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 greaterX(vector<int> &v, int x) {
  int l = 0;
  int r = v.size() - 1;
  while (l != r) {
    int mid = (l + r) / 2;
    if (v[mid] <= x)
      l = mid + 1;
    else
      r = mid;
  }
  if (v[r] <= x) r++;
  return v.size() - r;
}
int lowerX(vector<int> &v, int x) {
  int l = 0;
  int r = v.size() - 1;
  while (l != r) {
    int mid = (l + r + 1) / 2;
    if (v[mid] >= x)
      r = mid - 1;
    else
      l = mid;
  }
  if (v[l] >= x) return 0;
  return l + 1;
}
int main() {
  int TC;
  scanf("%d", &TC);
  int vet[200010];
  while (TC--) {
    int n, k;
    scanf("%d %d", &n, &k);
    vector<int> upper;
    vector<int> lower;
    for (int i = 0; i < n; i++) scanf("%d", &vet[i]);
    map<int, int> MAP;
    for (int i = 0; i < n / 2; i++) {
      MAP[vet[i] + vet[n - i - 1]]++;
      upper.push_back(max(vet[i], vet[n - i - 1]) + k);
      lower.push_back(min(vet[i], vet[n - i - 1]) + 1);
    }
    sort(upper.begin(), upper.end());
    sort(lower.begin(), lower.end());
    int best = n;
    for (int x = 1; x <= 2 * k; x++) {
      best = min(best, n / 2 - MAP[x] + greaterX(lower, x) + lowerX(upper, x));
    }
    printf("%d\n", best);
  }
}