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.
#include <bits/stdc++.h>
//For ordered_set
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define MOD 1000000007
#define test  \
    int t;    \
    cin >> t; \
    while (t--)
#define init(arr, val) memset(arr, val, sizeof(arr))
#define loop(i, a, b) for (int i = a; i < b; i++)
#define loopr(i, a, b) for (int i = a; i >= b; i--)
#define loops(i, a, b, step) for (int i = a; i < b; i += step)
#define looprs(i, a, b, step) for (int i = a; i >= b; i -= step)
#define ull unsigned long long int
#define ll long long int
#define P pair
#define PLL pair<long long, long long>
#define PII pair<int, int>
#define PUU pair<unsigned long long int, unsigned long long int>
#define L list
#define V vector
#define D deque
#define ST set
#define MS multiset
#define M map
#define UM unordered_map
#define mp make_pair
#define pb push_back
#define pf push_front
#define MM multimap
#define F first
#define S second
#define IT iterator
#define RIT reverse_iterator
#define FAST                          \
    ios_base::sync_with_stdio(false); \
    cin.tie();                        \
    cout.tie();
#define FILE_READ_IN freopen("input.txt", "r", stdin);
#define FILE_READ_OUT freopen("output.txt", "w", stdout);
#define all(a) a.begin(), a.end()
using namespace std;
// For ordered_set
using namespace __gnu_pbds;
template <typename T>
using ord_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const ll maxn = 1e5;
const ll inf = 1e9;
const double pi = acos(-1);

void solve()
{
    int n, k;
    cin >> n >> k;
    vector<vector<int>> need(n + 3, vector<int>(n + 3, 0));
    vector<vector<int>> dp(n + 3, vector<int>(n + 3, 0));

    vector<int> arr(n);
    loop(i, 0, n)
    {
        cin >> arr[i];
        arr[i]--;
        if (arr[i] <= i)
        {
            for (int j = i; j >= 0; j--)
                need[j][i - arr[i]]++;
        }
    }

    dp[0][0] = need[0][0];
    //cout << need[0][1] << endl;

    loop(pos, 0, n)
    {
        loop(rem, 0, pos + 2)
        {
            if (pos > 0)
                dp[pos][rem] = dp[pos - 1][rem];
            if (rem >= 1)
            {
                if (pos == 0)
                {
                    dp[pos][rem] = max(dp[pos][rem], 0 - need[pos][rem - 1] + need[pos + 1][rem]);
                }
                else
                {
                    dp[pos][rem] = max(dp[pos][rem], dp[pos - 1][rem - 1] - need[pos][rem - 1] + need[pos + 1][rem]);
                }
            }
        }
    }
    int ans = -1;
    loop(rem, 0, n + 1)
    {
        if (dp[n - 1][rem] >= k)
        {
            ans = rem;
            break;
        }
    }
    cout << ans << endl;
}

int main()
{
    int t = 0;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}