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>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using ll = long long;
using namespace std;
ll mod = 1e9 + 7;
void inout() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
int main() {
  inout();
  ll t;
  cin >> t;
  while (t--) {
    ll n, k;
    cin >> n >> k;
    vector<ll> a(n + 1);
    for (int i = 1; i <= n; i++) cin >> a[i];
    ll pre[2 * k + 15];
    memset(pre, 0, sizeof(pre));
    ll ans = n, l, r;
    ll tmp = n;
    map<ll, ll> mapi;
    for (int i = 1; i <= n / 2; i++) {
      mapi[a[i] + a[tmp]] += 1;
      l = min(a[i], a[tmp]) + 1;
      r = max(a[i], a[tmp]) + k;
      pre[l] += 1;
      pre[r + 1] -= 1;
      tmp -= 1;
    }
    for (int i = 1; i <= 2 * k + 14; i++) pre[i] = pre[i - 1] + pre[i];
    ll cnt0, cnt1, cnt2;
    for (int i = 2; i <= 2 * k; i++) {
      cnt0 = mapi[i];
      cnt1 = pre[i] - cnt0;
      cnt2 = (n / 2) - cnt1 - cnt0;
      tmp = cnt1 + cnt2 * 2;
      ans = min(ans, tmp);
    }
    cout << ans << "\n";
  }
  return 0;
}
