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.
#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using vb = vector<bool>;
using vs = vector<string>;
using vc = vector<char>;
using vd = vector<double>;
using pii = pair<int, int>;
using pss = pair<string, string>;
using ll = long long;
using tup = tuple<ll, int, int>;
const int INF = 1e9 + 7;
const int MOD = 1e9 + 7;
template <typename Os, class V>
Os& operator<<(Os& os, V const& v) {
  for (auto const& e : v) os << e << " ";
  return os;
}
vi diff;
int N, K, ans = INF;
const int MAXN = 2001;
pii memo[2001][2001];
pii dp(int i, int th, int have, int moves) {
  if (i > N) return {0, 0};
  pii& ret = memo[i][th];
  if (ret.first == -1) {
    if (diff[i] == th) {
      pii p = dp(i + 1, th, have + 1, moves);
      ret = {p.first + 1, p.second};
    } else {
      pii p1 = dp(i + 1, th + 1, have, moves + 1);
      pii p2 = dp(i + 1, th, have, moves);
      if (p1.first <= p2.first)
        ret = p2;
      else
        ret = {p1.first, p1.second + 1};
    }
  }
  if (ret.first + have >= K) ans = min(ans, ret.second + moves);
  return ret;
}
void solve() {
  cin >> N >> K;
  diff = vi(N + 1, 0);
  int tmp;
  for (int idx = (1); idx <= (N); ++idx) {
    cin >> tmp;
    diff[idx] = idx - tmp;
  }
  const pii INIT = {-1, -1};
  fill_n(&memo[0][0], MAXN * MAXN, INIT);
  ans = INF;
  dp(1, 0, 0, 0);
  cout << (ans == INF ? -1 : ans) << "\n";
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int T;
  cin >> T;
  for (int TC = (0); TC < (T); ++TC) {
    solve();
  }
}
