#include <bits/stdc++.h>
using namespace std;
void _print(long long t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(unsigned long long t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
bool sortm(pair<int, int> &a, pair<int, int> &b) {
  int x = abs(a.first) + abs(a.second);
  int y = abs(b.first) + abs(b.second);
  return x < y;
}
void init() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
vector<int> a(2005);
int dp[2001][2001];
int n, k;
void solve() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= n; j++) {
      if (a[i] == (i - j + 1))
        dp[i][j] = 1 + (i == 0 ? 0 : dp[i - 1][j]);
      else {
        if (i == 0)
          dp[i][j] = 0;
        else
          dp[i][j] = dp[i - 1][j];
        if (a[i] == (i - j + 1) && i > 0) {
          int jj = (i + 1 - j - a[i]);
          if (j - jj >= 0)
            dp[i][j] = max(dp[i][j], 1 + dp[i - 1][j - (i + 1 - j - a[i])]);
        } else if (i > 0 && j > 0)
          dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
      }
    }
  }
  int ans = INT_MAX;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= n; j++) {
      if (dp[i][j] >= k) {
        ans = min(ans, j);
      }
    }
  }
  cout << (ans == INT_MAX ? -1 : ans) << "\n";
  return;
}
int main() {
  init();
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Your task is to create 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;
void _print(long long t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(unsigned long long t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
bool sortm(pair<int, int> &a, pair<int, int> &b) {
  int x = abs(a.first) + abs(a.second);
  int y = abs(b.first) + abs(b.second);
  return x < y;
}
void init() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
vector<int> a(2005);
int dp[2001][2001];
int n, k;
void solve() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= n; j++) {
      if (a[i] == (i - j + 1))
        dp[i][j] = 1 + (i == 0 ? 0 : dp[i - 1][j]);
      else {
        if (i == 0)
          dp[i][j] = 0;
        else
          dp[i][j] = dp[i - 1][j];
        if (a[i] == (i - j + 1) && i > 0) {
          int jj = (i + 1 - j - a[i]);
          if (j - jj >= 0)
            dp[i][j] = max(dp[i][j], 1 + dp[i - 1][j - (i + 1 - j - a[i])]);
        } else if (i > 0 && j > 0)
          dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
      }
    }
  }
  int ans = INT_MAX;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= n; j++) {
      if (dp[i][j] >= k) {
        ans = min(ans, j);
      }
    }
  }
  cout << (ans == INT_MAX ? -1 : ans) << "\n";
  return;
}
int main() {
  init();
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

```