#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,popcnt")
using namespace std;
template <typename T>
void read(T &x) {
  x = 0;
  char c, f = 1;
  for (c = getchar(); c < '0' || c > '9'; c = getchar())
    if (c == '-') f = 0;
  for (; c >= '0' && c <= '9'; c = getchar())
    x = (((x << 2) + x) << 1) + (c ^ '0');
  x = f ? x : -x;
}
template <typename T, typename... Args>
void read(T &t, Args &...args) {
  read(t);
  read(args...);
}
long long read() {
  long long x;
  read(x);
  return x;
}
const int maxn = 2e3 + 5;
int num[maxn], dp[maxn];
void solve() {
  int n, k, ans = 1e9;
  read(n, k);
  for (int i = (1); i <= (n); i++) read(num[i]), num[i] = i - num[i];
  memset((dp), 0, sizeof(dp));
  for (int i = (1); i <= (n); i++)
    if (num[i] >= 0)
      for (int j = (0); j <= (i - 1); j++)
        if (num[i] >= num[j] && num[i] - num[j] < i - j)
          dp[i] = max(dp[i], dp[j] + 1);
  for (int i = (1); i <= (n); i++)
    if (dp[i] >= k) ans = min(ans, num[i]);
  printf("%d\n", ans == 1e9 ? -1 : ans);
}
int main() {
  int T;
  if (1)
    read(T);
  else
    T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,popcnt")
using namespace std;
template <typename T>
void read(T &x) {
  x = 0;
  char c, f = 1;
  for (c = getchar(); c < '0' || c > '9'; c = getchar())
    if (c == '-') f = 0;
  for (; c >= '0' && c <= '9'; c = getchar())
    x = (((x << 2) + x) << 1) + (c ^ '0');
  x = f ? x : -x;
}
template <typename T, typename... Args>
void read(T &t, Args &...args) {
  read(t);
  read(args...);
}
long long read() {
  long long x;
  read(x);
  return x;
}
const int maxn = 2e3 + 5;
int num[maxn], dp[maxn];
void solve() {
  int n, k, ans = 1e9;
  read(n, k);
  for (int i = (1); i <= (n); i++) read(num[i]), num[i] = i - num[i];
  memset((dp), 0, sizeof(dp));
  for (int i = (1); i <= (n); i++)
    if (num[i] >= 0)
      for (int j = (0); j <= (i - 1); j++)
        if (num[i] >= num[j] && num[i] - num[j] < i - j)
          dp[i] = max(dp[i], dp[j] + 1);
  for (int i = (1); i <= (n); i++)
    if (dp[i] >= k) ans = min(ans, num[i]);
  printf("%d\n", ans == 1e9 ? -1 : ans);
}
int main() {
  int T;
  if (1)
    read(T);
  else
    T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

```