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>
const int maxn = 2e5 + 10;
int vis[maxn << 1];
int maxx[maxn], minn[maxn];
int a[maxn];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    std::vector<int> vec;
    int n, k;
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++) {
      scanf("%d", a + i);
    }
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= n / 2; i++) {
      int tmp = a[i] + a[n - i + 1];
      if (a[i] >= a[n - i + 1]) {
        maxx[i] = a[i] + k;
        minn[i] = a[n - i + 1] + 1;
      } else
        maxx[i] = a[n - i + 1] + k, minn[i] = a[i] + 1;
      if (!vis[tmp]) vec.push_back(tmp);
      vis[tmp]++;
    }
    std::sort(a + 1, a + 1 + n);
    std::sort(vec.begin(), vec.end());
    std::sort(maxx + 1, maxx + n / 2 + 1);
    std::sort(minn + 1, minn + n / 2 + 1);
    int ans = n;
    for (int i = vec.size() - 1; i >= 0; i--) {
      int tmp = vec[i];
      int sum = n / 2;
      int pos1 = std::lower_bound(maxx + 1, maxx + n / 2 + 1, tmp) - maxx;
      int pos2 = std::upper_bound(minn + 1, minn + n / 2 + 1, tmp) - minn;
      sum -= vis[vec[i]];
      pos1--;
      pos2--;
      pos2 = n / 2 - pos2;
      sum += pos1 + pos2;
      ans = std::min(sum, ans);
    }
    if (ans > n / 2) ans = n / 2;
    printf("%d\n", ans);
  }
}