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>
double pi = acos(-1);
const long long mod = 1000000007;
using namespace std;
void optimizeIO() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
const long long N = 1000005;
long long fact[N], invfact[N];
long long power(long long x, long long y) {
  if (y <= 0) return 1LL;
  long long z = power(x, y / 2);
  if (y % 2) return (((z * z) % mod) * x) % mod;
  return (z * z) % mod;
}
void pre() {
  fact[0] = invfact[0] = invfact[1] = fact[1] = 1;
  for (long long i = 2; i < N; i++) fact[i] = (i * fact[i - 1]) % mod;
  for (long long i = 2; i < N; i++)
    invfact[i] = (invfact[i - 1] * power(i, mod - 2)) % mod;
}
long long nCr(long long n, long long k) {
  if (k < 0 || k > n) return 0;
  return (((fact[n] * invfact[n - k]) % mod) * invfact[k]) % mod;
}
string repeat(string second, int n) {
  string s1 = second;
  for (int i = 1; i < n; i++) second += s1;
  return second;
}
string getString(char x) {
  string second(1, x);
  return second;
}
const long long N1 = 1e6 + 1;
vector<long long> isprime(N1, 1), prime;
void seive() {
  for (long long i = 2; i < sqrt(N1) + 1; i++) {
    if (isprime[i]) {
      for (long long j = i * i; j < N1; j += i) isprime[j] = 0;
      prime.push_back(i);
    }
  }
  for (long long i = sqrt(N1) + 1; i < N1; i++)
    if (isprime[i]) prime.push_back(i);
}
struct dsu {
  vector<long long> par, rank;
  dsu(long long n) : par(n + 1), rank(n + 1) {
    for (long long i = 0; i <= n; i++) {
      par[i] = i;
      rank[i] = 1;
    }
  }
  long long root(long long a) {
    if (a == par[a]) return a;
    return par[a] = root(par[a]);
  }
  void merge(long long a, long long b) {
    a = root(a);
    b = root(b);
    if (a == b) return;
    if (rank[a] > rank[b]) swap(a, b);
    par[b] = a;
  }
  set<long long> parent(long long n) {
    set<long long> second;
    for (long long i = 1; i <= n; i++) {
      second.insert(root(i));
    }
    return second;
  }
};
void shift(vector<int> &a, int idx) {
  int x = a[idx - 2], y = a[idx - 1], z = a[idx];
  a[idx - 2] = z;
  a[idx - 1] = x;
  a[idx] = y;
}
bool check(vector<int> a) {
  for (long long i = 1; i < a.size(); i++)
    if (a[i] < a[i - 1]) return true;
  return false;
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  vector<int> ans;
  for (long long i = 0; i < n; i++) {
    int idx = min_element(a.begin() + i, a.end()) - a.begin();
    while (a[i] != a[idx]) {
      if (idx - 2 >= i) {
        ans.push_back(idx - 2);
        shift(a, idx);
      } else if (idx - 1 == i && idx + 1 < n) {
        ans.push_back(idx - 1);
        shift(a, idx + 1);
        ans.push_back(idx - 1);
        shift(a, idx + 1);
      } else
        break;
      idx = min_element(a.begin() + i, a.end()) - a.begin();
    }
  }
  bool flag = true;
  for (long long i = 1; i < n; i++)
    if (a[i] < a[i - 1]) flag = false;
  if (!flag) {
    map<int, int> m1;
    bool flag1 = true;
    for (long long i = 0; i < n; i++) m1[a[i]]++;
    for (auto i : m1)
      if (i.second > 1) flag1 = false;
    if (flag1) {
      cout << -1 << endl;
      return;
    }
  }
  if (!flag) {
    int idx = n - 1;
    while (check(a) && idx - 2 >= 0) {
      ans.push_back(idx - 2);
      shift(a, idx);
      if (a[idx - 2] <= a[idx - 1])
        idx -= 2;
      else
        idx -= 1;
    }
  }
  cout << ans.size() << endl;
  for (long long i = 0; i < ans.size(); i++) cout << ans[i] + 1 << " ";
  cout << endl;
}
int main() {
  optimizeIO();
  long long r = 1;
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}