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 N = 4e5 + 10;
struct node {
  int rt, l, r, sum;
  inline int mid() { return (l + r) >> 1; }
  inline void init(int l, int r) {
    this->l = l;
    this->r = r;
    sum = 0;
  }
  inline int d() { return (r - l + 1); }
} tree[N << 2];
struct info {
  int l, r, v;
};
int a[N];
void Build(int rt, int l, int r) {
  tree[rt].init(l, r);
  if (l == r) return;
  Build((rt << 1), l, tree[rt].mid());
  Build((rt << 1 | 1), tree[rt].mid() + 1, r);
}
inline void Push_up(int rt) {
  tree[rt].sum = tree[(rt << 1)].sum + tree[(rt << 1 | 1)].sum;
}
void Add(int rt, int l, int r, int v) {
  if (l > r) return;
  if (l <= tree[rt].l && tree[rt].r <= r) {
    tree[rt].sum += v;
    return;
  }
  if (tree[rt].mid() >= l) Add((rt << 1), l, r, v);
  if (tree[rt].mid() < r) Add((rt << 1 | 1), l, r, v);
}
int Query(int rt, int inx) {
  if (tree[rt].l == tree[rt].r) return tree[rt].sum;
  if (tree[rt].mid() >= inx) return tree[rt].sum + Query((rt << 1), inx);
  if (tree[rt].mid() < inx) return tree[rt].sum + Query((rt << 1 | 1), inx);
}
void solve() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int n, k;
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; ++i) scanf("%d", a + i);
    vector<info> vi;
    int m = n / 2;
    for (int i = 1; i <= m; ++i) {
      int Max = max(a[i], a[n - i + 1]);
      int Min = min(a[i], a[n - i + 1]);
      vi.push_back({Min + 1, Max + k, a[i] + a[n - i + 1]});
    }
    Build(1, 1, 2 * k);
    for (auto p : vi) {
      Add(1, p.l, p.r, 1);
      Add(1, 2, p.l - 1, 2);
      Add(1, p.r + 1, 2 * k, 2);
      Add(1, p.v, p.v, -1);
    }
    int Min_tot = 1e8;
    for (int i = 1; i <= m; ++i) {
      Min_tot = min(Min_tot, Query(1, a[i] + a[n - i + 1]));
    }
    printf("%d\n", min(Min_tot, m));
  }
}
int main() {
  solve();
  return 0;
}