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;
const int N = 505;
pair<long long, long long> a[N];
int b[N];
int n;
int n2;
int res[N * N];
bool ans = true;
int siz;
void call() {
  if (n2 == 1) {
    return;
  }
  if (n2 == 2) {
    if (b[0] == 1) {
      ans = false;
    }
    return;
  }
  for (int i = 0; i < n2; i++) {
    if (b[i] > b[0]) {
      b[i]--;
    }
  }
  if (b[0] % 2) {
    int temp = b[0];
    while (temp != 1) {
      res[siz] = temp - 2 + 1 + n - n2;
      temp -= 2;
      siz++;
    }
    res[siz] = 1 + n - n2;
    ;
    siz++;
    res[siz] = 1 + n - n2;
    siz++;
    for (int i = 0; i < n2; i++) {
      if (b[i] == 0) {
        b[i] = 1;
      } else if (b[i] == 1) {
        b[i] = 0;
      }
    }
  } else {
    int temp = b[0];
    while (temp) {
      res[siz] = temp - 2 + 1 + n - n2;
      temp -= 2;
      siz++;
    }
  }
  for (int i = 0; i < n2 - 1; i++) {
    b[i] = b[i + 1];
  }
  n2--;
  call();
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 0; i < n; i++) {
      int temp;
      cin >> temp;
      a[i] = {temp, i};
    }
    sort(a, a + n);
    siz = 0;
    for (int i = 0; i < n; i++) {
      b[i] = a[i].second;
    }
    ans = true;
    n2 = n;
    call();
    if (ans) {
      cout << siz << endl;
      for (int i = 0; i < siz; i++) {
        cout << res[i] << " ";
      }
      cout << endl;
      continue;
    }
    for (int i = 0; i < n - 1; i++) {
      if (a[i].first == a[i + 1].first) {
        swap(a[i], a[i + 1]);
        ans = true;
        break;
      }
    }
    siz = 0;
    for (int i = 0; i < n; i++) {
      b[i] = a[i].second;
    }
    n2 = n;
    call();
    if (ans) {
      cout << siz << endl;
      for (int i = 0; i < siz; i++) {
        cout << res[i] << " ";
      }
      cout << endl;
      continue;
    } else {
      cout << -1 << endl;
    }
  }
}