#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using vb = vector<bool>;
using vs = vector<string>;
using vc = vector<char>;
using vd = vector<double>;
using pii = pair<int, int>;
using pss = pair<string, string>;
using ll = long long;
using tup = tuple<ll, int, int>;
const int INF = 1e9 + 7;
const int MOD = 1e9 + 7;
template <typename Os, class V>
Os& operator<<(Os& os, V const& v) {
  for (auto const& e : v) os << e << " ";
  return os;
}
int N, K, ans = INF;
const int MAXN = 2001;
int memo[MAXN], history[MAXN];
vi nums;
void solve() {
  cin >> N >> K;
  nums = vi(N + 1, 0);
  for (int n = (1); n <= (N); ++n) cin >> nums[n];
  memset(memo, -1, sizeof memo);
  memset(history, -1, sizeof history);
  ans = INF;
  memo[0] = 0;
  for (int i = (1); i <= (N); ++i) {
    for (int j = (0); j < (i); ++j) {
      if (memo[j] == -1) continue;
      if (nums[j] < nums[i] && (nums[i] - nums[j] - 1 <= i - j - 1) &&
          memo[i] < memo[j] + 1) {
        memo[i] = memo[j] + 1;
        history[i] = j;
      }
    }
  }
  int ans = INF;
  vi deletion(N + 1, 0);
  for (int i = (1); i <= (N); ++i) {
    if (history[i] == -1) continue;
    int j = history[i];
    deletion[i] = deletion[j] + (i - j - 1) - (nums[i] - nums[j] - 1);
    if (memo[i] >= K) ans = min(ans, deletion[i]);
  }
  cout << (ans == INF ? -1 : ans) << "\n";
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int T;
  cin >> T;
  for (int TC = (0); TC < (T); ++TC) {
    solve();
  }
}

### 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;
using vi = vector<int>;
using vb = vector<bool>;
using vs = vector<string>;
using vc = vector<char>;
using vd = vector<double>;
using pii = pair<int, int>;
using pss = pair<string, string>;
using ll = long long;
using tup = tuple<ll, int, int>;
const int INF = 1e9 + 7;
const int MOD = 1e9 + 7;
template <typename Os, class V>
Os& operator<<(Os& os, V const& v) {
  for (auto const& e : v) os << e << " ";
  return os;
}
int N, K, ans = INF;
const int MAXN = 2001;
int memo[MAXN], history[MAXN];
vi nums;
void solve() {
  cin >> N >> K;
  nums = vi(N + 1, 0);
  for (int n = (1); n <= (N); ++n) cin >> nums[n];
  memset(memo, -1, sizeof memo);
  memset(history, -1, sizeof history);
  ans = INF;
  memo[0] = 0;
  for (int i = (1); i <= (N); ++i) {
    for (int j = (0); j < (i); ++j) {
      if (memo[j] == -1) continue;
      if (nums[j] < nums[i] && (nums[i] - nums[j] - 1 <= i - j - 1) &&
          memo[i] < memo[j] + 1) {
        memo[i] = memo[j] + 1;
        history[i] = j;
      }
    }
  }
  int ans = INF;
  vi deletion(N + 1, 0);
  for (int i = (1); i <= (N); ++i) {
    if (history[i] == -1) continue;
    int j = history[i];
    deletion[i] = deletion[j] + (i - j - 1) - (nums[i] - nums[j] - 1);
    if (memo[i] >= K) ans = min(ans, deletion[i]);
  }
  cout << (ans == INF ? -1 : ans) << "\n";
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int T;
  cin >> T;
  for (int TC = (0); TC < (T); ++TC) {
    solve();
  }
}

```