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 max(int a, int b) {
  if (a > b) {
    return a;
  } else {
    return b;
  }
}
int min(int a, int b) {
  if (a < b) {
    return a;
  } else {
    return b;
  }
}
void solution(vector<int> ar, int n, int k) {
  vector<long long> pref(2 * k + 2, 0);
  vector<long long> cnt(2 * k + 1, 0);
  for (int i = 0; i < n / 2; i++) {
    cnt[ar[i] + ar[n - 1 - i]]++;
  }
  for (int i = 0; i < n / 2; i++) {
    int lf1 = 1 + ar[i];
    int lf2 = 1 + ar[n - 1 - i];
    int rt1 = k + ar[i];
    int rt2 = k + ar[n - 1 - i];
    assert(max(lf1, lf2) <= min(rt1, rt2));
    pref[min(lf1, lf2)]++;
    pref[max(rt1, rt2) + 1]--;
  }
  for (int i = 1; i <= 2 * k + 1; i++) {
    pref[i] += pref[i - 1];
  }
  long long ans = 1e9;
  for (int sum = 2; sum <= 2 * k; sum++) {
    ans = min(ans, (pref[sum] - cnt[sum]) + (n / 2 - pref[sum]) * 2);
  }
  cout << ans;
}
int main() {
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    int n, k;
    cin >> n >> k;
    vector<int> ar(n);
    for (int j = 0; j < n; j++) {
      cin >> ar[j];
    }
    solution(ar, n, k);
    cout << "\n";
  }
}