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 long long MOD = 1e9 + 7;
const long long INF = 1e18;
const double PI = acos(-1.0);
const int MAXN = (int)3e5 + 7;
void solve() {
  int n, k;
  cin >> n >> k;
  int a[n], pref[2 * k + 1], temp[2 * k + 1];
  memset(temp, 0, sizeof pref);
  for (int i = 0; i < (int)n; ++i) cin >> a[i];
  int occ[2 * k + 1];
  memset(occ, 0, sizeof occ);
  for (int i = 0; i < (int)n / 2; ++i) {
    int s = a[i] + a[n - i - 1];
    ++occ[s];
    ++temp[min(a[i], a[n - i - 1]) + 1];
    --temp[max(a[i], a[n - i - 1]) + k + 1];
  }
  for (int i = 2; i <= 2 * k; ++i)
    pref[i] = temp[i] + (i == 2 ? 0 : pref[i - 1]);
  int ans = INT_MAX;
  for (int i = 2; i <= 2 * k; ++i)
    ans = min(ans, pref[i] - occ[i] + n - 2 * pref[i]);
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}