#include <bits/stdc++.h>
using namespace std;
long long mod1 = 998244353, mod2 = 1000000007, limit = 9223372036854775807;
long double pi = 3.1415926535897932;
long long modpow(long long x, long long n, long long m) {
  if (x > m) {
    x %= m;
  }
  if (n == 0) return 1 % m;
  long long u = modpow(x, n / 2, m);
  u = (u * u) % m;
  if (n % 2 == 1) u = (u * x) % m;
  return u;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool isprime(long long n) {
  if (n == 2) return true;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
long long power(long long x, long long n) {
  if (n < 0) {
    return 0;
  }
  long long x_n = 1;
  for (long long i = 0; i < n; i++) {
    x_n *= x;
  }
  return x_n;
}
int n, k, mid;
int arr[2005];
int dp[2005][2005];
int getans(int pos, int index) {
  if (pos == n) {
    if (n - index <= mid) {
      return 0;
    }
    return INT_MIN;
  }
  if (dp[pos][index] != -1) {
    return dp[pos][index];
  }
  int ans = INT_MIN;
  if (getans(pos + 1, index) != INT_MIN) {
    ans = getans(pos + 1, index);
  }
  int score = 0;
  if (arr[pos] == index + 1) {
    score = 1;
  }
  if (getans(pos + 1, index + 1) != INT_MIN) {
    ans = max(ans, score + getans(pos + 1, index + 1));
  }
  return dp[pos][index] = ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
    }
    int l = 0, r = n - 1;
    int ans = -1;
    while (l <= r) {
      mid = (l + r) / 2;
      for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= n; j++) {
          dp[i][j] = -1;
        }
      }
      if (getans(0, 0) >= k) {
        ans = mid;
        r = mid - 1;
      } else {
        l = mid + 1;
      }
    }
    cout << ans << endl;
  }
}

### 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;
long long mod1 = 998244353, mod2 = 1000000007, limit = 9223372036854775807;
long double pi = 3.1415926535897932;
long long modpow(long long x, long long n, long long m) {
  if (x > m) {
    x %= m;
  }
  if (n == 0) return 1 % m;
  long long u = modpow(x, n / 2, m);
  u = (u * u) % m;
  if (n % 2 == 1) u = (u * x) % m;
  return u;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool isprime(long long n) {
  if (n == 2) return true;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
long long power(long long x, long long n) {
  if (n < 0) {
    return 0;
  }
  long long x_n = 1;
  for (long long i = 0; i < n; i++) {
    x_n *= x;
  }
  return x_n;
}
int n, k, mid;
int arr[2005];
int dp[2005][2005];
int getans(int pos, int index) {
  if (pos == n) {
    if (n - index <= mid) {
      return 0;
    }
    return INT_MIN;
  }
  if (dp[pos][index] != -1) {
    return dp[pos][index];
  }
  int ans = INT_MIN;
  if (getans(pos + 1, index) != INT_MIN) {
    ans = getans(pos + 1, index);
  }
  int score = 0;
  if (arr[pos] == index + 1) {
    score = 1;
  }
  if (getans(pos + 1, index + 1) != INT_MIN) {
    ans = max(ans, score + getans(pos + 1, index + 1));
  }
  return dp[pos][index] = ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
    }
    int l = 0, r = n - 1;
    int ans = -1;
    while (l <= r) {
      mid = (l + r) / 2;
      for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= n; j++) {
          dp[i][j] = -1;
        }
      }
      if (getans(0, 0) >= k) {
        ans = mid;
        r = mid - 1;
      } else {
        l = mid + 1;
      }
    }
    cout << ans << endl;
  }
}

```