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
#include <bits/stdc++.h>
using namespace std;
const double pi = 3.141592653589;
const long long int zero = 0;
long long int mod = 998244353;
const long long int inf = 1e9;
const long long int inff = 9000000000000000000;
char al[26] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm',
               'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};
char capital[26] = {'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I',
                    'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R',
                    'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    int n, ind, a;
    cin >> n;
    int arr[n], arr1[n];
    for (int i = 0; i < n; i += 1) {
      cin >> arr[i];
      arr1[i] = arr[i];
    }
    int duplicate = -1;
    sort(arr1, arr1 + n);
    for (int i = 0; i < n - 1; i += 1) {
      if (arr1[i] == arr1[i + 1]) {
        duplicate = i;
      }
    }
    int ans = 0, x = 0;
    vector<int> v;
    for (int i = 0; i < n - 1; i += 1) {
      if (arr[i] != arr1[i]) {
        for (int j = i + 1; j < n; j += 1) {
          if (arr1[i] == arr[j]) {
            ind = j;
            break;
          }
        }
        while (ind - i >= 2) {
          ans++;
          v.push_back(ind - 2);
          a = arr[ind - 2];
          arr[ind - 2] = arr[ind];
          arr[ind] = arr[ind - 1];
          arr[ind - 1] = a;
          ind -= 2;
        }
        if (ind != i) {
          if (i < n - 2) {
            ans += 2;
            v.push_back(i);
            v.push_back(i);
            a = arr[i];
            arr[i] = arr[i + 1];
            arr[i + 1] = arr[i + 2];
            arr[i + 2] = a;
          } else {
            if (duplicate == -1) {
              x = 1;
              break;
            } else if (duplicate == n - 2) {
              continue;
            }
            ind = n - 1;
            if ((n - 1 - duplicate) % 2 == 1) {
              while (ind - duplicate > 2) {
                ans++;
                v.push_back(ind - 2);
                a = arr[ind - 2];
                arr[ind - 2] = arr[ind];
                arr[ind] = arr[ind - 1];
                arr[ind - 1] = a;
                ind -= 2;
              }
              v.push_back(duplicate);
              ans++;
              for (int ii = duplicate + 2; ii <= n - 3; ii += 2) {
                v.push_back(ii);
                v.push_back(ii);
                ans += 2;
              }
            } else {
              while (ind - duplicate > 2) {
                ans++;
                v.push_back(ind - 2);
                a = arr[ind - 2];
                arr[ind - 2] = arr[ind];
                arr[ind] = arr[ind - 1];
                arr[ind - 1] = a;
                ind -= 2;
              }
              if (duplicate + 2 <= n - 2) {
                v.push_back(duplicate + 1);
                ans++;
              }
              for (int ii = duplicate + 3; ii <= n - 3; ii += 2) {
                v.push_back(ii);
                v.push_back(ii);
                ans += 2;
              }
              v.push_back(duplicate);
              ans++;
            }
          }
        }
      }
    }
    if (x) {
      cout << "-1\n";
      continue;
    }
    cout << ans << "\n";
    for (int i = 0; i < ans; i += 1) {
      cout << v[i] + 1 << " ";
    }
    cout << "\n";
  }
}
