#include <bits/stdc++.h>
using namespace std;
const long long maxs = 1e6 + 5;
long long root[maxs];
class cmp {
  bool operator()(const long long &a, const long long &b) { return a < b; }
};
void init() { iota(root, root + maxs, 0); }
long long find(long long u) {
  long long f;
  if (root[u] == u)
    return u;
  else {
    long long f = find(root[u]);
    root[u] = f;
    return f;
  }
}
void Union(long long x, long long y) {
  long long u = find(x);
  long long v = find(y);
  root[v] = u;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void update(vector<long long> &BIT, long long ind, long long val) {
  while (ind < BIT.size()) {
    BIT[ind] += val;
    ind += (ind & (-ind));
  }
}
long long query(vector<long long> &BIT, long long ind) {
  long long sum = 0;
  while (ind > 0) {
    sum += BIT[ind];
    ind -= (ind & (-ind));
  }
  return sum;
}
void count_factors() {
  long long i, j;
  long long numOfDivisors[maxs] = {0};
  long long ans[maxs] = {0};
  for (i = 1; i < maxs; i++)
    for (j = i; j < maxs; j += i) {
      numOfDivisors[j]++;
      ans[j]++;
    }
  for (i = 1; i < maxs; i++)
    for (j = i; j < maxs; j += i)
      if (numOfDivisors[j / i] == 4) ans[j] = min(ans[j], ans[i]);
}
long long get(long long x, long long y) {
  if (x == 0) return 1;
  long long cnt = 0;
  while (x % y == 0) {
    cnt++;
    x /= y;
  }
  return cnt;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t, n, i, k, y, x, j, m, r, l;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    vector<long long> arr(n + 1);
    for (i = 1; i <= n; i++) {
      cin >> arr[i];
    }
    vector<vector<long long>> dp(n + 1, vector<long long>(n + 1, 0));
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= i; j++) {
        dp[i][j] = dp[i - 1][j];
        dp[i][j] = max(dp[i - 1][j - 1] + (arr[i] == j), dp[i][j]);
      }
    }
    long long ans = 0;
    for (i = 1; i <= n; i++) {
      if (dp[n][i] >= k) ans = i;
    }
    if (ans == 0)
      cout << -1 << '\n';
    else
      cout << n - ans << '\n';
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to 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;
const long long maxs = 1e6 + 5;
long long root[maxs];
class cmp {
  bool operator()(const long long &a, const long long &b) { return a < b; }
};
void init() { iota(root, root + maxs, 0); }
long long find(long long u) {
  long long f;
  if (root[u] == u)
    return u;
  else {
    long long f = find(root[u]);
    root[u] = f;
    return f;
  }
}
void Union(long long x, long long y) {
  long long u = find(x);
  long long v = find(y);
  root[v] = u;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void update(vector<long long> &BIT, long long ind, long long val) {
  while (ind < BIT.size()) {
    BIT[ind] += val;
    ind += (ind & (-ind));
  }
}
long long query(vector<long long> &BIT, long long ind) {
  long long sum = 0;
  while (ind > 0) {
    sum += BIT[ind];
    ind -= (ind & (-ind));
  }
  return sum;
}
void count_factors() {
  long long i, j;
  long long numOfDivisors[maxs] = {0};
  long long ans[maxs] = {0};
  for (i = 1; i < maxs; i++)
    for (j = i; j < maxs; j += i) {
      numOfDivisors[j]++;
      ans[j]++;
    }
  for (i = 1; i < maxs; i++)
    for (j = i; j < maxs; j += i)
      if (numOfDivisors[j / i] == 4) ans[j] = min(ans[j], ans[i]);
}
long long get(long long x, long long y) {
  if (x == 0) return 1;
  long long cnt = 0;
  while (x % y == 0) {
    cnt++;
    x /= y;
  }
  return cnt;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t, n, i, k, y, x, j, m, r, l;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    vector<long long> arr(n + 1);
    for (i = 1; i <= n; i++) {
      cin >> arr[i];
    }
    vector<vector<long long>> dp(n + 1, vector<long long>(n + 1, 0));
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= i; j++) {
        dp[i][j] = dp[i - 1][j];
        dp[i][j] = max(dp[i - 1][j - 1] + (arr[i] == j), dp[i][j]);
      }
    }
    long long ans = 0;
    for (i = 1; i <= n; i++) {
      if (dp[n][i] >= k) ans = i;
    }
    if (ans == 0)
      cout << -1 << '\n';
    else
      cout << n - ans << '\n';
  }
  return 0;
}

```