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;
int const inf = 1e8;
int n;
int a[1 << 10];
void shift(int pos) {
  int b[3];
  b[0] = a[pos + 2];
  b[1] = a[pos];
  b[2] = a[pos + 1];
  for (int j = 0; j < 3; ++j) a[pos + j] = b[j];
}
void print(vector<int> &ans) {
  printf("%d\n", int(ans.size()));
  for (int i : ans) {
    printf("%d ", i);
  }
  printf("\n");
}
void solve() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
  }
  int inv = 0;
  vector<int> ans;
  for (int i = 1; i < n - 1; ++i) {
    int mn = a[i];
    int pos = -1;
    for (int j = i + 1; j <= n; ++j) {
      if (a[j] < mn) {
        mn = a[j];
        pos = j;
      }
    }
    if (pos == -1) continue;
    while (pos - i >= 2) {
      ans.push_back(pos - 2);
      shift(pos - 2);
      pos -= 2;
    }
    if (pos > i) {
      ans.push_back(i);
      ans.push_back(i);
      shift(i);
      shift(i);
    }
  }
  if (is_sorted(a + 1, a + 1 + n)) {
    print(ans);
    return;
  }
  if (a[n] == a[n - 2]) {
    shift(n - 2);
    ans.push_back(n - 2);
    return print(ans);
  }
  int pos = -1;
  for (int i = n - 2; i >= 1; --i) {
    if (a[i] == a[i - 1]) {
      pos = i - 1;
      break;
    }
  }
  if (pos == -1) {
    puts("-1");
    return;
  }
  for (int i = n - 2; i >= pos; --i) {
    shift(i);
    ans.push_back(i);
  }
  assert(is_sorted(a + 1, a + 1 + n));
  print(ans);
}
int main() {
  int T;
  for (scanf("%d", &T); T--;) {
    solve();
  }
}