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;
inline long long read() {
  long long x = 0, w = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') w = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return w == 1 ? x : -x;
}
const int N = 5e2 + 10;
int n;
int a[N], vis[N];
int main() {
  int _ = read();
  while (_--) {
    n = read();
    for (int i = 1; i <= n; i++) a[i] = read();
    int base = 1;
    vector<int> ans;
    for (int k = 1; k < n - 1; ++k) {
      int mi = INT_MAX, id = 0;
      for (int i = base; i <= n; ++i) {
        if (a[i] < mi) mi = a[i], id = i;
      }
      while (id >= base + 2) {
        ans.push_back(id - 2);
        int tmp = a[id];
        a[id] = a[id - 1];
        a[id - 1] = a[id - 2];
        a[id - 2] = tmp;
        id = id - 2;
      }
      if (id == base + 1) {
        ans.push_back(base);
        ans.push_back(base);
        int tmp = a[id];
        a[id] = a[id + 1];
        a[id + 1] = a[base];
        a[base] = tmp;
      }
      base++;
    }
    int flag = 1;
    for (int i = 1; i + 1 <= n; ++i)
      if (a[i + 1] < a[i]) flag = 0;
    if (flag) {
      printf("%d\n", ans.size());
      for (int v : ans) printf("%d ", v);
      puts("");
    } else {
      int f = 0, mx = 0;
      for (int i = 1; i <= n; i++) mx = max(mx, a[i]);
      for (int i = 0; i <= mx; i++) vis[i] = 0;
      for (int i = 1; i <= n; i++) vis[a[i]]++;
      for (int i = 1; i <= n; i++)
        if (vis[a[i]] >= 2) f = 1;
      if (f) {
        int id = n;
        while (id - 1 >= 1 && a[id] < a[id - 1]) {
          ans.push_back(id - 2);
          int tmp = a[id];
          a[id] = a[id - 1];
          a[id - 1] = a[id - 2];
          a[id - 2] = tmp;
          id--;
        }
        printf("%d\n", ans.size());
        for (int v : ans) printf("%d ", v);
        puts("");
      } else
        puts("-1");
    }
  }
}
