#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
#pragma GCC optimize("-ffloat-store")
using namespace std;
const auto start_time = std::chrono::high_resolution_clock::now();
void aryanc403() {}
const long long int INF = 0xFFFFFFFFFFFFFFFL;
long long int seed;
mt19937 rng(seed = chrono::steady_clock::now().time_since_epoch().count());
inline long long int rnd(long long int l = 0, long long int r = INF) {
  return uniform_int_distribution<long long int>(l, r)(rng);
}
class CMP {
 public:
  bool operator()(pair<long long int, long long int> a,
                  pair<long long int, long long int> b) {
    return !(a.first < b.first || (a.first == b.first && a.second <= b.second));
  }
};
void add(map<long long int, long long int> &m, long long int x,
         long long int cnt = 1) {
  auto jt = m.find(x);
  if (jt == m.end())
    m.insert({x, cnt});
  else
    jt->second += cnt;
}
void del(map<long long int, long long int> &m, long long int x,
         long long int cnt = 1) {
  auto jt = m.find(x);
  if (jt->second <= cnt)
    m.erase(jt);
  else
    jt->second -= cnt;
}
bool cmp(const pair<long long int, long long int> &a,
         const pair<long long int, long long int> &b) {
  return a.first < b.first || (a.first == b.first && a.second < b.second);
}
const long long int mod = 1000000007L;
long long int T, n, i, j, k, in, cnt, l, r, u, v, x, y;
long long int m;
string s;
vector<long long int> a;
vector<vector<long long int> > dp;
long long int solve(long long int n, long long int k) {
  if (k == -1) return 0;
  if (n == -1) {
    return -INF;
  }
  42;
  ;
  long long int &ans = dp[n][k];
  if (ans != -1) return ans;
  ans = -INF;
  ans = max(ans, solve(n - 1, k));
  ans = max(ans, solve(n - 1, k - 1) + (a[n] == k));
  return ans;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> T;
  while (T--) {
    cin >> n >> k;
    a.clear();
    a.reserve(n);
    for (i = 0; i < (n); ++i) {
      cin >> in;
      in--;
      a.push_back(in);
    }
    dp.clear();
    dp.resize(n, vector<long long int>(n, -1));
    long long int ans = -1;
    for (long long int i = n - 1; i >= 0; --i) {
      const long long int c = solve(n - 1, i);
      if (c >= k) {
        ans = n - i - 1;
        break;
      }
    }
    cout << ans << "\n";
  }
  aryanc403();
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
#pragma GCC optimize("-ffloat-store")
using namespace std;
const auto start_time = std::chrono::high_resolution_clock::now();
void aryanc403() {}
const long long int INF = 0xFFFFFFFFFFFFFFFL;
long long int seed;
mt19937 rng(seed = chrono::steady_clock::now().time_since_epoch().count());
inline long long int rnd(long long int l = 0, long long int r = INF) {
  return uniform_int_distribution<long long int>(l, r)(rng);
}
class CMP {
 public:
  bool operator()(pair<long long int, long long int> a,
                  pair<long long int, long long int> b) {
    return !(a.first < b.first || (a.first == b.first && a.second <= b.second));
  }
};
void add(map<long long int, long long int> &m, long long int x,
         long long int cnt = 1) {
  auto jt = m.find(x);
  if (jt == m.end())
    m.insert({x, cnt});
  else
    jt->second += cnt;
}
void del(map<long long int, long long int> &m, long long int x,
         long long int cnt = 1) {
  auto jt = m.find(x);
  if (jt->second <= cnt)
    m.erase(jt);
  else
    jt->second -= cnt;
}
bool cmp(const pair<long long int, long long int> &a,
         const pair<long long int, long long int> &b) {
  return a.first < b.first || (a.first == b.first && a.second < b.second);
}
const long long int mod = 1000000007L;
long long int T, n, i, j, k, in, cnt, l, r, u, v, x, y;
long long int m;
string s;
vector<long long int> a;
vector<vector<long long int> > dp;
long long int solve(long long int n, long long int k) {
  if (k == -1) return 0;
  if (n == -1) {
    return -INF;
  }
  42;
  ;
  long long int &ans = dp[n][k];
  if (ans != -1) return ans;
  ans = -INF;
  ans = max(ans, solve(n - 1, k));
  ans = max(ans, solve(n - 1, k - 1) + (a[n] == k));
  return ans;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> T;
  while (T--) {
    cin >> n >> k;
    a.clear();
    a.reserve(n);
    for (i = 0; i < (n); ++i) {
      cin >> in;
      in--;
      a.push_back(in);
    }
    dp.clear();
    dp.resize(n, vector<long long int>(n, -1));
    long long int ans = -1;
    for (long long int i = n - 1; i >= 0; --i) {
      const long long int c = solve(n - 1, i);
      if (c >= k) {
        ans = n - i - 1;
        break;
      }
    }
    cout << ans << "\n";
  }
  aryanc403();
  return 0;
}

```