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>
using namespace std;

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T> using ordered_set =tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update> ;
#define int long long
#define LL long long
#define UL unsigned long long
#define BASE 1000000000000000000ull
#define ii pair < int , int >
#define iii pair < int , ii >
#define iv pair < ii , ii >
#define vi vector <int> 
#define vii vector <vi>
#define vp vector <ii>
#define vpp vector <vp>

#define FOR(i , x , n) for(int i = x ; i <= n ; ++i)
#define REP(i , n) for(int i = 0 ; i < n ; ++i)
#define FORD(i , n , x) for(int i = n ; i >= x ; --i)
#define FORC(v , a) for(auto v : a)

#define fi first
#define se second
#define pb push_back
#define pf push_front
#define pob pop_back
#define pof pop_front
#define lovethi ios::sync_with_stdio(NULL);cin.tie(NULL);cout.tie(NULL);
#define fill_a(x , i) memset(x , i ,sizeof x) ;
#define ar array
const int oo = 1e9 + 7 ;
const int N = 2e5 + 5 ;
const int MOD = 1e9 + 7 ;

int n, k, numTest;
signed main() {
    lovethi  
    cin >> numTest;
    while(numTest --) {
        cin >> n >> k;
        vi dp(n + 1, 0);
        FOR(i, 1, n) {
            int x; cin >> x;
            FORD(j, n, 1) 
                dp[j] = max(dp[j], dp[j - 1] + (x == j));
        }
        int ans = 0;
        FORD(i, n, 1) {
            if(dp[i] >= k) {
                ans = n - i;
                break;
            }
        }
        cout << (ans > 0 ? ans : -1) << '\n';
 
    }  
                           
}

                                                                                                                                                             
/*
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