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 maxn = 505;
const int maxm = 100;
const int inf = 2147483647;
const double eps = 1e-9;
const long long INF = 9223372036854775807ll;
long long qpow(long long a, long long b, long long c) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % c;
    a = a * a % c;
    b >>= 1;
  }
  return ans;
}
int n, a[maxn];
bool move(int i, int j, vector<int> &as) {
  while (i > j) {
    if (i - 2 >= j) {
      as.push_back(i - 2);
      swap(a[i - 1], a[i]);
      swap(a[i - 2], a[i - 1]);
      i -= 2;
    } else {
      if (i + 1 >= n) {
        if (a[i] == a[i - 2]) {
          swap(a[i], a[i - 1]);
          swap(a[i - 1], a[i - 2]);
          as.push_back(i - 2);
          i--;
        } else
          return false;
      } else {
        as.push_back(i - 1);
        as.push_back(i - 1);
        swap(a[i - 1], a[i]);
        swap(a[i], a[i + 1]);
      }
      i--;
    }
  }
  return true;
}
int find_(int b) {
  int minn = inf, id = -1;
  for (register int i = (b); i < (n); ++i) {
    if (minn > a[i]) minn = a[i], id = i;
  }
  return id;
}
bool sort_(vector<int> &ans) {
  if (n == 1) return false;
  if (n == 2) {
    if (a[0] < a[1]) return false;
    return true;
  }
  for (register int i = (0); i < (n); ++i) {
    int sb = find_(i);
    if (!move(sb, i, ans)) {
      return true;
    }
  }
  return false;
}
void swap_(int i, vector<int> &as) {
  if (n == 3) return;
  as.push_back(i + 1);
  as.push_back(i);
  swap(a[i + 2], a[i + 3]);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    bool flag = true;
    for (register int i = (0); i < (n); ++i) {
      scanf("%d", &a[i]);
      if (i && a[i] < a[i - 1]) flag = false;
    }
    if (flag)
      printf("0\n\n");
    else {
      vector<int> ans;
      bool failed = sort_(ans);
      if (failed) {
        for (register int i = (n - 4); i >= (0); --i) {
          if (a[i] == a[i + 1] && a[i + 1] != a[i + 2] &&
              a[i + 2] != a[i + 3]) {
            swap_(i, ans);
            failed = sort_(ans);
            break;
          }
        }
      }
      if (failed)
        printf("-1\n");
      else {
        printf("%d\n", (int)ans.size());
        for (register int i = (0); i < (ans.size()); ++i)
          printf("%d ", ans[i] + 1);
        printf("\n");
      }
    }
  }
}