Problem: You are given an array a consisting of n integers (it is guaranteed that n is even, i.e. divisible by 2). All a_i does not exceed some integer k.

Your task is to replace the minimum number of elements (replacement is the following operation: choose some index i from 1 to n and replace a_i with some integer in range [1; k]) to satisfy the following conditions:

  * after all replacements, all a_i are positive integers not greater than k; 
  * for all i from 1 to n/2 the following equation is true: a_i + a_{n - i + 1} = x, where x should be the same for all n/2 pairs of elements. 



You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

The first line of the test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ k ≤ 2 ⋅ 10^5) — the length of a and the maximum possible value of some a_i correspondingly. It is guratanteed that n is even (i.e. divisible by 2). The second line of the test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the i-th element of a.

It is guaranteed that the sum of n (as well as the sum of k) over all test cases does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5, ∑ k ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer — the minimum number of elements you have to replace in a to satisfy the conditions from the problem statement.

Example

Input


4
4 2
1 2 1 2
4 3
1 2 2 1
8 7
6 1 1 7 6 3 4 6
6 6
5 2 6 1 3 4


Output


0
1
4
2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void announce(long long int ans, vector<long long int> &op) {
  cout << ans << "\n";
  for (long long int j = 0; j < op.size(); j++) {
    cout << op[j] + 1 << " ";
  }
  cout << "\n";
}
bool perm(vector<long long int> &v) {
  long long int mp[501] = {0};
  long long int n = v.size();
  for (long long int j = 0; j < n; j++) {
    mp[v[j]]++;
    if (mp[v[j]] >= 2) {
      return false;
    }
  }
  return true;
}
void operation(long long int i, vector<long long int> &v) {
  long long int s1 = v[i];
  long long int s2 = v[i + 1];
  long long int s3 = v[i + 2];
  v[i] = s3;
  v[i + 1] = s1;
  v[i + 2] = s2;
}
long long int solve() {
  long long int n;
  vector<long long int> v;
  cin >> n;
  for (long long int j = 0; j < n; j++) {
    long long int c;
    cin >> c;
    v.push_back(c);
  }
  vector<long long int> d;
  vector<long long int> p = v;
  sort(p.begin(), p.end());
  long long int ans = 0;
  long long int ptr = 0;
  for (true; true; ptr++) {
    if (ptr == n - 3) {
      break;
    }
    for (long long int i = ptr; i < n; i++) {
      if (v[i] == p[ptr]) {
        while (i - ptr >= 2) {
          operation(i - 2, v);
          d.push_back(i - 2);
          i -= 2;
          ans++;
        }
        if (i - ptr == 1) {
          ans += 2;
          d.push_back(ptr);
          d.push_back(ptr);
          operation(ptr, v);
          operation(ptr, v);
        }
        break;
      }
    }
  }
  if (perm(v)) {
    long long int a = v[n - 3];
    long long int b = v[n - 2];
    long long int c = v[n - 1];
    if (a <= b && b <= c) {
      announce(ans, d);
    } else if ((a >= b && a >= c) && (c >= b)) {
      ans += 2;
      d.push_back(n - 3);
      d.push_back(n - 3);
      operation(n - 3, v);
      operation(n - 3, v);
      announce(ans, d);
      return 0;
    } else if ((b >= a && b >= c) && (a >= c)) {
      ans++;
      d.push_back(n - 3);
      operation(n - 3, v);
      announce(ans, d);
      return 0;
    } else {
      cout << -1 << "\n";
    }
  } else {
    long long int a = v[n - 3];
    long long int b = v[n - 2];
    long long int c = v[n - 1];
    if (a <= b && b <= c) {
      announce(ans, d);
    } else if ((a >= b && a >= c) && (c >= b)) {
      ans += 2;
      d.push_back(n - 3);
      d.push_back(n - 3);
      operation(n - 3, v);
      operation(n - 3, v);
      announce(ans, d);
      return 0;
    } else if ((b >= a && b >= c) && (a >= c)) {
      ans++;
      d.push_back(n - 3);
      operation(n - 3, v);
      announce(ans, d);
      return 0;
    } else {
      if (a == min(a, min(b, c))) {
        true;
      } else if (b == min(a, min(b, c))) {
        ans += 2;
        operation(n - 3, v);
        operation(n - 3, v);
        d.push_back(n - 3);
        d.push_back(n - 3);
      } else {
        ans++;
        operation(n - 3, v);
        d.push_back(n - 3);
      }
      long long int i;
      for (long long int j = n - 3; j > 0; j--) {
        if (v[j] == v[j - 1]) {
          i = j - 1;
        }
      }
      for (long long int j = n - 3; j >= i; j--) {
        ans++;
        operation(j, v);
        d.push_back(j);
      }
      announce(ans, d);
    }
  }
  return 0;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}