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;
const int maxn = 505;
int a[maxn];
int t;
int n;
int tot = 0;
vector<int> ans;
void R(int u) {
  ans.push_back(u);
  tot++;
  int t1 = a[u];
  int t2 = a[u + 1];
  int t3 = a[u + 2];
  a[u] = t3;
  a[u + 1] = t1;
  a[u + 2] = t2;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    ans.clear();
    tot = 0;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= n - 2; i++) {
      int u;
      int Min = 1e9;
      for (int j = i; j <= n; j++) {
        if (a[j] < Min) Min = a[j], u = j;
      }
      while (u >= i + 2) R(u - 2), u -= 2;
      if (u == i + 1) {
        R(i);
        R(i);
        continue;
      }
    }
    for (int i = n; i >= 3; i--) {
      int u;
      int Max = -1;
      for (int j = i; j >= 1; j--) {
        if (a[j] > Max) Max = a[j], u = j;
      }
      while (u <= i - 2) R(u), R(u), u += 2;
      if (u == i - 1) {
        R(i - 2);
        continue;
      }
    }
    for (int i = 1; i < n; i++)
      if (a[i] > a[i + 1]) tot = -1;
    printf("%d\n", tot);
    if (tot != -1) {
      for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
      printf("\n");
    }
  }
}