#include <bits/stdc++.h>
using namespace std;
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) {
  return cout << "(" << p.first << ", " << p.second << ")";
}
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  cout << "[";
  for (int i = 0; i < v.size(); i++) {
    if (i) cout << ", ";
    cout << v[i];
  }
  return cout << "]";
}
const long long INF = 1e18;
long long T, n, k, a[2005], pf[2005][2005], dp[2005][2005], v[2005][2005];
void dfs(long long in, long long rem) {
  if (v[in][rem]) return;
  v[in][rem] = 1;
  dp[in][rem] = -INF;
  if (rem > in) return;
  if (in == 0 && rem == 0) {
    dp[in][rem] = pf[0][n];
    return;
  }
  if (!v[in - 1][rem]) dfs(in - 1, rem);
  dp[in][rem] = max(dp[in][rem], dp[in - 1][rem]);
  if (rem == 0) return;
  if (!v[in - 1][rem - 1]) dfs(in - 1, rem - 1);
  dp[in][rem] =
      max(dp[in][rem], dp[in - 1][rem - 1] + (pf[rem][n] - pf[rem][in]) -
                           (pf[rem - 1][n] - pf[rem - 1][in - 1]));
}
void solve() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 0; i < n + 2; i++)
    for (int j = 0; j < n + 2; j++) {
      dp[i][j] = -INF;
      pf[i][j] = 0;
      v[i][j] = 0;
    }
  for (int i = 1; i <= n; i++) a[i] = i - a[i];
  for (int i = 1; i <= n; i++) {
    if (a[i] < 0) continue;
    pf[a[i]][i] = 1;
  }
  for (int i = 0; i < n + 1; i++)
    for (int j = 1; j <= n; j++) pf[i][j] += pf[i][j - 1];
  for (int j = 0; j < n + 1; j++) {
    dfs(n, j);
    if (dp[n][j] >= k) {
      cout << j << '\n';
      return;
    }
  }
  cout << -1 << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  T = 1;
  cin >> T;
  for (int t = 1; t <= T; t++) solve();
  cout.flush();
  return 0;
}

### Prompt

Develop a solution in Cpp 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>
using namespace std;
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) {
  return cout << "(" << p.first << ", " << p.second << ")";
}
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  cout << "[";
  for (int i = 0; i < v.size(); i++) {
    if (i) cout << ", ";
    cout << v[i];
  }
  return cout << "]";
}
const long long INF = 1e18;
long long T, n, k, a[2005], pf[2005][2005], dp[2005][2005], v[2005][2005];
void dfs(long long in, long long rem) {
  if (v[in][rem]) return;
  v[in][rem] = 1;
  dp[in][rem] = -INF;
  if (rem > in) return;
  if (in == 0 && rem == 0) {
    dp[in][rem] = pf[0][n];
    return;
  }
  if (!v[in - 1][rem]) dfs(in - 1, rem);
  dp[in][rem] = max(dp[in][rem], dp[in - 1][rem]);
  if (rem == 0) return;
  if (!v[in - 1][rem - 1]) dfs(in - 1, rem - 1);
  dp[in][rem] =
      max(dp[in][rem], dp[in - 1][rem - 1] + (pf[rem][n] - pf[rem][in]) -
                           (pf[rem - 1][n] - pf[rem - 1][in - 1]));
}
void solve() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 0; i < n + 2; i++)
    for (int j = 0; j < n + 2; j++) {
      dp[i][j] = -INF;
      pf[i][j] = 0;
      v[i][j] = 0;
    }
  for (int i = 1; i <= n; i++) a[i] = i - a[i];
  for (int i = 1; i <= n; i++) {
    if (a[i] < 0) continue;
    pf[a[i]][i] = 1;
  }
  for (int i = 0; i < n + 1; i++)
    for (int j = 1; j <= n; j++) pf[i][j] += pf[i][j - 1];
  for (int j = 0; j < n + 1; j++) {
    dfs(n, j);
    if (dp[n][j] >= k) {
      cout << j << '\n';
      return;
    }
  }
  cout << -1 << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  T = 1;
  cin >> T;
  for (int t = 1; t <= T; t++) solve();
  cout.flush();
  return 0;
}

```