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;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    int a[n];
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    vector<int> ans;
    int pr = -1;
    auto transpose = [&ans, &pr, &a](int i) {
      swap(a[i], a[i + 1]);
      if (pr == -1) {
        pr = i;
        return;
      }
      if (i == pr - 1)
        ans.push_back(i);
      else
        for (int j = min(i, pr); j < max(i, pr); j++) {
          ans.push_back(j);
          ans.push_back(j);
        }
      pr = -1;
    };
    for (int i = 0; i < n; i++)
      for (int j = 0; j + i + 1 < n; j++)
        if (a[j] > a[j + 1]) transpose(j);
    if (pr != -1)
      for (int i = 0; i + 1 < n; i++)
        if (a[i] == a[i + 1]) {
          transpose(i);
          break;
        }
    if (pr != -1)
      printf("-1\n");
    else {
      assert(ans.size() <= n * n);
      printf("%d\n", ans.size());
      for (int i = 0; i < ans.size(); i++) {
        if (i) printf(" ");
        printf("%d", ans[i] + 1);
      }
      printf("\n");
    }
  }
}