#include <bits/stdc++.h>
using namespace std;
const long long Mod = 1e9 + 7;
long long poww(long long a, long long b, long long mod) {
  long long res = 1;
  if (b < 0) b = (b % (mod - 1) + mod - 1) % (mod - 1);
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) res = 1ll * res * a % mod;
  return res;
}
void Max(long long& x, long long y) { x = max(x, y); }
void Min(long long& x, long long y) { x = min(x, y); }
void OK(bool yes) { cout << (yes ? "Yes\n" : "No\n"); }
const long long N = 200500, M = 2010, LN = (1 << 24), K = 17, Mx = 4e6 + 50,
                inf = 3e18, Mod1 = 1e9 + 7, Mod2 = 999997457;
const long double pi = acos(-1), eps = 1e-12;
long long Fact[N];
void init() {
  Fact[0] = 1ll;
  for (int i = 1; i < N; i++) Fact[i] = (1ll * Fact[i - 1] * i) % Mod;
}
int Cnk(long long x, long long y) {
  if (y > x) return 0;
  return (((Fact[x] * poww(Fact[y], -1, Mod)) % Mod) *
          poww(Fact[x - y], -1, Mod)) %
         Mod;
}
void Add(long long& x, long long y, int j = 0) {
  long long mod = Mod;
  if (j == 1) mod = Mod1;
  if (j == 2) mod = Mod2;
  x %= mod;
  y %= mod;
  if (x < 0) x += mod;
  if (y < 0) y += mod;
  x = (x + y >= mod ? x + y - mod : x + y);
}
long long dp[2020], a[2020];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int T = 1;
  int Case = 1;
  init();
  cin >> T;
  while (T--) {
    int n, k;
    cin >> n >> k;
    for (int i = 0; i <= n; i++) dp[i] = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      if (i - a[i] >= 0) dp[i] = 1;
    }
    long long Ans = inf;
    if (k == 1) {
      for (int i = 1; i <= n; i++)
        if (a[i] <= i) Min(Ans, i - a[i]);
      if (Ans == inf)
        cout << -1 << '\n';
      else
        cout << Ans << '\n';
      continue;
    }
    for (int i = 1; i <= n; i++) {
      if (i - a[i] < 0) continue;
      for (int j = 1; j < i; j++) {
        if (j - a[j] < 0) continue;
        if (i - a[i] >= j - a[j] && a[i] > a[j]) {
          Max(dp[i], dp[j] + 1);
          if (dp[i] >= k) {
            Min(Ans, i - a[i]);
          }
        }
      }
    }
    if (Ans < inf)
      cout << Ans << '\n';
    else
      cout << -1 << '\n';
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 Mod = 1e9 + 7;
long long poww(long long a, long long b, long long mod) {
  long long res = 1;
  if (b < 0) b = (b % (mod - 1) + mod - 1) % (mod - 1);
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) res = 1ll * res * a % mod;
  return res;
}
void Max(long long& x, long long y) { x = max(x, y); }
void Min(long long& x, long long y) { x = min(x, y); }
void OK(bool yes) { cout << (yes ? "Yes\n" : "No\n"); }
const long long N = 200500, M = 2010, LN = (1 << 24), K = 17, Mx = 4e6 + 50,
                inf = 3e18, Mod1 = 1e9 + 7, Mod2 = 999997457;
const long double pi = acos(-1), eps = 1e-12;
long long Fact[N];
void init() {
  Fact[0] = 1ll;
  for (int i = 1; i < N; i++) Fact[i] = (1ll * Fact[i - 1] * i) % Mod;
}
int Cnk(long long x, long long y) {
  if (y > x) return 0;
  return (((Fact[x] * poww(Fact[y], -1, Mod)) % Mod) *
          poww(Fact[x - y], -1, Mod)) %
         Mod;
}
void Add(long long& x, long long y, int j = 0) {
  long long mod = Mod;
  if (j == 1) mod = Mod1;
  if (j == 2) mod = Mod2;
  x %= mod;
  y %= mod;
  if (x < 0) x += mod;
  if (y < 0) y += mod;
  x = (x + y >= mod ? x + y - mod : x + y);
}
long long dp[2020], a[2020];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int T = 1;
  int Case = 1;
  init();
  cin >> T;
  while (T--) {
    int n, k;
    cin >> n >> k;
    for (int i = 0; i <= n; i++) dp[i] = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      if (i - a[i] >= 0) dp[i] = 1;
    }
    long long Ans = inf;
    if (k == 1) {
      for (int i = 1; i <= n; i++)
        if (a[i] <= i) Min(Ans, i - a[i]);
      if (Ans == inf)
        cout << -1 << '\n';
      else
        cout << Ans << '\n';
      continue;
    }
    for (int i = 1; i <= n; i++) {
      if (i - a[i] < 0) continue;
      for (int j = 1; j < i; j++) {
        if (j - a[j] < 0) continue;
        if (i - a[i] >= j - a[j] && a[i] > a[j]) {
          Max(dp[i], dp[j] + 1);
          if (dp[i] >= k) {
            Min(Ans, i - a[i]);
          }
        }
      }
    }
    if (Ans < inf)
      cout << Ans << '\n';
    else
      cout << -1 << '\n';
  }
  return 0;
}

```