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 = 2e5 + 7;
long long t, n, k, a[maxn], v[maxn << 1], kk, ans;
inline void add(long long x, long long t) {
  for (; x <= kk; x += x & -x) v[x] += t;
  return;
}
inline long long query(long long x) {
  long long ans = 0;
  for (; x > 0; x ^= x & -x) ans += v[x];
  return ans;
}
int main() {
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld%lld", &n, &k);
    kk = k << 1;
    for (register int i = 1; i <= n; i++) scanf("%lld", &a[i]);
    for (register int i = 2; i <= kk; i++) v[i] = 0;
    for (register int i = 1; (i << 1) <= n; i++) {
      add(2, 2);
      add(min(a[i], a[n - i + 1]) + 1, -1);
      add(a[i] + a[n - i + 1], -1);
      add(a[i] + a[n - i + 1] + 1, 1);
      add(max(a[i], a[n - i + 1]) + k + 1, 1);
    }
    ans = query(2);
    for (register int i = 3; i <= kk; i++) ans = min(ans, query(i));
    printf("%lld\n", ans);
  }
  return 0;
}