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>
using namespace std;
const int Maxn = 1e5 + 10;
const int Inf = 0x7f7f7f7f;
const long long Inf_ll = 1ll * Inf * Inf;
const int Mod = 1e9 + 7;
const double eps = 1e-7;
pair<int, int> p[Maxn];
vector<int> ans;
int n, a[Maxn], b[Maxn];
void init() { ans.clear(); }
void Turn(int pos) {
  int t = a[pos];
  a[pos] = a[pos + 2], a[pos + 2] = a[pos + 1], a[pos + 1] = t;
  ans.push_back(pos);
}
void solve() {
  init();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    scanf("%d", &b[i]), p[i] = make_pair(b[i], i), a[i] = i;
  sort(p + 1, p + 1 + n);
  for (int i = 1; i <= n - 2; i++) {
    int pos = p[i].second, posa = i;
    while (a[posa] != pos) posa++;
    if (posa == i) continue;
    while (posa - 2 >= i) {
      posa -= 2;
      Turn(posa);
    }
    if (posa == i + 1) Turn(i), posa++;
    if (posa == i + 2) Turn(i), posa = i;
  }
  if (p[n].second != a[n]) {
    if (p[n].first == p[n - 1].first)
      ;
    else if (p[n - 1].first == p[n - 2].first)
      Turn(n - 2);
    else {
      int pos = -1;
      for (int i = 1; i <= n - 2; i++)
        if (p[i].first == p[i + 1].first) {
          pos = i;
          break;
        }
      if (pos == -1) return (void)puts("-1");
      for (int i = pos; i <= n - 2; i++) Turn(i), Turn(i);
    }
  }
  printf("%d\n", ((int)ans.size()));
  for (auto x : ans) printf("%d ", x);
  puts("");
}
int main() {
  int T;
  scanf("%d", &T);
  for (int _ = 1; _ <= T; _++) {
    solve();
  }
  return 0;
}
