#include <bits/stdc++.h>
using namespace std;
void print() { cout << "\n"; }
template <typename T>
void print(const T& x) {
  cout << x;
}
template <typename T, typename... R>
void print(const T& x, const R&... r) {
  cout << x;
  print(r...);
}
template <typename T>
void _prints(const T& x) {
  cout << x;
}
template <typename T, typename... R>
void _prints(const T& x, const R&... r) {
  cout << x << " ";
  _prints(r...);
}
template <typename T>
void prints(const T& x) {
  cout << x << " ";
}
template <typename T, typename... R>
void prints(const T& x, const R&... r) {
  _prints(x, r...);
  print();
}
template <typename T>
void read(T& x) {
  cin >> x;
}
template <typename T, typename... R>
void read(T& x, R&... r) {
  cin >> x;
  read(r...);
}
template <typename _T>
void _debug(const char* s, _T x) {
  cerr << s << " = " << x << "\n";
}
template <typename _T, typename... R>
void _debug(const char* s, _T x, R... r) {
  while (*s != ',') cerr << *s++;
  cerr << " = " << x << ", ";
  _debug(s + 1, r...);
}
int _int(const char& c) {
  if (c >= 'a' && c <= 'z') return c - 'a';
  if (c >= 'A' && c <= 'Z') return c - 'A';
  if (c >= '0' && c <= '9') return c - '0';
  return c;
}
using ll = long long;
using ld = long double;
const int N = 2e3 + 5, INF = 1e9, MOD = 1e9 + 7;
int dp[N][N];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  read(t);
  for (int tt = 0; tt < t; tt++) {
    int n, k;
    read(n, k);
    vector<int> arr(n + 1);
    for (int i = 1; i <= n; i++) {
      read(arr[i]);
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j <= i; j++) {
        bool flag = (arr[i + 1] == j + 1);
        dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
        dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j] + flag);
      }
    }
    int ans = INF;
    for (int i = 0; i <= n; i++) {
      if (dp[n][i] >= k) ans = min(ans, n - i);
    }
    memset(dp, 0, sizeof(dp));
    print(ans != INF ? ans : -1);
    print();
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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() { cout << "\n"; }
template <typename T>
void print(const T& x) {
  cout << x;
}
template <typename T, typename... R>
void print(const T& x, const R&... r) {
  cout << x;
  print(r...);
}
template <typename T>
void _prints(const T& x) {
  cout << x;
}
template <typename T, typename... R>
void _prints(const T& x, const R&... r) {
  cout << x << " ";
  _prints(r...);
}
template <typename T>
void prints(const T& x) {
  cout << x << " ";
}
template <typename T, typename... R>
void prints(const T& x, const R&... r) {
  _prints(x, r...);
  print();
}
template <typename T>
void read(T& x) {
  cin >> x;
}
template <typename T, typename... R>
void read(T& x, R&... r) {
  cin >> x;
  read(r...);
}
template <typename _T>
void _debug(const char* s, _T x) {
  cerr << s << " = " << x << "\n";
}
template <typename _T, typename... R>
void _debug(const char* s, _T x, R... r) {
  while (*s != ',') cerr << *s++;
  cerr << " = " << x << ", ";
  _debug(s + 1, r...);
}
int _int(const char& c) {
  if (c >= 'a' && c <= 'z') return c - 'a';
  if (c >= 'A' && c <= 'Z') return c - 'A';
  if (c >= '0' && c <= '9') return c - '0';
  return c;
}
using ll = long long;
using ld = long double;
const int N = 2e3 + 5, INF = 1e9, MOD = 1e9 + 7;
int dp[N][N];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  read(t);
  for (int tt = 0; tt < t; tt++) {
    int n, k;
    read(n, k);
    vector<int> arr(n + 1);
    for (int i = 1; i <= n; i++) {
      read(arr[i]);
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j <= i; j++) {
        bool flag = (arr[i + 1] == j + 1);
        dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
        dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j] + flag);
      }
    }
    int ans = INF;
    for (int i = 0; i <= n; i++) {
      if (dp[n][i] >= k) ans = min(ans, n - i);
    }
    memset(dp, 0, sizeof(dp));
    print(ans != INF ? ans : -1);
    print();
  }
  return 0;
}

```