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>
#include<string.h>
using namespace std;
#define fastIO ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define int long long int
#define fi first
#define se second
#define pub push_back
#define mkp make_pair
#define pi pair<int,int>
#define push push
#define all(v) (v).begin(), (v).end()
#define rep(i, l, r) for(int i=(int)(l);i<(int)(r);i++)
#define repd(i, l, r) for (int i=(int)(l);i>=(int)(r);i--)
int power(int x, unsigned int y){int res = 1;while (y > 0){ if (y & 1){res = res*x;} y = y>>1;x = x*x;}return res;}
int powermod(int x, unsigned int y, int p){int res = 1;x = x % p;while (y > 0){if (y & 1){res = (res*x) % p;}y = y>>1; x = (x*x) % p;}return res;}
#define print2d(mat,n,m){for(int i=0;i<(int)(n);i++){for(int j=0;j<(m);j++){cout<<mat[i][j]<<" "<< "endl";}}}
#define print1d(mat,n){for(int i=0;i<(int)(n);i++)cout<<mat[i]<<" ";cout<<endl;}
#define clr(a,x) memset(a,x,sizeof(a))
#define rr(v) for(auto &val:v)
#define print(v) for (const auto itr : v){cout<<itr<<' ';}cout<<endl;
#define init(v,x) for (auto &itr:v){itr=x;}
#define minpq(x) x,vector<x>,greater<x>
#define ln length()
#define sz size()
const int  inf = 1e18;
void solve(){
    int n,k;
    cin>>n>>k;
    vector<int> arr(n);
    rep(i,0,n){
        cin>>arr[i];
    }
    vector<vector<int> > dp(n+1, vector<int>(n+1,0));
    int ans = inf;
    rep(i,1,n+1){
        rep(j,1,n+1){
            dp[i][j]= max(dp[i-1][j],dp[i-1][j-1]);
            if(arr[i-1]==j){
                dp[i][j]=max(dp[i][j],dp[i-1][j-1]+1);
            }
            if(dp[i][j]>=k){
                ans = min(ans,i-j);
            }
        }
    }
    if(ans==inf){
        cout<<"-1"<<endl;
    }
    else{
        cout<<ans<<endl;
    }
}
main()
{
    int t;
    cin>>t;
    while(t--){
        solve();
    }
}
