#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using matr = vector<vector<ll>>;
using dbl = long double;
using ui = unsigned int;
using ld = long double;
mt19937 rnd;
ll random(ll limit) {
  uniform_int_distribution<ll> uniform_dist(0, limit);
  return uniform_dist(rnd);
}
ll mod = 998244353;
ll bpow(ll a, ll p) {
  ll res = 1;
  while (p > 0) {
    if (p % 2) res *= a;
    p /= 2;
    a *= a;
    res %= mod;
    a %= mod;
  }
  return res;
}
ll inv(ll a) { return bpow(a, mod - 2); }
ll gcd(ll a, ll b) {
  if (!b) return a;
  return gcd(b, a % b);
}
vector<int> ans;
void lol(vector<int>& x, int a) {
  for (int i = 0; i < (a - 1) / 2; i++) {
    swap(x[i], x[a - i - 1]);
  }
  ans.push_back(a);
}
void mem(vector<int>& x, int& pos1, int& pos2, int cur) {
  pos1 = -1, pos2 = -1;
  for (int i = 0; i < x.size(); i++) {
    if (x[i] == cur) pos1 = i;
    if (x[i] == cur - 1) pos2 = i;
  }
}
void solve() {
  int n;
  cin >> n;
  vector<int> x(n);
  bool f = 1;
  for (int i = 0; i < n; i++) {
    cin >> x[i];
    if (x[i] % 2 == i % 2) {
      f = 0;
    }
  }
  if (!f) {
    cout << "-1\n";
    return;
  }
  ans = {};
  for (int cur = n; cur >= 3; cur -= 2) {
    int pos1, pos2;
    mem(x, pos1, pos2, cur);
    lol(x, pos1 + 1);
    mem(x, pos1, pos2, cur);
    lol(x, pos2);
    mem(x, pos1, pos2, cur);
    lol(x, pos2 + 2);
    mem(x, pos1, pos2, cur);
    lol(x, 3);
    lol(x, cur);
  }
  cout << ans.size() << '\n';
  for (auto v : ans) cout << v << ' ';
  cout << '\n';
}
int main() {
  int t;
  t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

You have a permutation: an array a = [a_1, a_2, …, a_n] of distinct integers from 1 to n. The length of the permutation n is odd.

You need to sort the permutation in increasing order.

In one step, you can choose any prefix of the permutation with an odd length and reverse it. Formally, if a = [a_1, a_2, …, a_n], you can choose any odd integer p between 1 and n, inclusive, and set a to [a_p, a_{p-1}, …, a_1, a_{p+1}, a_{p+2}, …, a_n].

Find a way to sort a using no more than 5n/2 reversals of the above kind, or determine that such a way doesn't exist. The number of reversals doesn't have to be minimized.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.

The first line of each test case contains a single integer n (3 ≤ n ≤ 2021; n is odd) — the length of the permutation.

The second line contains n distinct integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — the permutation itself. 

It is guaranteed that the sum of n over all test cases does not exceed 2021.

Output

For each test case, if it's impossible to sort the given permutation in at most 5n/2 reversals, print a single integer -1.

Otherwise, print an integer m (0 ≤ m ≤ 5n/2), denoting the number of reversals in your sequence of steps, followed by m integers p_i (1 ≤ p_i ≤ n; p_i is odd), denoting the lengths of the prefixes of a to be reversed, in chronological order.

Note that m doesn't have to be minimized. If there are multiple answers, print any.

Example

Input


3
3
1 2 3
5
3 4 5 2 1
3
2 1 3


Output


4
3 3 3 3
2
3 5
-1

Note

In the first test case, the permutation is already sorted. Any even number of reversals of the length 3 prefix doesn't change that fact.

In the second test case, after reversing the prefix of length 3 the permutation will change to [5, 4, 3, 2, 1], and then after reversing the prefix of length 5 the permutation will change to [1, 2, 3, 4, 5].

In the third test case, it's impossible to sort the permutation.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using matr = vector<vector<ll>>;
using dbl = long double;
using ui = unsigned int;
using ld = long double;
mt19937 rnd;
ll random(ll limit) {
  uniform_int_distribution<ll> uniform_dist(0, limit);
  return uniform_dist(rnd);
}
ll mod = 998244353;
ll bpow(ll a, ll p) {
  ll res = 1;
  while (p > 0) {
    if (p % 2) res *= a;
    p /= 2;
    a *= a;
    res %= mod;
    a %= mod;
  }
  return res;
}
ll inv(ll a) { return bpow(a, mod - 2); }
ll gcd(ll a, ll b) {
  if (!b) return a;
  return gcd(b, a % b);
}
vector<int> ans;
void lol(vector<int>& x, int a) {
  for (int i = 0; i < (a - 1) / 2; i++) {
    swap(x[i], x[a - i - 1]);
  }
  ans.push_back(a);
}
void mem(vector<int>& x, int& pos1, int& pos2, int cur) {
  pos1 = -1, pos2 = -1;
  for (int i = 0; i < x.size(); i++) {
    if (x[i] == cur) pos1 = i;
    if (x[i] == cur - 1) pos2 = i;
  }
}
void solve() {
  int n;
  cin >> n;
  vector<int> x(n);
  bool f = 1;
  for (int i = 0; i < n; i++) {
    cin >> x[i];
    if (x[i] % 2 == i % 2) {
      f = 0;
    }
  }
  if (!f) {
    cout << "-1\n";
    return;
  }
  ans = {};
  for (int cur = n; cur >= 3; cur -= 2) {
    int pos1, pos2;
    mem(x, pos1, pos2, cur);
    lol(x, pos1 + 1);
    mem(x, pos1, pos2, cur);
    lol(x, pos2);
    mem(x, pos1, pos2, cur);
    lol(x, pos2 + 2);
    mem(x, pos1, pos2, cur);
    lol(x, 3);
    lol(x, cur);
  }
  cout << ans.size() << '\n';
  for (auto v : ans) cout << v << ' ';
  cout << '\n';
}
int main() {
  int t;
  t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```