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;
int main() {
  int T, num, n, pos, ans, a[501], b[501];
  bool flag;
  scanf("%d", &T);
  vector<int> id;
  while (T--) {
    id.clear();
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      b[i] = a[i];
    }
    sort(a + 1, a + n + 1);
    pos = 0;
    ans = 0;
    flag = false;
    while (pos < n) {
      if (ans > n * n) {
        flag = true;
        break;
      }
      if (pos == n - 2 && a[pos + 1] != b[pos + 1]) {
        if (b[pos] == b[pos + 2]) {
          id.push_back(pos);
          ans += 1;
          break;
        }
        int k = 0;
        for (int i = pos - 1; i >= 1; i--)
          if (b[i] == b[i + 1]) {
            k = i;
            break;
          }
        if (k == 0) flag = true;
        for (int i = k; i <= pos; i++) {
          id.push_back(i);
          id.push_back(i);
          ans += 2;
        }
        break;
      }
      int t = 0, g;
      for (int i = pos + 1; i <= n; i++)
        if (b[i] == a[pos + 1] && (t == 0 || (i - pos - 1) % 2 == 0)) t = i;
      while (t != pos + 1)
        if (t >= pos + 3) {
          ans++;
          id.push_back(t - 2);
          g = b[t];
          b[t] = b[t - 1];
          b[t - 1] = b[t - 2];
          b[t - 2] = g;
          t = t - 2;
        } else {
          ans += 2;
          id.push_back(t - 1);
          id.push_back(t - 1);
          g = b[t];
          b[t] = b[t + 1];
          b[t + 1] = b[t - 1];
          b[t - 1] = g;
          t = t - 1;
        }
      pos++;
    }
    if (flag)
      printf("-1\n");
    else {
      printf("%d\n", ans);
      for (auto it : id) printf("%d ", it);
      printf("\n");
    }
  }
  return 0;
}
