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 int N = 3e5 + 5;
const double pi = acos(-1.0);
const long long int inf = 0x3f3f3f3f3f3f3f3f;
const long long int mod = 998244353;
bool isPowerOfTwo(int x) { return x && (!(x & (x - 1))); }
void fast() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int modInverse(long long int n, long long int p) {
  return power(n, p - 2, p);
}
int main() {
  fast();
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    int x = 2 * k;
    x += 2;
    int a[x];
    for (int i = 0; i < x; i++) {
      a[i] = 0;
    }
    int b[n];
    for (int i = 0; i < n; i++) {
      cin >> b[i];
    }
    for (int i = 0; i < n / 2; i++) {
      a[min(b[i], b[n - i - 1]) + 1] += -1;
      a[b[i] + b[n - i - 1]] += -1;
      a[b[i] + b[n - i - 1] + 1] += 1;
      a[max(b[i], b[n - i - 1]) + k + 1] += 1;
    }
    long long int sum = n;
    long long int ans = INT_MAX;
    for (int i = 0; i < 2 * k + 1; i++) {
      sum += a[i];
      ans = min(ans, sum);
    }
    cout << ans << endl;
  }
}