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;
  using ll = long long;
  using ld = long double;
  using db = double;
  using str = string;
  using pi = pair<int, int>;
  using pl = pair<ll, ll>;
  using vi = vector<int>;
  using vl = vector<ll>;
  using vs = vector<str>;
  using vc = vector<char>;
  using vpi = vector<pi>;
  using vpl = vector<pl>;
  template <class T>
  using V = vector<T>;
  template <class T, size_t SZ>
  using AR = array<T, SZ>;
  template <class T, class U>
  using PR = pair<T, U>;
  template <class T, class U>
  using umap = unordered_map<T, U>;
  template <class T>
  using uset = unordered_set<T>;
  template <class T>
  using mset = multiset<T>;
  const int MOD = 1e9 + 7;
  const int MX = 2e5 + 5;
  const ll INFL = ll(3e18) + 10;
  const int INF = int(1e9) + 10;
  const ld PI = acos((ld)-1);
  const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
  template <class T>
  using pqg = priority_queue<T, vector<T>, greater<T>>;
  template <class T>
  using pql = priority_queue<T, vector<T>, less<T>>;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  constexpr int pct(int x) { return __builtin_popcount(x); }
  constexpr int bits(int x) { return x == 0 ? 0 : 31 - __builtin_clz(x); }
  constexpr int p2(int x) { return 1 << x; }
  constexpr int msk2(int x) { return p2(x) - 1; }
  ll cdiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); }
  ll fdiv(ll a, ll b) { return a / b - ((a ^ b) < 0 && a % b); }
  template <class T>
  bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  template <class T, class U>
  T fstTrue(T lo, T hi, U first) {
    hi++;
    assert(lo <= hi);
    while (lo < hi) {
      T mid = lo + (hi - lo) / 2;
      first(mid) ? hi = mid : lo = mid + 1;
    }
    return lo;
  }
  template <class T, class U>
  T lstTrue(T lo, T hi, U first) {
    lo--;
    assert(lo <= hi);
    while (lo < hi) {
      T mid = lo + (hi - lo + 1) / 2;
      first(mid) ? lo = mid : hi = mid - 1;
    }
    return lo;
  }
  template <class T>
  void remDup(vector<T>& v) {
    sort(begin(v), end(v));
    v.erase(unique(begin(v), end(v)), end(v));
  }
  template <class T, class U>
  void erase(T& t, const U& u) {
    auto it = t.find(u);
    assert(it != end(t));
    t.erase(it);
  }
  str to_string(char c) { return str(1, c); }
  str to_string(const char* second) { return (str)second; }
  str to_string(str second) { return second; }
  str to_string(bool b) { return to_string((int)b); }
  template <class T, class U>
  str to_string(pair<T, U> p) {
    return to_string(p.first) + " " + to_string(p.second);
  }
  template <class T, class U>
  str to_string(V<pair<T, U>> v) {
    bool fst = 1;
    str res = "";
    for (const auto& x : v) {
      if (!fst) res += " ";
      fst = 0;
      res += to_string(x);
    }
    return res;
  }
  template <class T>
  str to_string(T v) {
    bool fst = 1;
    str res = "";
    for (const auto& x : v) {
      if (!fst) res += " ";
      fst = 0;
      res += to_string(x);
    }
    return res;
  }
  void DBG() { cerr << "]" << endl; }
  template <class T, class... U>
  void DBG(const T& t, const U&... u) {
    cerr << to_string(t);
    if (sizeof...(u)) cerr << ", ";
    DBG(u...);
  }
  void unsyncIO() { cin.tie(0)->sync_with_stdio(0); }
  void setPrec() { cout << fixed << setprecision(15); }
  void setIn(str second) { freopen(second.c_str(), "r", stdin); }
  void setOut(str second) { freopen(second.c_str(), "w", stdout); }
  void setIO(str second = "") {
    unsyncIO();
    setPrec();
    if (int((second).size())) setIn(second + ".in"), setOut(second + ".out");
  }
  const int nax = 2005;
  int dp[nax][nax];
  int n, k;
  void reset() {
    for (int i = (0); i < (n + 5); ++i)
      for (int j = (0); j < (n + 5); ++j) dp[i][j] = 0;
  }
  void solve() {
    cin >> n >> k;
    reset();
    vi a(n);
    for (auto& x : a) cin >> x;
    for (int i = (0); i < (n); ++i) {
      for (int j = (0); j < (n); ++j) {
        ckmax(dp[i + 1][j + 1], dp[i][j]);
        int ni = i - j + 1;
        ckmax(dp[i + 1][j], dp[i][j] + (ni == a[i]));
      }
    }
    for (int i = (0); i < (n); ++i) {
      if (dp[n][i] >= k) {
        cout << i << '\n';
        return;
      }
    }
    cout << -1 << '\n';
  }
  int main() {
    setIO();
    int TT = 1;
    cin >> TT;
    for (int _ = (0); _ < (TT); ++_) solve();
    exit(0);
  }
  