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;
const int MOD = (int)1e9 + 7;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
const int N = (int)2 * 1e5 + 10;
void TimeExecution() {
  cout << "\n\n===>> Time Execution: " << clock() / (double)1000 << " sec(s).";
}
long long sqr(long long a) { return ((a % MOD) * (a % MOD)) % MOD; }
long long msub(long long a, long long b) { return ((a - b) % MOD + MOD) % MOD; }
long long madd(long long a, long long b) { return (a + b) % MOD; }
long long mmul(long long a, long long b) {
  return ((a % MOD) * (b % MOD)) % MOD;
}
struct Node {
  int val;
  int lazy;
};
int t;
int a[N];
Node st[4 * 2 * N];
void down(int id) {
  int d = st[id].lazy;
  if (d == 0) return;
  st[2 * id].val += d;
  st[2 * id + 1].val += d;
  st[2 * id].lazy += d;
  st[2 * id + 1].lazy += d;
  st[id].lazy = 0;
}
void update(int id, int l, int r, int u, int v, int x) {
  if (v < l || u > r) return;
  if (u <= l && r <= v) {
    st[id].val += x;
    st[id].lazy += x;
    return;
  }
  int mid = (l + r) / 2;
  down(id);
  update(2 * id, l, mid, u, v, x);
  update(2 * id + 1, mid + 1, r, u, v, x);
  st[id].val = min(st[2 * id].val, st[2 * id + 1].val);
}
int get(int id, int l, int r, int u, int v) {
  if (v < l || u > r) return INF;
  if (u <= l && r <= v) return st[id].val;
  int mid = (l + r) / 2;
  down(id);
  return min(get(2 * id, l, mid, u, v), get(2 * id + 1, mid + 1, r, u, v));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    for (int i = (1); i <= (n); i++) cin >> a[i];
    for (int i = (1); i <= (4 * 2 * k); i++) st[i].lazy = st[i].val = 0;
    for (int i = (1); i <= (n / 2); i++) {
      int M = a[i] + a[n - i + 1];
      int R = M + max(k - a[i], k - a[n - i + 1]);
      int L = M + min(1 - a[i], 1 - a[n - i + 1]);
      if (M > L) update(1, 1, 2 * k, L, M - 1, 1);
      if (R > M) update(1, 1, 2 * k, M + 1, R, 1);
      if (L > 1) update(1, 1, 2 * k, 1, L - 1, 2);
      if (R < 2 * k) update(1, 1, 2 * k, R + 1, 2 * k, 2);
    }
    int res = get(1, 1, 2 * k, 1, 2 * k);
    cout << res << '\n';
  }
  return 0;
}