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 optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
  #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,popcnt")
  using namespace std;
  template <typename T>
  void read(T &x) {
    x = 0;
    char c, f = 1;
    for (c = getchar(); c < '0' || c > '9'; c = getchar())
      if (c == '-') f = 0;
    for (; c >= '0' && c <= '9'; c = getchar())
      x = (((x << 2) + x) << 1) + (c ^ '0');
    x = f ? x : -x;
  }
  template <typename T, typename... Args>
  void read(T &t, Args &...args) {
    read(t);
    read(args...);
  }
  long long read() {
    long long x;
    read(x);
    return x;
  }
  const int maxn = 2e3 + 5;
  int num[maxn], dp[maxn];
  void solve() {
    int n, k, ans = 1e9;
    read(n, k);
    for (int i = (1); i <= (n); i++) read(num[i]), num[i] = i - num[i];
    memset((dp), 0, sizeof(dp));
    for (int i = (1); i <= (n); i++)
      if (num[i] >= 0)
        for (int j = (0); j <= (i - 1); j++)
          if (num[i] >= num[j] && num[i] - num[j] < i - j)
            dp[i] = max(dp[i], dp[j] + 1);
    for (int i = (1); i <= (n); i++)
      if (dp[i] >= k) ans = min(ans, num[i]);
    printf("%d\n", ans == 1e9 ? -1 : ans);
  }
  int main() {
    int T;
    if (1)
      read(T);
    else
      T = 1;
    while (T--) {
      solve();
    }
    return 0;
  }
  