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;
long long n, a[569], sq[250069], zs, inf = 1e18;
void cyc(long long x) {
  swap(a[x + 1], a[x + 2]);
  swap(a[x], a[x + 1]);
  zs++;
  sq[zs] = x;
}
int main() {
  long long t, rr, i, j, ii, e;
  pair<long long, long long> mne;
  scanf("%lld", &t);
  for (rr = 0; rr < t; rr++) {
    scanf("%lld", &n);
    for (i = 1; i <= n; i++) {
      scanf("%lld", a + i);
    }
    zs = 0;
    for (i = 1; i <= n - 2; i++) {
      mne = {inf, -1};
      for (j = i; j <= n; j++) {
        mne = min(mne, {a[j], j});
      }
      e = mne.second;
      for (j = e - 2; j >= i; j -= 2) {
        cyc(j);
      }
      if (j == i - 1) {
        for (ii = 0; ii < 2; ii++) {
          cyc(i);
        }
      }
    }
    if (a[n - 1] > a[n]) {
      for (i = n - 2; i; i--) {
        cyc(i);
        if (a[i] <= a[i + 1]) {
          break;
        }
      }
      if (!i) {
        zs = -1;
      }
    }
    printf("%lld\n", zs);
    for (i = 1; i <= zs; i++) {
      printf("%lld%c", sq[i], " \n"[i == zs]);
    }
  }
}