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 long long MOD = (long long)1e9 + 7;
const long double PI = 3.141592653589793238462643383279502884197;
long long fac[1] = {1}, inv[1] = {1};
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long mp(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return ret;
}
long long cmb(long long r, long long c) {
  return fac[r] * inv[c] % MOD * inv[r - c] % MOD;
}
priority_queue<int, vector<int>, greater<int>> pq;
int n;
vector<int> v, ord;
vector<int> ans;
inline void swa(int i) {
  swap(v[i + 2], v[i + 1]);
  swap(v[i + 1], v[i]);
  ans.push_back(i);
}
int main() {
  int tc;
  scanf("%d", &tc);
  while (tc--) {
    scanf("%d", &n);
    int tt;
    ord.clear();
    v.clear();
    ans.clear();
    int re = 0, flag = 0;
    for (int i = 0; i < n; i++) {
      scanf("%d", &tt);
      v.push_back(tt);
      for (int j = 0; j < i; j++)
        if (v[j] > v[i]) re++;
    }
    ord = v;
    sort((ord).begin(), (ord).end());
    for (int j = 0; j < n - 1; j++)
      if (ord[j] == ord[j + 1]) flag = 1;
    if (!flag && re % 2) {
      printf("-1\n");
      continue;
    }
    for (int i = 0; i < n - 2; i++) {
      int j;
      for (j = i; j < n; j++)
        if (v[j] == ord[i]) break;
      if (i == j) continue;
      if (j % 2 != i % 2) {
        if (j == n - 1) {
          swa(j - 2);
          swa(j - 2);
          j--;
        } else {
          swa(j - 1), j++;
        }
      }
      for (; j != i; j -= 2) swa(j - 2);
    }
    if (ord == v) {
      printf("%d\n", (int)(ans).size());
      for (int i : ans) printf("%d ", i + 1);
      printf("\n");
      continue;
    } else {
      int j;
      for (j = n - 2; j >= 0; j--) {
        if (v[j] == v[j + 1]) {
          swa(j);
          swa(j);
          swa(j + 1);
          swa(j + 1);
          break;
        }
      }
      if (j == -1) swa(n - 3);
      for (int i = 0; i < n - 2; i++) {
        for (j = i; j < n; j++)
          if (v[j] == ord[i]) break;
        if (i == j) continue;
        if (j % 2 != i % 2) {
          if (j == n - 1) {
            swa(j - 2);
            swa(j - 2);
            j--;
          } else
            swa(j - 1), j++;
        }
        for (; j != i; j -= 2) swa(j - 2);
      }
      printf("%d\n", (int)(ans).size());
      for (int i : ans) printf("%d ", i + 1);
      printf("\n");
    }
  }
}