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>
  #pragma GCC target("avx")
  #pragma GCC optimize("O3")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  using ll = long long;
  using datas = pair<ll, ll>;
  using ddatas = pair<long double, long double>;
  using tdata = pair<ll, datas>;
  using vec = vector<ll>;
  using mat = vector<vec>;
  using pvec = vector<datas>;
  using pmat = vector<pvec>;
  constexpr ll mod = 1000000007;
  constexpr ll inf = 1LL << 60;
  constexpr long double eps = 1e-9;
  const long double PI = acosl(-1);
  template <class T, class E>
  ostream& operator<<(ostream& os, const pair<T, E>& p) {
    return os << "(" << p.first << "," << p.second << ")";
  }
  template <class T>
  ostream& operator<<(ostream& os, const vector<T>& v) {
    os << "{";
    bool f = false;
    for (auto& x : v) {
      if (f) os << ",";
      os << x;
      f = true;
    }
    os << "}";
    return os;
  }
  template <class T>
  ostream& operator<<(ostream& os, const set<T>& v) {
    os << "{";
    bool f = false;
    for (auto& x : v) {
      if (f) os << ",";
      os << x;
      f = true;
    }
    os << "}";
    return os;
  }
  template <class T>
  ostream& operator<<(ostream& os, const multiset<T>& v) {
    os << "{";
    bool f = false;
    for (auto& x : v) {
      if (f) os << ",";
      os << x;
      f = true;
    }
    os << "}";
    return os;
  }
  template <class T, class E>
  ostream& operator<<(ostream& os, const map<T, E>& v) {
    os << "{";
    bool f = false;
    for (auto& x : v) {
      if (f) os << ",";
      os << x;
      f = true;
    }
    os << "}";
    return os;
  }
  template <class T>
  inline bool chmax(T& a, const T b) {
    bool x = a < b;
    if (x) a = b;
    return x;
  }
  template <class T>
  inline bool chmin(T& a, const T b) {
    bool x = a > b;
    if (x) a = b;
    return x;
  }
  inline void startupcpp(void) noexcept {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
  }
  ll modinv(ll a, const ll m = mod) noexcept {
    ll b = m, u = 1, v = 0, t;
    while (b) {
      t = a / b;
      a -= t * b;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    return (u + m) % m;
  }
  ll moddevide(const ll a, const ll b) noexcept { return (a * modinv(b)) % mod; }
  vec modncrlistp, modncrlistm;
  ll modncr(const ll n, const ll r) noexcept {
    if (n < r) return 0;
    ll i, size = modncrlistp.size();
    if (size <= n) {
      modncrlistp.resize(n + 1);
      modncrlistm.resize(n + 1);
      if (!size) {
        modncrlistp[0] = modncrlistm[0] = 1;
        size++;
      }
      for (i = size; i < (ll)n + 1; ++i)
        modncrlistp[i] = modncrlistp[i - 1] * i % mod;
      modncrlistm[n] = modinv(modncrlistp[n]);
      for (i = n; i > size; --i) modncrlistm[i - 1] = modncrlistm[i] * i % mod;
    }
    return modncrlistp[n] * modncrlistm[r] % mod * modncrlistm[n - r] % mod;
  }
  ll modpow(ll a, ll n, const ll m = mod) {
    if (n < 0) return 0;
    ll res = 1;
    while (n > 0) {
      if (n & 1) res = res * a % m;
      a = a * a % m;
      n >>= 1;
    }
    return res;
  }
  constexpr ll gcd(const ll a, const ll b) noexcept {
    return (!b) ? abs(a) : (a % b == 0) ? abs(b) : gcd(b, a % b);
  }
  constexpr ll lcm(const ll a, const ll b) noexcept { return a / gcd(a, b) * b; }
  ll N, K;
  int main() {
    startupcpp();
    int codeforces;
    cin >> codeforces;
    while (codeforces--) {
      ll i, j;
      cin >> N >> K;
      mat dp(2, vec(N, 0));
      for (i = 0; i < (ll)N; ++i) {
        ll x;
        cin >> x;
        x = i + 1 - x;
        for (j = 0; j < (ll)N; ++j) dp[1][j] = dp[0][j];
        if (x >= 0 && x < N) chmax(dp[1][x], dp[0][x] + 1);
        for (j = 0; j < (ll)N - 1; ++j) chmax(dp[1][j + 1], dp[0][j]);
        swap(dp[0], dp[1]);
      }
      for (i = 0; i < (ll)N; ++i)
        if (dp[0][i] >= K) break;
      if (i == N)
        cout << -1 << "\n";
      else
        cout << i << "\n";
    }
  }
  