#include <bits/stdc++.h>
using namespace std;
vector<long long int> Tree;
void make_Tree(vector<long long int> &v) {
  long long int i;
  long long int N = v.size();
  while (N & (N - 1)) N++;
  v.resize(N);
  Tree.resize(2 * N);
  for (i = 0; i < N; i++) {
    Tree[N + i] = v[i];
  }
  for (i = N - 1; i >= 1; i--) {
    Tree[i] = Tree[2 * i] + Tree[2 * i + 1];
  }
}
long long int segment(long long int node, long long int node_left,
                      long long int node_right, long long int query_left,
                      long long int query_right) {
  if (query_left <= node_left && node_right <= query_right) {
    return Tree[node];
  }
  if (query_right < node_left || node_right < query_left) {
    return 0;
  }
  long long int last_left = (node_left + node_right) / 2;
  return segment(2 * node, node_left, last_left, query_left, query_right) +
         segment(2 * node + 1, last_left + 1, node_right, query_left,
                 query_right);
}
void update_Tree(long long int Index, long long int Val) {
  long long int N = Tree.size() / 2;
  long long int Diff = Val - Tree[N + Index - 1];
  for (long long int i = N + Index - 1; i >= 1; i /= 2) {
    Tree[i] += Diff;
  }
}
const long long int maxN = 100001;
vector<long long int> inF(maxN), F(maxN), isPrime(maxN), Ad[maxN];
map<long long int, long long int> Vis;
void dfs(long long int node) {
  if (Vis[node]) return;
  Vis[node] = 1;
  for (auto child : Ad[node]) {
    dfs(child);
  }
  return;
}
long long int pow1(long long int n, long long int k) {
  if (k == 0) return 1;
  if (k == 1) return n;
  long long int X = 1;
  if (k % 2 == 1) X = n % 1000000007;
  long long int Y = pow1(n, k / 2) % 1000000007;
  return (((Y * Y) % 1000000007) * X) % 1000000007;
}
void fact_compute() {
  long long int N = maxN - 1;
  F[0] = 1;
  F[1] = 1;
  for (long long int i = 2; i <= N; i++) {
    F[i] = (F[i - 1] * (i)) % 1000000007;
  }
  inF[N] = pow1(F[N], 1000000007 - 2) % 1000000007;
  for (long long int i = N - 1; i >= 0; i--) {
    inF[i] = (inF[i + 1] * (i + 1)) % 1000000007;
  }
}
void sieve() {
  for (long long int i = 2; i < maxN; i++) {
    isPrime[i] = 1;
  }
  for (long long int i = 2; i < maxN; i++) {
    for (long long int j = 2 * i; j < maxN; j += i) {
      isPrime[j] = 0;
    }
  }
}
long long int gcd(long long int n, long long int m) {
  if (!m) return n;
  return gcd(m, n % m);
}
long long int lcm(long long int n, long long int m) {
  return (n / gcd(n, m)) * m;
}
void set_io() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
}
void solve() {
  long long int n, m, j, k, l, i;
  cin >> n >> k;
  vector<long long int> v(n);
  long long int isOne = 0;
  vector<vector<long long int>> dp(n + 1, vector<long long int>(n + 1, n));
  vector<long long int> cnt(n + 1, 0);
  for (long long int i = 0; i < n; i++) {
    cin >> v[i];
  }
  long long int f = -1;
  long long int ans = n;
  long long int maxi = 0;
  for (i = 0; i < n; i++) {
    long long int val = v[i];
    if (i != 0) {
      if (i + 1 >= v[i]) {
        dp[i][v[i]] = i + 1 - v[i];
        cnt[i] = 1;
      }
      for (j = 0; j < i; j++) {
        if (v[i] > v[j]) {
          if (i - j >= v[i] - v[j]) {
            if (cnt[i] <= cnt[j] + 1) {
              cnt[i] = max(cnt[i], cnt[j] + 1);
              maxi = max(maxi, cnt[i]);
              dp[i][v[i]] = min(dp[i][v[i]], dp[j][v[j]] + i - j);
            }
          }
        }
      }
    } else {
      if (v[i] == 1) {
        cnt[i] = 1;
        maxi = max(maxi, cnt[i]);
        dp[i][1] = 0;
      }
    }
  }
  if (maxi < k) {
    cout << -1;
    cout << "\n";
    return;
  }
  for (i = 0; i < n; i++) {
    if (cnt[i] >= k) {
      ans = min(ans, dp[i][v[i]]);
    }
  }
  cout << ans;
  cout << "\n";
  return;
}
int main() {
  set_io();
  long long int totalTestCase = 1;
  cin >> totalTestCase;
  for (long long int testCase = 1; testCase <= totalTestCase; testCase++) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
vector<long long int> Tree;
void make_Tree(vector<long long int> &v) {
  long long int i;
  long long int N = v.size();
  while (N & (N - 1)) N++;
  v.resize(N);
  Tree.resize(2 * N);
  for (i = 0; i < N; i++) {
    Tree[N + i] = v[i];
  }
  for (i = N - 1; i >= 1; i--) {
    Tree[i] = Tree[2 * i] + Tree[2 * i + 1];
  }
}
long long int segment(long long int node, long long int node_left,
                      long long int node_right, long long int query_left,
                      long long int query_right) {
  if (query_left <= node_left && node_right <= query_right) {
    return Tree[node];
  }
  if (query_right < node_left || node_right < query_left) {
    return 0;
  }
  long long int last_left = (node_left + node_right) / 2;
  return segment(2 * node, node_left, last_left, query_left, query_right) +
         segment(2 * node + 1, last_left + 1, node_right, query_left,
                 query_right);
}
void update_Tree(long long int Index, long long int Val) {
  long long int N = Tree.size() / 2;
  long long int Diff = Val - Tree[N + Index - 1];
  for (long long int i = N + Index - 1; i >= 1; i /= 2) {
    Tree[i] += Diff;
  }
}
const long long int maxN = 100001;
vector<long long int> inF(maxN), F(maxN), isPrime(maxN), Ad[maxN];
map<long long int, long long int> Vis;
void dfs(long long int node) {
  if (Vis[node]) return;
  Vis[node] = 1;
  for (auto child : Ad[node]) {
    dfs(child);
  }
  return;
}
long long int pow1(long long int n, long long int k) {
  if (k == 0) return 1;
  if (k == 1) return n;
  long long int X = 1;
  if (k % 2 == 1) X = n % 1000000007;
  long long int Y = pow1(n, k / 2) % 1000000007;
  return (((Y * Y) % 1000000007) * X) % 1000000007;
}
void fact_compute() {
  long long int N = maxN - 1;
  F[0] = 1;
  F[1] = 1;
  for (long long int i = 2; i <= N; i++) {
    F[i] = (F[i - 1] * (i)) % 1000000007;
  }
  inF[N] = pow1(F[N], 1000000007 - 2) % 1000000007;
  for (long long int i = N - 1; i >= 0; i--) {
    inF[i] = (inF[i + 1] * (i + 1)) % 1000000007;
  }
}
void sieve() {
  for (long long int i = 2; i < maxN; i++) {
    isPrime[i] = 1;
  }
  for (long long int i = 2; i < maxN; i++) {
    for (long long int j = 2 * i; j < maxN; j += i) {
      isPrime[j] = 0;
    }
  }
}
long long int gcd(long long int n, long long int m) {
  if (!m) return n;
  return gcd(m, n % m);
}
long long int lcm(long long int n, long long int m) {
  return (n / gcd(n, m)) * m;
}
void set_io() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
}
void solve() {
  long long int n, m, j, k, l, i;
  cin >> n >> k;
  vector<long long int> v(n);
  long long int isOne = 0;
  vector<vector<long long int>> dp(n + 1, vector<long long int>(n + 1, n));
  vector<long long int> cnt(n + 1, 0);
  for (long long int i = 0; i < n; i++) {
    cin >> v[i];
  }
  long long int f = -1;
  long long int ans = n;
  long long int maxi = 0;
  for (i = 0; i < n; i++) {
    long long int val = v[i];
    if (i != 0) {
      if (i + 1 >= v[i]) {
        dp[i][v[i]] = i + 1 - v[i];
        cnt[i] = 1;
      }
      for (j = 0; j < i; j++) {
        if (v[i] > v[j]) {
          if (i - j >= v[i] - v[j]) {
            if (cnt[i] <= cnt[j] + 1) {
              cnt[i] = max(cnt[i], cnt[j] + 1);
              maxi = max(maxi, cnt[i]);
              dp[i][v[i]] = min(dp[i][v[i]], dp[j][v[j]] + i - j);
            }
          }
        }
      }
    } else {
      if (v[i] == 1) {
        cnt[i] = 1;
        maxi = max(maxi, cnt[i]);
        dp[i][1] = 0;
      }
    }
  }
  if (maxi < k) {
    cout << -1;
    cout << "\n";
    return;
  }
  for (i = 0; i < n; i++) {
    if (cnt[i] >= k) {
      ans = min(ans, dp[i][v[i]]);
    }
  }
  cout << ans;
  cout << "\n";
  return;
}
int main() {
  set_io();
  long long int totalTestCase = 1;
  cin >> totalTestCase;
  for (long long int testCase = 1; testCase <= totalTestCase; testCase++) {
    solve();
  }
  return 0;
}

```