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>
#pragma GCC optimize(2)
#pragma G++ optimize(2)
using namespace std;
const int MAX_N = 5e2 + 5;
vector<int> ans;
int a[MAX_N];
void op(int k) {
  ans.push_back(k);
  int x = a[k], y = a[k + 1], z = a[k + 2];
  a[k] = z, a[k + 1] = x, a[k + 2] = y;
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    ans.clear();
    int n;
    scanf("%d", &n);
    for (register int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    for (register int k = 1; k <= n - 2; ++k) {
      int mn = 501, id;
      for (register int i = k; i <= n; ++i)
        if (a[i] < mn) id = i, mn = a[i];
      while (id - 2 >= k) op(id - 2), id -= 2;
      if (id != k) op(id - 1), op(id - 1);
    }
    if (a[n - 1] > a[n]) {
      if (a[n - 2] == a[n])
        op(n - 2);
      else {
        bool flag = false;
        for (register int i = 1; i <= n - 2; ++i)
          if (a[i] == a[i + 1]) {
            for (register int j = i; j <= n - 2; ++j) op(j), op(j);
            flag = true;
            break;
          }
        if (!flag) {
          puts("-1");
          continue;
        }
      }
    }
    int len = ans.size();
    printf("%d\n", len);
    for (register int i = 0; i <= len - 1; ++i)
      printf("%d%c", ans[i], i == len - 1 ? '\n' : ' ');
  }
  return 0;
}
