Problem: You are given a positive integer n, it is guaranteed that n is even (i.e. divisible by 2).

You want to construct the array a of length n such that: 

  * The first n/2 elements of a are even (divisible by 2); 
  * the second n/2 elements of a are odd (not divisible by 2); 
  * all elements of a are distinct and positive; 
  * the sum of the first half equals to the sum of the second half (∑_{i=1}^{n/2} a_i = ∑_{i=n/2 + 1}^{n} a_i). 



If there are multiple answers, you can print any. It is not guaranteed that the answer exists.

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 only line of the test case contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of the array. It is guaranteed that that n is even (i.e. divisible by 2).

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer — "NO" (without quotes), if there is no suitable answer for the given test case or "YES" in the first line and any suitable array a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) satisfying conditions from the problem statement on the second line.

Example

Input


5
2
4
6
8
10


Output


NO
YES
2 4 1 5
NO
YES
2 4 6 8 1 3 5 11
NO

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef unsigned int UI;
typedef long int LI;
typedef unsigned long int ULI;
typedef long long int LL;
typedef unsigned long long int ULL;
const LL MOD = 1e9 + 7;
inline int scanInt() {
  int n = 0;
  char ch = getchar();
  int sign = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    n = (n << 1) + (n << 3) + (int)(ch - '0');
    ch = getchar();
  }
  return n * sign;
}
inline LL scanLong() {
  LL n = 0;
  char ch = getchar();
  LL sign = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    n = (n << 1) + (n << 3) + (LL)(ch - '0');
    ch = getchar();
  }
  return n * sign;
}
void solve() {
  LL n = scanLong();
  ;
  LL k = scanLong();
  ;
  LL arr[n + 1];
  for (__typeof(n + 1) i = (1) - ((1) > (n + 1));
       i != (n + 1) - ((1) > (n + 1)); i += 1 - 2 * ((1) > (n + 1)))
    arr[i] = scanLong();
  LL cnt[2 * k + 1];
  memset(cnt, 0, sizeof(cnt));
  for (__typeof((n / 2) + 1) i = (1) - ((1) > ((n / 2) + 1));
       i != ((n / 2) + 1) - ((1) > ((n / 2) + 1));
       i += 1 - 2 * ((1) > ((n / 2) + 1))) {
    LL sum = arr[i] + arr[n - i + 1];
    cnt[sum]++;
  }
  LL pref[2 * k + 1];
  memset(pref, 0, sizeof(pref));
  for (__typeof((n / 2) + 1) i = (1) - ((1) > ((n / 2) + 1));
       i != ((n / 2) + 1) - ((1) > ((n / 2) + 1));
       i += 1 - 2 * ((1) > ((n / 2) + 1))) {
    LL low = min(arr[i], arr[n - i + 1]) + 1;
    LL high = max(arr[i], arr[n - i + 1]) + k;
    pref[low]++;
    if (high < 2 * k) pref[high + 1]--;
  }
  for (__typeof(2 * k + 1) i = (1) - ((1) > (2 * k + 1));
       i != (2 * k + 1) - ((1) > (2 * k + 1)); i += 1 - 2 * ((1) > (2 * k + 1)))
    pref[i] += pref[i - 1];
  LL ans = n / 2;
  for (__typeof(2 * k + 1) x = (2) - ((2) > (2 * k + 1));
       x != (2 * k + 1) - ((2) > (2 * k + 1));
       x += 1 - 2 * ((2) > (2 * k + 1))) {
    LL changes = (pref[x] - cnt[x]) + 2 * ((n / 2) - pref[x]);
    ans = min(ans, changes);
  }
  printf("%lld ", ans);
  printf("\n");
}
int main() {
  LL t = scanLong();
  ;
  for (__typeof(t) _ = (0) - ((0) > (t)); _ != (t) - ((0) > (t));
       _ += 1 - 2 * ((0) > (t)))
    solve();
}