A permutation of length n is an array p=[p_1,p_2,...,p_n], which contains every integer from 1 to n (inclusive) and, moreover, each number appears exactly once. For example, p=[3,1,4,2,5] is a permutation of length 5.

For a given number n (n ≥ 2), find a permutation p in which absolute difference (that is, the absolute value of difference) of any two neighboring (adjacent) elements is between 2 and 4, inclusive. Formally, find such permutation p that 2 ≤ |p_i - p_{i+1}| ≤ 4 for each i (1 ≤ i < n).

Print any such permutation for the given integer n or determine that it does not exist.

Input

The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t test cases follow.

Each test case is described by a single line containing an integer n (2 ≤ n ≤ 1000).

Output

Print t lines. Print a permutation that meets the given requirements. If there are several such permutations, then print any of them. If no such permutation exists, print -1.

Example

Input


6
10
2
4
6
7
13


Output


9 6 10 8 4 7 3 1 5 2 
-1
3 1 4 2 
5 3 6 2 4 1 
5 1 3 6 2 4 7 
13 9 7 11 8 4 1 3 5 2 6 10 12 
#include <bits/stdc++.h>
using namespace std;
int dx[] = {0, 0, 1, -1, -1, -1, 1, 1};
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
template <class T>
inline T biton(T n, T pos) {
  return n | ((T)1 << pos);
}
template <class T>
inline T bitoff(T n, T pos) {
  return n & ~((T)1 << pos);
}
template <class T>
inline T ison(T n, T pos) {
  return (bool)(n & ((T)1 << pos));
}
template <class T>
inline T gcd(T a, T b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
inline int nxt() {
  int aaa;
  scanf("%d", &aaa);
  return aaa;
}
inline long long lxt() {
  long long aaa;
  scanf("%lld", &aaa);
  return aaa;
}
inline double dxt() {
  double aaa;
  scanf("%lf", &aaa);
  return aaa;
}
template <class T>
inline T bigmod(T p, T e, T m) {
  T ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % m;
    p = (p * p) % m;
  }
  return (T)ret;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& input) {
  bool isFirst = true;
  for (T out : input) {
    if (isFirst) {
      os << out;
      isFirst = false;
    } else
      os << " " << out;
  }
  return os;
}
vector<int> ans, vis;
int n;
int dp[1001][1001];
int F(int pos, int num) {
  if (pos >= n) {
    return true;
  }
  int& ret = dp[pos][num];
  ret = 0;
  for (int x = max(1, num - 4); x <= min(num + 4, n); x++) {
    if (vis[x] == 1 || abs(num - x) < 2 || abs(num - x) > 4) continue;
    vis[x] = 1;
    if (F(pos + 1, x)) {
      ans.push_back(x);
      return ret = 1;
    }
    vis[x] = 0;
  }
  return ret;
}
void solve() {
  n = nxt();
  for (int i = 1; i < min(200, n); i++) {
    vis.assign(n + 1, 0);
    memset(dp, -1, sizeof(dp));
    ans.clear();
    if (F(0, 1) && (int)(ans).size() == n) {
      cout << ans << '\n';
      return;
    }
  }
  cout << -1 << '\n';
}
int main() {
  cout.setf(ios::fixed);
  cout.precision(10);
  int tc = nxt();
  while (tc--) {
    solve();
  }
  return 0;
}
