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
#include <bits/stdc++.h>
using namespace std;
const int64_t M1 = 998244353;
const int64_t M2 = 1000000007;
int64_t mp[400001];
int64_t b[400001];
int64_t a[200001];
void solve() {
  int64_t n, k;
  cin >> n >> k;
  for (int64_t i = 0; i <= 2 * k; i++) mp[i] = b[i] = 0;
  for (int64_t i = 0; i < n; i++) cin >> a[i];
  for (int64_t i = 0; i < (n / 2); i++) {
    mp[a[i] + a[n - 1 - i]]++;
    b[min(a[i] + 1, a[n - 1 - i] + 1)]++;
    b[max(a[i] + k, a[n - 1 - i] + k) + 1]--;
  }
  for (int64_t i = 1; i <= 2 * k; i++) b[i] += b[i - 1];
  int64_t ans = n;
  for (int64_t i = 2; i <= 2 * k; i++) ans = min(ans, n - b[i] - mp[i]);
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int64_t t;
  cin >> t;
  while (t--) solve();
}
