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;
  const long double eps = 1e-8;
  const long long INF = 1e9;
  const long long N = 2e3 + 5;
  const long long mod = 1e9 + 7;
  int n, k;
  int arr[N];
  int dp[N][N], vis[N][N], id;
  int solve(int idx, int shift, int& mid) {
    if (idx == n + 1) return 0;
    if (vis[idx][shift] == id) return dp[idx][shift];
    int ans = 0;
    if (arr[idx] == idx - shift) ans = solve(idx + 1, shift, mid) + 1;
    if (shift + 1 <= mid) ans = max(ans, solve(idx + 1, shift + 1, mid));
    ans = max(ans, solve(idx + 1, shift, mid));
    vis[idx][shift] = id;
    return dp[idx][shift] = ans;
  }
  void testCase() {
    cin >> n >> k;
    for (int i = 1; i <= n; i++) cin >> arr[i];
    int s = 0, e = n, mid, res = INF;
    while (s <= e) {
      mid = (s + e) >> 1;
      id++;
      int mx = solve(1, 0, mid);
      if (mx >= k)
        e = mid - 1, res = min(res, mid);
      else
        s = mid + 1;
    }
    cout << (res == INF ? -1 : res) << '\n';
  }
  int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int tests = 1;
    cin >> tests;
    while (tests--) testCase();
  }
  