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<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;

template<typename T>
using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;

const int infI=1e9+5;
const long long infL=1e16;
const int MOD=1e9+7;
const int MAX=2e5+5;

#define mp make_pair
#define pb push_back
#define ii pair<int,int>
#define iii pair<int,ii>
#define ll long long
#define pll pair<ll,ll>
#define vi vector<int>
#define vvi vector<vector<int> >
#define vl vector<ll>
#define forn(i,a,b) for(ll i=a;i<b;i++)
#define fastio ios_base::sync_with_stdio(0);cin.tie(0)
#define endl '\n'
#define popb pop_back()
#define se second
#define fi first
#define test(t) int t;cin>>t;while(t--)
#define debug(x) cerr<<"=>"<<#x<<" = "<<x<<endl

ll power(ll a,ll b){
    if(b==0){
        return 1;
    }
    if(b==1){
        return a;
    }
    ll temp=power(a,b/2);
    temp=(temp*temp)%MOD;
    if(b%2){
        temp=(temp*a)%MOD;
    }
    return temp;
}

//only use if mod is prime
ll mod_inv(ll a,ll mod){
    a%=mod;
    return power(a,mod-2);
}

void extended_gcd(ll a,ll b,ll &x,ll &y){
    if(b==0){
        x=1;
        y=0;
        return;
    }
    ll x1,y1;
    extended_gcd(b,a%b,x1,y1);
    x=y1;
    y=x1-((a/b)*y1);
    return;
}

ll ceiling(ll a,ll b){
    return (a+b-1)/b;
}

//TLE can also occur if vectors are not passed by reference, debug statements can also cause TLE.
//priority_queue implements max heap by default

int a[2005];
int dp[2005][2005];

int main(){
    fastio;
    // think about edge cases like n=1(especially graph questions).
    // If you need to check for overflows, add the -1 condition in power function.
    test(zz){
        int n,k;
        cin>>n>>k;
        forn(i,1,n+1){
            cin>>a[i];
        }
        forn(i,0,n+1){
            forn(j,0,n+1){
                dp[i][j] = 0;
            }
        }
        forn(i,0,n){
            forn(j,0,n){
                dp[i+1][j] = max(dp[i+1][j], dp[i][j]);//delete this element
                dp[i+1][j+1] = max(dp[i+1][j+1], (a[i+1] == j+1) + dp[i][j]);
            }
        }
        int ans = -1;
        forn(j,0,n+1){
            if(dp[n][j]>=k){
                ans = n-j;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}