#include <bits/stdc++.h>
const long long int mod = 1000000000 + 7;
const long long int INF = 9000000000000000000;
using namespace std;
void yes() { cout << "YES" << endl; }
void no() { cout << "NO" << endl; }
long long int gcd(long long int a, long long int b) {
  return (a == 0 ? b : gcd(b % a, a));
}
long long int lcm(long long int a, long long int b) {
  return (a * b / (gcd(a, b)));
}
long long int add(long long int a, long long int b, long long int c = 0) {
  return ((a % mod + b % mod) % mod + c % mod) % mod;
}
long long int mul(long long int a, long long int b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long int sub(long long int a, long long int b) {
  return (a % mod - b % mod + mod) % mod;
}
void print(vector<long long int>& ans) {
  for (auto x : ans) cout << x << " ";
  cout << endl;
}
void print(long long int a) { cout << a << endl; }
long long int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1};
long long int n, w, h, d, k, m, mx, timer, realans, cc, start, queries,
    startindex;
long long int arr[2005], dp[2005][2005];
long long int go(long long int id, long long int del) {
  if (id < 0) return 0;
  if (del == 0) {
    long long int cnt = 0;
    for (long long int i = id; i >= 1; i--) {
      if (arr[i] == i) cnt++;
    }
    return cnt;
  }
  if (dp[id][del] != -1) return dp[id][del];
  long long int fst = 0, scd = 0;
  if (arr[id] == id - del) {
    fst = 1 + go(id - 1, del);
  } else {
    fst = go(id - 1, del);
  }
  scd = go(id - 1, del - 1);
  return dp[id][del] = max(fst, scd);
}
void solve() {
  cin >> n >> k;
  for (long long int i = 1; i <= n; i++) cin >> arr[i];
  long long int ans = -1;
  memset(dp, -1, sizeof(dp));
  for (long long int i = 0; i <= n; i++) {
    if (go(n, i) >= k) {
      ans = i;
      break;
    }
  }
  cout << ans << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int tests = 1;
  cin >> tests;
  while (tests--) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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>
const long long int mod = 1000000000 + 7;
const long long int INF = 9000000000000000000;
using namespace std;
void yes() { cout << "YES" << endl; }
void no() { cout << "NO" << endl; }
long long int gcd(long long int a, long long int b) {
  return (a == 0 ? b : gcd(b % a, a));
}
long long int lcm(long long int a, long long int b) {
  return (a * b / (gcd(a, b)));
}
long long int add(long long int a, long long int b, long long int c = 0) {
  return ((a % mod + b % mod) % mod + c % mod) % mod;
}
long long int mul(long long int a, long long int b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long int sub(long long int a, long long int b) {
  return (a % mod - b % mod + mod) % mod;
}
void print(vector<long long int>& ans) {
  for (auto x : ans) cout << x << " ";
  cout << endl;
}
void print(long long int a) { cout << a << endl; }
long long int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1};
long long int n, w, h, d, k, m, mx, timer, realans, cc, start, queries,
    startindex;
long long int arr[2005], dp[2005][2005];
long long int go(long long int id, long long int del) {
  if (id < 0) return 0;
  if (del == 0) {
    long long int cnt = 0;
    for (long long int i = id; i >= 1; i--) {
      if (arr[i] == i) cnt++;
    }
    return cnt;
  }
  if (dp[id][del] != -1) return dp[id][del];
  long long int fst = 0, scd = 0;
  if (arr[id] == id - del) {
    fst = 1 + go(id - 1, del);
  } else {
    fst = go(id - 1, del);
  }
  scd = go(id - 1, del - 1);
  return dp[id][del] = max(fst, scd);
}
void solve() {
  cin >> n >> k;
  for (long long int i = 1; i <= n; i++) cin >> arr[i];
  long long int ans = -1;
  memset(dp, -1, sizeof(dp));
  for (long long int i = 0; i <= n; i++) {
    if (go(n, i) >= k) {
      ans = i;
      break;
    }
  }
  cout << ans << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int tests = 1;
  cin >> tests;
  while (tests--) {
    solve();
  }
  return 0;
}

```