#include <bits/stdc++.h>
using namespace std;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
const long long INF = 1e18;
const int mod = 1e9 + 7;
inline int add(int x, int y) {
  x += y;
  if (x >= mod) return x - mod;
  return x;
}
inline int sub(int x, int y) {
  x -= y;
  if (x < 0) return x + mod;
  return x;
}
inline int mul(int x, int y) { return (x * 1ll * y) % mod; }
inline int power(int x, int y) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = mul(ans, x);
    x = mul(x, x);
    y >>= 1;
  }
  return ans;
}
inline int inv(int x) { return power(x, mod - 2); }
const long long FACTORIAL_SIZE = 1.1e6;
long long fact[FACTORIAL_SIZE], ifact[FACTORIAL_SIZE];
void gen_factorial(long long n) {
  fact[0] = fact[1] = ifact[0] = ifact[1] = 1;
  for (long long i = 2; i <= n; i++) {
    fact[i] = (i * fact[i - 1]) % mod;
  }
  ifact[n] = inv(fact[n]);
  for (long long i = n - 1; i >= 2; i--) {
    ifact[i] = ((i + 1) * ifact[i + 1]) % mod;
  }
}
long long nck(long long n, long long k) {
  long long den = (ifact[k] * ifact[n - k]) % mod;
  return (den * fact[n]) % mod;
}
const int N = 2e5 + 5;
void solve() {
  long long n, k;
  cin >> n >> k;
  long long dp[n + 1][n + 1];
  memset(dp, 0, sizeof dp);
  long long a[n + 1];
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 0; j <= i; j++) {
      dp[i][j] = dp[i - 1][j] + (a[i] == i - j);
      if (j - 1 >= 0) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
    }
  }
  long long ans = INF;
  for (long long i = 0; i < n; i++) {
    if (dp[n][i] >= k) {
      ans = min(ans, i);
    }
  }
  if (ans == INF) {
    cout << -1 << "\n";
    return;
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long tc = 1;
  cin >> tc;
  long long i = 1;
  while (tc--) {
    solve();
    i++;
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
const long long INF = 1e18;
const int mod = 1e9 + 7;
inline int add(int x, int y) {
  x += y;
  if (x >= mod) return x - mod;
  return x;
}
inline int sub(int x, int y) {
  x -= y;
  if (x < 0) return x + mod;
  return x;
}
inline int mul(int x, int y) { return (x * 1ll * y) % mod; }
inline int power(int x, int y) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = mul(ans, x);
    x = mul(x, x);
    y >>= 1;
  }
  return ans;
}
inline int inv(int x) { return power(x, mod - 2); }
const long long FACTORIAL_SIZE = 1.1e6;
long long fact[FACTORIAL_SIZE], ifact[FACTORIAL_SIZE];
void gen_factorial(long long n) {
  fact[0] = fact[1] = ifact[0] = ifact[1] = 1;
  for (long long i = 2; i <= n; i++) {
    fact[i] = (i * fact[i - 1]) % mod;
  }
  ifact[n] = inv(fact[n]);
  for (long long i = n - 1; i >= 2; i--) {
    ifact[i] = ((i + 1) * ifact[i + 1]) % mod;
  }
}
long long nck(long long n, long long k) {
  long long den = (ifact[k] * ifact[n - k]) % mod;
  return (den * fact[n]) % mod;
}
const int N = 2e5 + 5;
void solve() {
  long long n, k;
  cin >> n >> k;
  long long dp[n + 1][n + 1];
  memset(dp, 0, sizeof dp);
  long long a[n + 1];
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 0; j <= i; j++) {
      dp[i][j] = dp[i - 1][j] + (a[i] == i - j);
      if (j - 1 >= 0) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
    }
  }
  long long ans = INF;
  for (long long i = 0; i < n; i++) {
    if (dp[n][i] >= k) {
      ans = min(ans, i);
    }
  }
  if (ans == INF) {
    cout << -1 << "\n";
    return;
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long tc = 1;
  cin >> tc;
  long long i = 1;
  while (tc--) {
    solve();
    i++;
  }
  return 0;
}

```