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 MAXN = 505;
vector<int> ans;
struct node {
  int val, id;
  bool operator<(const node &o) const {
    if (val == o.val) return id < o.id;
    return val < o.val;
  }
};
node a[MAXN], b[MAXN];
void doit(int n) {
  for (int i = 1; i <= n - 2; i++) {
    int id = i;
    for (int j = i; j <= n; j++)
      if (a[j] < a[id]) id = j;
    for (int j = id; j >= i; j--) {
      if (j - 2 >= i) {
        ans.push_back(j - 2);
        swap(a[j], a[j - 1]);
        swap(a[j - 1], a[j - 2]);
        j--;
      } else if (j - 1 >= i) {
        ans.push_back(j - 1);
        ans.push_back(j - 1);
        swap(a[j], a[j - 1]);
        swap(a[j], a[j + 1]);
      }
    }
  }
}
void solve() {
  ans.clear();
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i].val);
    a[i].id = i;
  }
  memcpy(b, a, sizeof(a));
  doit(n);
  if (a[n - 1].val > a[n].val) {
    memcpy(a, b, sizeof(a));
    bool flag = false;
    for (int i = 1; i <= n && !flag; i++)
      for (int j = i + 1; j <= n && !flag; j++)
        if (a[i].val == a[j].val) {
          swap(a[i], a[j]);
          flag = true;
        }
    ans.clear();
    doit(n);
    if (a[n - 1].val > a[n].val) {
      puts("-1");
      return;
    }
  }
  printf("%d\n", ans.size());
  for (auto v : ans) printf("%d ", v);
  puts("");
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    solve();
  }
  return 0;
}