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
#include <bits/stdc++.h>
char *fs, *ft, buf[1 << 20];
inline long long read() {
  long long x = 0, f = 1;
  char ch =
      (fs == ft && (ft = (fs = buf) + fread(buf, 1, 1 << 20, stdin), fs == ft))
          ? 0
          : *fs++;
  ;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = (fs == ft &&
          (ft = (fs = buf) + fread(buf, 1, 1 << 20, stdin), fs == ft))
             ? 0
             : *fs++;
    ;
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = (fs == ft &&
          (ft = (fs = buf) + fread(buf, 1, 1 << 20, stdin), fs == ft))
             ? 0
             : *fs++;
    ;
  }
  return x * f;
}
using namespace std;
const long long N = 2e5 + 5;
const long long inf = 0x7fffffff;
const long long mod = 1e9 + 7;
const double eps = 1e-6;
long long a[N], sum[N << 1];
void add(long long l, long long r, long long x) {
  if (r < l) return;
  sum[l] += x;
  sum[r + 1] -= x;
}
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    for (long long i = 0; i <= 2 * k; i++) sum[i] = 0;
    for (long long i = 1; i <= n; i++) cin >> a[i];
    for (long long i = 1; i <= n / 2; i++) {
      long long aa = a[i], bb = a[n - i + 1];
      add(2, min(aa, bb), 2);
      add(min(aa, bb) + 1, aa + bb - 1, 1);
      add(aa + bb + 1, max(aa, bb) + k, 1);
      add(max(aa, bb) + k + 1, 2 * k, 2);
    }
    for (long long i = 1; i <= 2 * k; i++) sum[i] = sum[i] + sum[i - 1];
    long long res = inf;
    for (long long i = 2; i <= 2 * k; i++) res = min(sum[i], res);
    cout << res << '\n';
  }
}
