vector<int>(n + 2, 0)),
        dp(n + 1, vector<int>(n + 1, 0)), mx(n + 1, vector<int>(n + 1, 0));
    for (int diff = 0; diff < n; diff++) {
      for (int i = n; i >= 0; i--) {
        if (i - a[i] == diff)
          df[diff][i] = df[diff][i + 1] + 1;
        else
          df[diff][i] = df[diff][i + 1];
      }
    }
    for (int i = 1; i < n + 1; i++) dp[0][i] = df[0][1], mx[0][i] = df[0][1];
    for (int i = 1; i < n + 1; i++)
      dp[1][i] = df[1][i + 1] + df[0][1] - df[0][i],
      mx[1][i] = max(mx[1][i - 1], dp[1][i]);
    for (int mov = 2; mov < n; mov++) {
      for (int i = 1; i < n + 1; i++) {
        dp[mov][i] = df[mov][i + 1] - df[mov - 1][i];
        dp[mov][i] += mx[mov - 1][i - 1];
        mx[mov][i] = dp[mov][i];
        mx[mov][i] = max(mx[mov][i], mx[mov][i - 1]);
      }
    }
    if (df[0][1] >= k) {
      cout << 0 << '\n';
      continue;
    }
    int ans = -1;
    for (int i = 1; i < n; i++) {
      int f = 0;
      for (int j = 1; j < n + 1; j++) {
        if (dp[i][j] >= k) {
          f = 1;
          break;
        }
      }
      if (f) {
        ans = i;
        break;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

```