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;
using ll = long long;
using pii = pair<ll, ll>;
const ll MOD = 1e9 + 7, N = 1e5 + 10;
void test() {
  ll n, k;
  cin >> n >> k;
  vector<ll> a(n);
  for (ll i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<ll> upd(2 * k + 2);
  for (ll i = 0; i < n / 2; i++) {
    ll sm = a[i] + a[n - i - 1];
    ll mn = min(a[i], a[n - i - 1]);
    ll d = k - mn;
    upd[sm + 1]++;
    upd[sm + d + 1]--;
    upd[sm + d + 1] += 2;
    upd[2 * k + 1] -= 2;
    ll mx = max(a[i], a[n - i - 1]);
    d = mx - 1;
    upd[sm]--;
    upd[sm - d]++;
    upd[2] += 2;
    upd[sm - d] -= 2;
  }
  ll cur = 0;
  ll ans = 1e18;
  for (ll i = 2; i < 2 * k + 1; i++) {
    cur += upd[i];
    ans = min(ans, cur);
  }
  cout << ans << '\n';
  return;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  ll tt = 1;
  cin >> tt;
  for (ll i = 0; i < tt; i++) test();
  return 0;
}