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 pd = pair<db, db>;
  using vi = vector<int>;
  using vb = vector<bool>;
  using vl = vector<ll>;
  using vd = vector<db>;
  using vs = vector<str>;
  using vpi = vector<pi>;
  using vpl = vector<pl>;
  using vpd = vector<pd>;
  template <class T>
  using V = vector<T>;
  template <class T, size_t SZ>
  using AR = array<T, SZ>;
  template <class T>
  using PR = pair<T, T>;
  const int MOD = 1e9 + 7;
  const int mxN = 2e3 + 5;
  const ll INF = 1e18;
  const ld PI = acos((ld)-1);
  const int tSZ = (1 << 21);
  const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
  mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
  constexpr int pcnt(int x) { return __builtin_popcount(x); }
  constexpr int bits(int x) { return 31 - __builtin_clz(x); }
  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); }
  ll add(ll first, ll second) {
    first += second;
    if (first >= MOD) first %= MOD;
    if (first < 0) first += MOD;
    return first;
  }
  ll mul(ll first, ll second) {
    first *= second;
    if (first >= MOD) first %= MOD;
    return first;
  }
  ll pw(ll x, ll y) {
    ll ans = 1, ml = x;
    while (y) {
      if (y & 1) ans = mul(ans, ml);
      ml = mul(ml, ml);
      y >>= 1;
    }
    return ans;
  }
  ll inv(ll x) { return pw(x, MOD - 2); }
  template <class T>
  void remDup(vector<T>& x) {
    sort(begin(x), end(x));
    x.erase(unique(begin(x), end(x)), end(x));
  }
  template <class T, class U>
  void erase(T& x, const U& v) {
    auto it = x.find(v);
    assert(it != end(x));
    x.erase(v);
  }
  template <class T>
  void re(complex<T>& c);
  template <class T, class U>
  void re(pair<T, U>& p);
  template <class T>
  void re(vector<T>& v);
  template <class T, size_t SZ>
  void re(array<T, SZ>& a);
  template <class T>
  void re(T& x) {
    cin >> x;
  }
  void re(db& d) {
    str t;
    re(t);
    d = stod(t);
  }
  void re(ld& d) {
    str t;
    re(t);
    d = stold(t);
  }
  template <class T, class... U>
  void re(T& t, U&... u) {
    re(t);
    re(u...);
  }
  template <class T>
  void re(complex<T>& c) {
    T a, b;
    re(a, b);
    c = {a, b};
  }
  template <class T, class U>
  void re(pair<T, U>& p) {
    re(p.first, p.second);
  }
  template <class T>
  void re(vector<T>& x) {
    for (auto& a : x) re(a);
  }
  template <class T, size_t SZ>
  void re(array<T, SZ>& x) {
    for (auto& a : x) re(a);
  }
  template <class T>
  void rv(int& n, vector<T>& x) {
    re(n);
    x.resize(n);
    for (auto& a : x) re(a);
  }
  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>
  str to_string(complex<T> c) {
    stringstream ss;
    ss << c;
    return ss.str();
  }
  str to_string(vector<bool> v) {
    str res = "{";
    for (int i = (0); i < ((int)(v).size()); i++) res += char('0' + v[i]);
    res += "}";
    return res;
  }
  template <size_t SZ>
  str to_string(bitset<SZ> b) {
    str res = "";
    for (int i = (0); i < (SZ); i++) res += char('0' + b[i]);
    return res;
  }
  template <class T, class U>
  str to_string(pair<T, U> p);
  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;
  }
  template <class T, class U>
  str to_string(pair<T, U> p) {
    return to_string(p.first) + " " + to_string(p.second);
  }
  template <class T>
  void pr(T v) {
    cout << to_string(v);
  }
  template <class T, class... U>
  void pr(const T& t, const U&... u) {
    pr(t);
    pr(u...);
  }
  void ps() { pr("\n"); }
  template <class T, class... U>
  void ps(const T& t, const U&... u) {
    pr(t);
    if (sizeof...(u)) pr(" ");
    ps(u...);
  }
  void setIn(str second) { freopen(second.c_str(), "r", stdin); }
  void setOut(str second) { freopen(second.c_str(), "w", stdout); }
  void unsyncIO() { cin.tie(0)->sync_with_stdio(0); }
  void setIO(str st = "") {
    unsyncIO();
    if ((int)(st).size()) {
      setIn(st + ".in"), setOut(st + ".out");
    }
  }
  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...);
  }
  int n, k, a[mxN], T[mxN][mxN];
  int solve() {
    for (int i = (0); i < (n); i++)
      for (int j = (0); j < (n + 1); j++) T[i][j] = -n;
    T[0][0] = 0;
    T[0][1] = (a[0] == 1);
    for (int i = (1); i < (n); i++) {
      for (int j = (0); j < (i + 2); j++) {
        T[i][j] =
            max((j - 1 >= 0 ? (T[i - 1][j - 1] + (j == a[i])) : 0), T[i - 1][j]);
      }
    }
    int ans = -1;
    for (int i = (0); i < (n + 1); i++) {
      if (T[n - 1][i] >= k) ans = (n - i);
    }
    return ans;
  }
  int main() {
    setIO();
    int t;
    re(t);
    while (t--) {
      re(n, k);
      for (int i = (0); i < (n); i++) re(a[i]);
      ps(solve());
    }
    return 0;
  }
  