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;
const long long N = 998244353;
const long long mod = 1e9 + 7;
long long binpow(long long b, long long p) {
  long long ans = 1;
  while (p > 0) {
    if (p & 1) ans = (ans * b) % mod;
    b = (b * b) % mod;
    p >>= 1;
  }
  return ans;
}
long long inv(long long x) { return binpow(x, mod - 2); }
long long divide(long long x, long long y) {
  return ((x % mod) * (inv(y))) % mod;
}
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long tt, i, n, k;
  cin >> tt;
  while (tt--) {
    cin >> n >> k;
    vector<long long> v1(n);
    for (i = 0; i < n; i++) cin >> v1[i];
    vector<long long> v2(2 * k + 2, 0);
    v2[0] = LLONG_MAX;
    v2[1] = v2[0];
    for (i = 0; i < n / 2; i++) {
      long long small = min(v1[i], v1[n - 1 - i]);
      long long big = max(v1[i], v1[n - 1 - i]);
      v2[2] += 2;
      v2[small + 1] -= 1;
      v2[small + big] -= 1;
      v2[small + big + 1] += 1;
      v2[big + k + 1] += 1;
    }
    for (i = 3; i < 2 * k + 1; i++) v2[i] += v2[i - 1];
    long long min1 = v2[2];
    for (i = 2; i < 2 * k + 1; i++) min1 = min(min1, v2[i]);
    cout << min1 << '\n';
  }
  return 0;
}