#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using db = double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vs = vector<str>;
using vc = vector<char>;
using vpi = vector<pi>;
using vpl = vector<pl>;
template <class T>
using V = vector<T>;
template <class T, size_t SZ>
using AR = array<T, SZ>;
template <class T, class U>
using PR = pair<T, U>;
template <class T, class U>
using umap = unordered_map<T, U>;
template <class T>
using uset = unordered_set<T>;
template <class T>
using mset = multiset<T>;
const int MOD = 1e9 + 7;
const int MX = 2e5 + 5;
const ll INFL = ll(3e18) + 10;
const int INF = int(1e9) + 10;
const ld PI = acos((ld)-1);
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
template <class T>
using pqg = priority_queue<T, vector<T>, greater<T>>;
template <class T>
using pql = priority_queue<T, vector<T>, less<T>>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
constexpr int pct(int x) { return __builtin_popcount(x); }
constexpr int bits(int x) { return x == 0 ? 0 : 31 - __builtin_clz(x); }
constexpr int p2(int x) { return 1 << x; }
constexpr int msk2(int x) { return p2(x) - 1; }
ll cdiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); }
ll fdiv(ll a, ll b) { return a / b - ((a ^ b) < 0 && a % b); }
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
template <class T, class U>
T fstTrue(T lo, T hi, U first) {
  hi++;
  assert(lo <= hi);
  while (lo < hi) {
    T mid = lo + (hi - lo) / 2;
    first(mid) ? hi = mid : lo = mid + 1;
  }
  return lo;
}
template <class T, class U>
T lstTrue(T lo, T hi, U first) {
  lo--;
  assert(lo <= hi);
  while (lo < hi) {
    T mid = lo + (hi - lo + 1) / 2;
    first(mid) ? lo = mid : hi = mid - 1;
  }
  return lo;
}
template <class T>
void remDup(vector<T>& v) {
  sort(begin(v), end(v));
  v.erase(unique(begin(v), end(v)), end(v));
}
template <class T, class U>
void erase(T& t, const U& u) {
  auto it = t.find(u);
  assert(it != end(t));
  t.erase(it);
}
str to_string(char c) { return str(1, c); }
str to_string(const char* second) { return (str)second; }
str to_string(str second) { return second; }
str to_string(bool b) { return to_string((int)b); }
template <class T, class U>
str to_string(pair<T, U> p) {
  return to_string(p.first) + " " + to_string(p.second);
}
template <class T, class U>
str to_string(V<pair<T, U>> v) {
  bool fst = 1;
  str res = "";
  for (const auto& x : v) {
    if (!fst) res += " ";
    fst = 0;
    res += to_string(x);
  }
  return res;
}
template <class T>
str to_string(T v) {
  bool fst = 1;
  str res = "";
  for (const auto& x : v) {
    if (!fst) res += " ";
    fst = 0;
    res += to_string(x);
  }
  return res;
}
void DBG() { cerr << "]" << endl; }
template <class T, class... U>
void DBG(const T& t, const U&... u) {
  cerr << to_string(t);
  if (sizeof...(u)) cerr << ", ";
  DBG(u...);
}
void unsyncIO() { cin.tie(0)->sync_with_stdio(0); }
void setPrec() { cout << fixed << setprecision(15); }
void setIn(str second) { freopen(second.c_str(), "r", stdin); }
void setOut(str second) { freopen(second.c_str(), "w", stdout); }
void setIO(str second = "") {
  unsyncIO();
  setPrec();
  if (int((second).size())) setIn(second + ".in"), setOut(second + ".out");
}
const int nax = 2005;
int dp[nax][nax];
int n, k;
void reset() {
  for (int i = (0); i < (n + 5); ++i)
    for (int j = (0); j < (n + 5); ++j) dp[i][j] = 0;
}
void solve() {
  cin >> n >> k;
  reset();
  vi a(n);
  for (auto& x : a) cin >> x;
  for (int i = (0); i < (n); ++i) {
    for (int j = (0); j < (n); ++j) {
      ckmax(dp[i + 1][j + 1], dp[i][j]);
      int ni = i - j + 1;
      ckmax(dp[i + 1][j], dp[i][j] + (ni == a[i]));
    }
  }
  for (int i = (0); i < (n); ++i) {
    if (dp[n][i] >= k) {
      cout << i << '\n';
      return;
    }
  }
  cout << -1 << '\n';
}
int main() {
  setIO();
  int TT = 1;
  cin >> TT;
  for (int _ = (0); _ < (TT); ++_) solve();
  exit(0);
}

### Prompt

Construct a CPP code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using db = double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vs = vector<str>;
using vc = vector<char>;
using vpi = vector<pi>;
using vpl = vector<pl>;
template <class T>
using V = vector<T>;
template <class T, size_t SZ>
using AR = array<T, SZ>;
template <class T, class U>
using PR = pair<T, U>;
template <class T, class U>
using umap = unordered_map<T, U>;
template <class T>
using uset = unordered_set<T>;
template <class T>
using mset = multiset<T>;
const int MOD = 1e9 + 7;
const int MX = 2e5 + 5;
const ll INFL = ll(3e18) + 10;
const int INF = int(1e9) + 10;
const ld PI = acos((ld)-1);
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
template <class T>
using pqg = priority_queue<T, vector<T>, greater<T>>;
template <class T>
using pql = priority_queue<T, vector<T>, less<T>>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
constexpr int pct(int x) { return __builtin_popcount(x); }
constexpr int bits(int x) { return x == 0 ? 0 : 31 - __builtin_clz(x); }
constexpr int p2(int x) { return 1 << x; }
constexpr int msk2(int x) { return p2(x) - 1; }
ll cdiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); }
ll fdiv(ll a, ll b) { return a / b - ((a ^ b) < 0 && a % b); }
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
template <class T, class U>
T fstTrue(T lo, T hi, U first) {
  hi++;
  assert(lo <= hi);
  while (lo < hi) {
    T mid = lo + (hi - lo) / 2;
    first(mid) ? hi = mid : lo = mid + 1;
  }
  return lo;
}
template <class T, class U>
T lstTrue(T lo, T hi, U first) {
  lo--;
  assert(lo <= hi);
  while (lo < hi) {
    T mid = lo + (hi - lo + 1) / 2;
    first(mid) ? lo = mid : hi = mid - 1;
  }
  return lo;
}
template <class T>
void remDup(vector<T>& v) {
  sort(begin(v), end(v));
  v.erase(unique(begin(v), end(v)), end(v));
}
template <class T, class U>
void erase(T& t, const U& u) {
  auto it = t.find(u);
  assert(it != end(t));
  t.erase(it);
}
str to_string(char c) { return str(1, c); }
str to_string(const char* second) { return (str)second; }
str to_string(str second) { return second; }
str to_string(bool b) { return to_string((int)b); }
template <class T, class U>
str to_string(pair<T, U> p) {
  return to_string(p.first) + " " + to_string(p.second);
}
template <class T, class U>
str to_string(V<pair<T, U>> v) {
  bool fst = 1;
  str res = "";
  for (const auto& x : v) {
    if (!fst) res += " ";
    fst = 0;
    res += to_string(x);
  }
  return res;
}
template <class T>
str to_string(T v) {
  bool fst = 1;
  str res = "";
  for (const auto& x : v) {
    if (!fst) res += " ";
    fst = 0;
    res += to_string(x);
  }
  return res;
}
void DBG() { cerr << "]" << endl; }
template <class T, class... U>
void DBG(const T& t, const U&... u) {
  cerr << to_string(t);
  if (sizeof...(u)) cerr << ", ";
  DBG(u...);
}
void unsyncIO() { cin.tie(0)->sync_with_stdio(0); }
void setPrec() { cout << fixed << setprecision(15); }
void setIn(str second) { freopen(second.c_str(), "r", stdin); }
void setOut(str second) { freopen(second.c_str(), "w", stdout); }
void setIO(str second = "") {
  unsyncIO();
  setPrec();
  if (int((second).size())) setIn(second + ".in"), setOut(second + ".out");
}
const int nax = 2005;
int dp[nax][nax];
int n, k;
void reset() {
  for (int i = (0); i < (n + 5); ++i)
    for (int j = (0); j < (n + 5); ++j) dp[i][j] = 0;
}
void solve() {
  cin >> n >> k;
  reset();
  vi a(n);
  for (auto& x : a) cin >> x;
  for (int i = (0); i < (n); ++i) {
    for (int j = (0); j < (n); ++j) {
      ckmax(dp[i + 1][j + 1], dp[i][j]);
      int ni = i - j + 1;
      ckmax(dp[i + 1][j], dp[i][j] + (ni == a[i]));
    }
  }
  for (int i = (0); i < (n); ++i) {
    if (dp[n][i] >= k) {
      cout << i << '\n';
      return;
    }
  }
  cout << -1 << '\n';
}
int main() {
  setIO();
  int TT = 1;
  cin >> TT;
  for (int _ = (0); _ < (TT); ++_) solve();
  exit(0);
}

```