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 int max(long long int a, long long int b) {
  if (a > b) {
    return a;
  }
  return b;
}
long long int min(long long int a, long long int b) {
  if (a < b) {
    return a;
  }
  return b;
}
long long int n, k;
long long int arr[200005];
long long int power(long long int x, long long int n) {
  long long int res = 1;
  while (n > 0) {
    if ((long long)n % 2 == 1) {
      res = (res * x);
    }
    x = (x * x);
    n = n / 2;
  }
  return res;
}
void solve() {
  long long int ctr[2 * k + 1];
  long long int prefix[2 * k + 1];
  memset(prefix, 0, sizeof(prefix));
  memset(ctr, 0, sizeof(ctr));
  for (int i = 0; i < n / 2; i++) {
    ctr[arr[i] + arr[n - i - 1]]++;
    long long int mn = min(arr[i], arr[n - i - 1]);
    long long int mx = max(arr[i], arr[n - i - 1]);
    mn += 1;
    mx += k;
    prefix[mn]++;
    prefix[mx + 1]--;
  }
  for (int i = 1; i < 2 * k + 1; i++) {
    prefix[i] += prefix[i - 1];
  }
  long long int ans = LONG_MAX;
  for (int sum = 2; sum <= 2 * k; sum++) {
    ans = min(ans, (prefix[sum] - ctr[sum]) + (n / 2 - prefix[sum]) * 2);
  }
  cout << ans << "\n";
}
int main() {
  int tc;
  cin >> tc;
  while (tc--) {
    cin >> n >> k;
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
    }
    solve();
  }
  return 0;
}