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 maxn = 400010;
const long long mod = 1000000007;
const long double eps = 1e-9;
const long long inf = ((1ll << 31ll) - 1ll);
const long long INF = 2000000000000000000ll;
const long double pi = acos(-1);
long long qpow(long long b, long long e) {
  if (!e) return 1;
  if (e & 1) return qpow(b, e - 1) * b % mod;
  long long pwur = qpow(b, e >> 1);
  return pwur * pwur % mod;
}
long long modinv(long long x) { return qpow(x, mod - 2); }
long long tc, n, a[maxn];
vector<long long> ans;
bool sorted() {
  for (long long i = 1; i < n; i++)
    if (a[i] > a[i + 1]) return 0;
  return 1;
}
void shft(long long x) {
  swap(a[x + 2], a[x + 1]);
  swap(a[x + 1], a[x]);
}
void tryl() {
  for (long long i = 1; i <= n - 2; i++) {
    pair<long long, long long> mn = {mod, -1};
    for (long long j = i; j <= n; j++) {
      if (mn.first > a[j]) {
        mn = {a[j], j};
      }
    }
    while (mn.second != i) {
      if (mn.second - 2 >= i) {
        shft(mn.second - 2);
        mn.second -= 2;
        ans.push_back(mn.second);
      } else {
        shft(mn.second - 1);
        shft(mn.second - 1);
        mn.second -= 1;
        ans.push_back(mn.second);
        ans.push_back(mn.second);
      }
    }
  }
}
void tryr() {
  for (long long i = n; i >= 3; i--) {
    pair<long long, long long> mx = {-1, -1};
    for (long long j = 1; j <= i; j++) {
      if (mx.first <= a[j]) {
        mx = {a[j], j};
      }
    }
    while (mx.second != i) {
      if (mx.second + 2 <= i) {
        shft(mx.second);
        ans.push_back(mx.second);
        mx.second += 2;
      } else {
        shft(mx.second - 1);
        ans.push_back(mx.second - 1);
        mx.second += 1;
      }
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.setf(ios::fixed);
  cout.precision(0);
  srand(time(NULL));
  cin >> tc;
  while (tc--) {
    ans.clear();
    cin >> n;
    for (long long i = 1; i <= n; i++) cin >> a[i];
    long long bu = 10;
    while (bu--) {
      tryl();
      tryr();
    }
    for (long long i = 0; i <= 3; i++) {
      if (i) shft(n - 2);
      if (a[n - 2] <= a[n - 1] && a[n - 1] <= a[n]) {
        for (long long j = 1; j <= i; j++) ans.push_back(n - 2);
        break;
      }
    }
    if (!sorted()) {
      cout << -1 << '\n';
      continue;
    }
    cout << ans.size() << '\n';
    for (auto i : ans) cout << i << ' ';
    cout << '\n';
  }
  return 0;
}