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.
//HEADER FILES AND NAMESPACES

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>  
#include <ext/pb_ds/tree_policy.hpp>   
using namespace std;
using namespace __gnu_pbds;

#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")

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

template <typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>; 

// DEFINE STATEMENTS
const long long infty = 1e18;
#define num1 1000000007
#define num2 998244353
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define REPd(i,a,n) for(ll i=a; i>=n; i--)
#define pb push_back
#define pob pop_back
#define f first
#define s second
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define all(x) x.begin(), x.end()
#define M_PI 3.14159265358979323846
#define epsilon (double)(0.000000001)
#define popcount __builtin_popcountll
#define fileio(x) freopen("input.txt", "r", stdin); freopen(x, "w", stdout);
#define out(x) cout << ((x) ? "Yes\n" : "No\n")
#define sz(x) x.size()


typedef long long ll;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;

// DEBUG FUNCTIONS 
#ifdef LOCALY

template<typename T>
void __p(T a) {
	cout<<a;
}
template<typename T, typename F>
void __p(pair<T, F> a) {
	cout<<"{";
	__p(a.first);
	cout<<",";
	__p(a.second);
	cout<<"}";
}
template<typename T>
void __p(std::vector<T> a) {
	cout<<"{";
	for(auto it=a.begin(); it<a.end(); it++)
		__p(*it),cout<<",}"[it+1==a.end()];
}
template<typename T>
void __p(std::set<T> a) {
	cout<<"{";
	for(auto it=a.begin(); it!=a.end();){
		__p(*it); 
		cout<<",}"[++it==a.end()];
	}

}
template<typename T>
void __p(std::multiset<T> a) {
	cout<<"{";
	for(auto it=a.begin(); it!=a.end();){
		__p(*it); 
		cout<<",}"[++it==a.end()];
	}
}
template<typename T, typename F>
void __p(std::map<T,F> a) {
	cout<<"{\n";
	for(auto it=a.begin(); it!=a.end();++it)
	{
		__p(it->first);
		cout << ": ";
		__p(it->second);
		cout<<"\n";
	}
	cout << "}\n";
}

template<typename T, typename ...Arg>
void __p(T a1, Arg ...a) {
	__p(a1);
	__p(a...);
}
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
	cout<<name<<" : ";
	__p(arg1);
	cout<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
	int bracket=0,i=0;
	for(;; i++)
		if(names[i]==','&&bracket==0)
			break;
		else if(names[i]=='(')
			bracket++;
		else if(names[i]==')')
			bracket--;
	const char *comma=names+i;
	cout.write(names,comma-names)<<" : ";
	__p(arg1);
	cout<<" | ";
	__f(comma+1,args...);
}
#define trace(...) cout<<"Line:"<<__LINE__<<" ", __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...)
#define error(...)
#endif

// DEBUG FUNCTIONS END 

// CUSTOM HASH TO SPEED UP UNORDERED MAP AND TO AVOID FORCED CLASHES
struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }
 
    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());  // FOR RANDOM NUMBER GENERATION

ll mod_exp(ll a, ll b, ll c)
{
	ll res=1;   a=a%c;
	while(b>0)
	{
		if(b%2==1)
			res=(res*a)%c;
		b/=2;
		a=(a*a)%c;
	}
	return res;
}
ll mymod(ll a,ll b)
{
	return (((a = a%b) < 0) ? a + b : a);
}
ll gcdExtended(ll,ll,ll *,ll *);
ll modInverse(ll a, ll m) 
{ 
		ll x, y; 
		ll g = gcdExtended(a, m, &x, &y); 
		g++;   //this line was added just to remove compiler warning
		ll res = (x%m + m) % m;
		return res;
} 
ll gcdExtended(ll a, ll b, ll *x, ll *y) 
{ 
		if (a == 0) 
		{ 
				*x = 0, *y = 1; 
				return b; 
		} 
		ll x1, y1; 
		ll gcd = gcdExtended(b%a, a, &x1, &y1);
		*x = y1 - (b/a) * x1; 
		*y = x1; 
		return gcd; 
}

struct Graph
{
	vector<vector<int>> adj;
	Graph(int n)
	{
		adj.resize(n+1);
	}	
	void add_edge(int a, int b, bool directed = false)
	{
		adj[a].pb(b);
		if(!directed) adj[b].pb(a);
	}
};

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
    ll t;
	cin >> t;
	REP(i, 0, t)
	{
		ll n, k;
		cin >> n >> k;
		vll A(n);
		REP(i, 0, n)
		{
			cin >> A[i];
		}
		vector<vll> dp(n, vll(n+1, 0));    //dp[i][j]: max fixed point if we choose some j out of the first i indices
		dp[0][0] = 0;
		dp[0][1] = (A[0] == 1);
		REP(i, 1, n)
		{
			dp[i][0] = 0;
			REP(j, 1, n+1)
			{
				dp[i][j] = max(dp[i-1][j], dp[i-1][j-1] + (A[i] == j));
			}
		}
		ll ans = -1;
		REPd(i, n, 0)
		{
			if(dp[n-1][i] >= k)
			{
				ans = (n-i);
				break;
			}
		}
		cout << ans << "\n";
	}
	return 0;
}

/*
1. Check borderline constraints. Can a variable you are dividing by be 0?
2. Use ll while using bitshifts
3. Do not erase from set while iterating it
4. Initialise everything
5. Read the task carefully, is something unique, sorted, adjacent, guaranteed??
6. DO NOT use if(!mp[x]) if you want to iterate the map later
7. Are you using i in all loops? Are the i's conflicting?
*/ 



