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 pairs[100010][2];
int cnt[400040];
int pre[400040];
int main() {
  int t, n, k, ans;
  scanf("%d", &t);
  for (int w = 0; w < t; w++) {
    scanf("%d %d", &n, &k);
    ans = 2000000000;
    memset(pre, 0, sizeof(pre));
    memset(cnt, 0, sizeof(cnt));
    for (int i = 0; i < n / 2; i++) {
      scanf("%d", &pairs[i][0]);
    }
    for (int i = n / 2; i < n; i++) {
      scanf("%d", &pairs[n - i - 1][1]);
    }
    for (int i = 0; i < n / 2; i++) {
      cnt[pairs[i][0] + pairs[i][1]]++;
      pre[min(pairs[i][0], pairs[i][1]) + 1]++;
      pre[max(pairs[i][0], pairs[i][1]) + k + 1]--;
    }
    for (int i = 0; i < 2 * k; i++) {
      pre[i + 1] += pre[i];
    }
    for (int i = 2; i <= 2 * k; i++) {
      ans = min(ans, (pre[i] - cnt[i]) + (n / 2 - pre[i]) * 2);
    }
    printf("%d\n", ans);
  }
  return 0;
}