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;
#define pb emplace_back
#define mp make_pair
#define fs first
#define sn second
#define cor(x) (((x%mod)+mod)%mod)
#define vec(T) vector<T>
#define printv(v) for(auto i: v)cout<<i<<" ";cout<<"\n";
#define scanv(v) for(auto& i: v)cin>>i;
 
typedef long long int ll;
 
template<typename T, typename U> static inline void amin(T &x, U y){if (y < x)x = y;}  
template<typename T, typename U> static inline void amax(T &x, U y){if (x < y)x = y;} 

const int mod = 1e9+7;
 
//void precomp(){}
 
void solve(){
	int n,k;cin>>n>>k;
	int a[n],dp[n]={0},t[n];
	for(int i=0;i<n;i++){
		cin>>a[i];t[i]=i+1-a[i];
	}
	if(t[0]==0)dp[0]=1;
	for(int i=1;i<n;i++){
		if(t[i]>=0)dp[i]=1;
		for(int j=0;j<i;j++){
			if(t[j]<0)continue;
			else if(t[j]>t[i])continue;
			else if(t[i]+dp[j]>i)continue;
			else {
				amax(dp[i],dp[j]+1);
			}
		}
	}

	printv(t);
	printv(dp);

	int res=n;
	for(int i=0;i<n;i++){
		if(t[i]<0)continue;
		else if(dp[i]<k)continue;
		else {
			amin(res,t[i]);
		}
	}

	if(res!=n)cout<<res<<"\n";
	else cout<<"-1\n";
}
 
int main(){
	ios_base::sync_with_stdio(false); 
	cin.tie(NULL);cout.tie(NULL);
    
    //precomp();
    
    int tc;cin>>tc;while(tc--)
    solve();
    
    return 0;
}