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 = 502;
int t, n, a[N], b[N];
vector<int> ans;
void op(int i) {
  ans.push_back(i);
  int tmp = a[i + 2];
  a[i + 2] = a[i + 1], a[i + 1] = a[i];
  a[i] = tmp;
}
void solve(int l, int r) {
  int mn = 1000000;
  for (int i = l; i <= r; i++) mn = min(mn, a[i]);
  if (a[l] == mn) return;
  int pos = 0;
  for (int i = l; i <= r; i++)
    if (a[i] == mn) pos = i;
  if ((pos - l) % 2 == 1) {
    if (pos == r) op(pos - 2), pos = pos - 2;
    op(pos - 1);
    pos++;
  }
  for (int i = pos - 2; i >= l; i -= 2) {
    op(i);
  }
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    ans.clear();
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i], a[i] *= 1000;
    int inv = 0;
    for (int i = 1; i <= n; i++)
      for (int j = i + 1; j <= n; j++)
        if (a[i] > a[j]) inv++;
    for (int i = 1; i <= n; i++) {
      int cnt = 0;
      for (int j = 1; j < i; j++)
        if (b[i] == b[j]) a[i]++;
    }
    if (inv % 2) {
      bool fg = 0;
      for (int i = 1; i <= n; i++) {
        if (fg) break;
        for (int j = i + 1; j <= n; j++) {
          if (fg) break;
          if (b[i] == b[j]) {
            swap(a[i], a[j]);
            fg = 1;
          }
        }
      }
      if (!fg) {
        printf("-1\n");
        continue;
      }
    }
    for (int i = 1; i <= n; i++) {
      solve(i, n);
    }
    bool fg = 1;
    for (int i = 1; i < n; i++)
      if (a[i] > a[i + 1]) fg = 0;
    if (!fg) {
      printf("-1\n");
      continue;
    }
    printf("%d\n", ans.size());
    for (auto x : ans) printf("%d ", x);
    printf("\n");
    for (int i = 1; i < n; i++) assert(a[i] <= a[i + 1]);
  }
}