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 n, k;
long long arr[200005];
map<long long, long long> mp;
map<long long, long long> mp2;
vector<long long> vec1;
vector<long long> vec2;
long long count_greater(long long val) {
  long long n = vec1.size();
  long long l = 0;
  long long r = n - 1;
  long long idx = n;
  while (l <= r) {
    long long mid = (l + r) / 2;
    if (vec1[mid] <= val)
      l = mid + 1;
    else {
      r = mid - 1;
      idx = mid;
    }
  }
  return (n - idx);
}
long long count_smaller_or_equal(long long key) {
  long long idx = 0;
  long long l = 0;
  long long r = vec2.size() - 1;
  while (l <= r) {
    long long mid = (l + r) / 2;
    if (vec2[mid] > key)
      r = mid - 1;
    else {
      l = mid + 1;
      idx = mid + 1;
    }
  }
  return idx;
}
void solve2() {
  sort(vec1.begin(), vec1.end());
  sort(vec2.begin(), vec2.end());
  long long ans = n / 2;
  for (long long x = 2; x <= 2 * k; x++) {
    long long temp_ans = n / 2;
    long long a = count_greater(x);
    long long b = count_smaller_or_equal(x);
    b -= mp2[x];
    temp_ans += (a + b);
    temp_ans -= mp[x];
    ans = min(ans, temp_ans);
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    mp.clear();
    mp2.clear();
    vec1.clear();
    vec2.clear();
    cin >> n >> k;
    for (long long i = 0; i < n; i++) cin >> arr[i];
    vector<long long> min_vec;
    for (long long i = 0; i < n / 2; i++) {
      long long val1 = arr[i];
      long long val2 = arr[n - 1 - i];
      min_vec.push_back(min(val1, val2));
      long long a = min(val1, val2);
      long long b = max(val1, val2);
      vec1.push_back(a + 1);
      vec2.push_back(b + k);
      mp[val1 + val2]++;
      mp2[b + k]++;
    }
    solve2();
  }
  return 0;
}