description: 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.
solution: #include <bits/stdc++.h>
  using namespace std;
  vector<long long int> Tree;
  void make_Tree(vector<long long int> &v) {
    long long int i;
    long long int N = v.size();
    while (N & (N - 1)) N++;
    v.resize(N);
    Tree.resize(2 * N);
    for (i = 0; i < N; i++) {
      Tree[N + i] = v[i];
    }
    for (i = N - 1; i >= 1; i--) {
      Tree[i] = Tree[2 * i] + Tree[2 * i + 1];
    }
  }
  long long int segment(long long int node, long long int node_left,
                        long long int node_right, long long int query_left,
                        long long int query_right) {
    if (query_left <= node_left && node_right <= query_right) {
      return Tree[node];
    }
    if (query_right < node_left || node_right < query_left) {
      return 0;
    }
    long long int last_left = (node_left + node_right) / 2;
    return segment(2 * node, node_left, last_left, query_left, query_right) +
           segment(2 * node + 1, last_left + 1, node_right, query_left,
                   query_right);
  }
  void update_Tree(long long int Index, long long int Val) {
    long long int N = Tree.size() / 2;
    long long int Diff = Val - Tree[N + Index - 1];
    for (long long int i = N + Index - 1; i >= 1; i /= 2) {
      Tree[i] += Diff;
    }
  }
  const long long int maxN = 100001;
  vector<long long int> inF(maxN), F(maxN), isPrime(maxN), Ad[maxN];
  map<long long int, long long int> Vis;
  void dfs(long long int node) {
    if (Vis[node]) return;
    Vis[node] = 1;
    for (auto child : Ad[node]) {
      dfs(child);
    }
    return;
  }
  long long int pow1(long long int n, long long int k) {
    if (k == 0) return 1;
    if (k == 1) return n;
    long long int X = 1;
    if (k % 2 == 1) X = n % 1000000007;
    long long int Y = pow1(n, k / 2) % 1000000007;
    return (((Y * Y) % 1000000007) * X) % 1000000007;
  }
  void fact_compute() {
    long long int N = maxN - 1;
    F[0] = 1;
    F[1] = 1;
    for (long long int i = 2; i <= N; i++) {
      F[i] = (F[i - 1] * (i)) % 1000000007;
    }
    inF[N] = pow1(F[N], 1000000007 - 2) % 1000000007;
    for (long long int i = N - 1; i >= 0; i--) {
      inF[i] = (inF[i + 1] * (i + 1)) % 1000000007;
    }
  }
  void sieve() {
    for (long long int i = 2; i < maxN; i++) {
      isPrime[i] = 1;
    }
    for (long long int i = 2; i < maxN; i++) {
      for (long long int j = 2 * i; j < maxN; j += i) {
        isPrime[j] = 0;
      }
    }
  }
  long long int gcd(long long int n, long long int m) {
    if (!m) return n;
    return gcd(m, n % m);
  }
  long long int lcm(long long int n, long long int m) {
    return (n / gcd(n, m)) * m;
  }
  void set_io() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
  }
  void solve() {
    long long int n, m, j, k, l, i;
    cin >> n >> k;
    vector<long long int> v(n);
    long long int isOne = 0;
    vector<vector<long long int>> dp(n + 1, vector<long long int>(n + 1, n));
    vector<long long int> cnt(n + 1, 0);
    for (long long int i = 0; i < n; i++) {
      cin >> v[i];
    }
    long long int f = -1;
    long long int ans = n;
    long long int maxi = 0;
    for (i = 0; i < n; i++) {
      long long int val = v[i];
      if (i != 0) {
        if (i + 1 >= v[i]) {
          dp[i][v[i]] = i + 1 - v[i];
          cnt[i] = 1;
        }
        for (j = 0; j < i; j++) {
          if (v[i] > v[j]) {
            if (i - j >= v[i] - v[j]) {
              if (cnt[i] <= cnt[j] + 1) {
                cnt[i] = max(cnt[i], cnt[j] + 1);
                maxi = max(maxi, cnt[i]);
                dp[i][v[i]] = min(dp[i][v[i]], dp[j][v[j]] + i - j);
              }
            }
          }
        }
      } else {
        if (v[i] == 1) {
          cnt[i] = 1;
          maxi = max(maxi, cnt[i]);
          dp[i][1] = 0;
        }
      }
    }
    if (maxi < k) {
      cout << -1;
      cout << "\n";
      return;
    }
    for (i = 0; i < n; i++) {
      if (cnt[i] >= k) {
        ans = min(ans, dp[i][v[i]]);
      }
    }
    cout << ans;
    cout << "\n";
    return;
  }
  int main() {
    set_io();
    long long int totalTestCase = 1;
    cin >> totalTestCase;
    for (long long int testCase = 1; testCase <= totalTestCase; testCase++) {
      solve();
    }
    return 0;
  }
  