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;
int read() {
  int x;
  scanf("%d", &x);
  return x;
}
const int maxn = 2e5 + 10;
const int INF = 1e9;
int a[maxn], c[maxn], d[maxn];
pair<int, int> b[maxn];
vector<int> ans;
void Move(int x, int bef) {
  while (x > bef + 1) {
    ans.push_back(x - 2);
    c[d[x]] -= 2;
    c[d[x - 1]]++;
    c[d[x - 2]]++;
    int t = d[x];
    d[x] = d[x - 1];
    d[x - 1] = d[x - 2];
    d[x - 2] = t;
    x -= 2;
  }
  if (x == bef + 1) {
    ans.push_back(bef);
    ans.push_back(bef);
    c[d[x]]--;
    c[d[x + 1]]--;
    c[d[x - 1]] += 2;
    int t = d[x];
    d[x] = d[x + 1];
    d[x + 1] = d[x - 1];
    d[x - 1] = t;
  }
}
void Solve(int n) {
  for (int i = 1; i <= n; ++i) Move(c[b[i].second], i);
}
int main() {
  int t = read();
  while (t--) {
    int n = read();
    ans.clear();
    for (int i = 1; i <= n; ++i)
      a[i] = read(), b[i] = make_pair(a[i], i), c[i] = i, d[i] = i;
    sort(b + 1, b + 1 + n);
    int cnt = 0, sm = 0, b1 = 0, b2 = 0;
    for (int i = 1; i <= n; ++i) {
      for (int j = i + 1; j <= n; ++j) {
        if (a[i] > a[j]) {
          cnt++;
          b1 = i;
          b2 = j;
        }
        if (a[i] == a[j]) sm = 1;
      }
    }
    if (!sm && cnt & 1) {
      puts("-1");
      continue;
    } else {
      if (cnt & 1) {
        for (int i = 2; i <= n; ++i) {
          if (b[i].first == b[i - 1].first) {
            swap(b[i].second, b[i - 1].second);
            break;
          }
        }
      }
      Solve(n);
      printf("%d\n", ans.size());
      for (int i = 0; i < ans.size(); ++i) {
        printf("%d ", ans[i]);
      }
      puts("");
    }
  }
  return 0;
}