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 = 1e9 + 7;
long long fast(long long x, long long n) {
  long long ans = 1;
  for (; n; n >>= 1, x = x * x % mod)
    if (n & 1) ans = ans * x % mod;
  return ans;
}
const int N = 1e6 + 100;
long long t[N << 2];
long long col[N << 2];
void build(int l, int r, int pos) {
  col[pos] = 0;
  t[pos] = 0;
  if (l == r) {
    return;
  }
  int m = l + r >> 1;
  build(l, m, pos << 1);
  build(m + 1, r, pos << 1 | 1);
}
void down(int pos) {
  if (!col[pos]) return;
  col[pos << 1] += col[pos];
  col[pos << 1 | 1] += col[pos];
  t[pos << 1] += col[pos];
  t[pos << 1 | 1] += col[pos];
  col[pos] = 0;
}
void update(int L, int R, int v, int l, int r, int pos) {
  if (L <= l && r <= R) {
    t[pos] += v;
    col[pos] += v;
    return;
  }
  int m = l + r >> 1;
  down(pos);
  if (L <= m) update(L, R, v, l, m, pos << 1);
  if (R > m) update(L, R, v, m + 1, r, pos << 1 | 1);
  t[pos] = min(t[pos << 1], t[pos << 1 | 1]);
}
long long a[N];
int main() {
  int cas;
  cin >> cas;
  while (cas--) {
    long long n, k;
    cin >> n >> k;
    build(2, 2 * k, 1);
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
    }
    for (int i = 1; i <= n / 2; i++) {
      long long x = a[i], y = a[n - i + 1];
      long long up = max(k - x, k - y);
      if (up) {
        update(x + y + 1, x + y + up, 1, 2, 2 * k, 1);
      }
      if (x + y + up + 1 <= 2 * k)
        update(x + y + up + 1, 2 * k, 2, 2, 2 * k, 1);
      up = max(x - 1, y - 1);
      if (up) {
        update(x + y - up, x + y - 1, 1, 2, 2 * k, 1);
      }
      if (x + y - up - 1 >= 2) update(2, x + y - up - 1, 2, 2, 2 * k, 1);
    }
    printf("%lld\n", t[1]);
  }
}