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 <bits/extc++.h>

#define sz(v) int (v.size())
#define ff first
#define ss second
#define pb push_back
#define all(a) a.begin(),a.end()
typedef long long ll;
typedef long double ld;
using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const int MOD = 1000000009;
mt19937 rnd(std::chrono::high_resolution_clock::now().time_since_epoch().count());

double GetTime() { return clock() / (double) CLOCKS_PER_SEC; };
/// Actual code starts here
int n, k;
const int N = 2005;
int dp[N][N], TT = 1;
int used[N][N];
vector<pair<int, int>> g[N];
int v[N];

int go(int p, int del, int ri, int tt) {
    if (p == n) return 0;
    if (used[p][del] == tt) return dp[p][del];
    used[p][del] = tt;
    int res = go(p + 1, del, ri, tt);
    for (auto i : g[p])
        if (i.ss >= del && i.ss <= ri)
            res = max(res, go(i.ff, i.ss, ri, tt) + 1);
    dp[p][del] = res;
    return res;
}



bool check(int cnt) {
    TT++;
    int res = go(0, 0, cnt, TT);
    return res >= k;
}

void solve() {
    cin >> n >> k;
    for(int i = 0;i <= n;i++) {
        g[i].clear();
        v[i] = MOD;
    }
    for (int i = 0; i < n; i++) {
        cin >> v[i];
        v[i]--;
    }
    for (int i = 0; i < n; i++) {
        if (i >= v[i] && i != v[i])
            g[v[i]].pb({i, i - v[i]});
        else if(i == v[i])
            g[i].pb({i + 1, 0});
    }
    int l = -1, r = n + 5;
    while (r - l > 1) {
        int c = (l + r) / 2;
        if (check(c)) r = c;
        else l = c;
    }
    while (r > -1 && check(r)) r--;
    r++;
    if(r == n + 6) r = -1;
    cout << r << '\n';
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
//    freopen("input.txt", "r", stdin);
    int tt = 1;
    cin >> tt;
    while (tt--)
        solve();
}
