#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <class T>
using V = vector<T>;
template <class T>
using VV = V<V<T>>;
template <typename T>
void view(T e) {
  std::cerr << e << std::endl;
}
template <typename T>
void view(const std::vector<T> &v) {
  for (const auto &e : v) {
    std::cerr << e << " ";
  }
  std::cerr << std::endl;
}
template <typename T>
void view(const std::vector<std::vector<T>> &vv) {
  cerr << endl;
  int cnt = 0;
  for (const auto &v : vv) {
    cerr << cnt << "th : ";
    view(v);
    cnt++;
  }
  cerr << endl;
}
std::ostream &operator<<(std::ostream &dest, __int128_t value) {
  std::ostream::sentry s(dest);
  if (s) {
    __uint128_t tmp = value < 0 ? -value : value;
    char buffer[128];
    char *d = std::end(buffer);
    do {
      --d;
      *d = "0123456789"[tmp % 10];
      tmp /= 10;
    } while (tmp != 0);
    if (value < 0) {
      --d;
      *d = '-';
    }
    int len = std::end(buffer) - d;
    if (dest.rdbuf()->sputn(d, len) != len) {
      dest.setstate(std::ios_base::badbit);
    }
  }
  return dest;
}
ll power(ll a, ll p) {
  ll ret = 1;
  while (p) {
    if (p & 1) {
      ret = ret * a;
    }
    a = a * a;
    p >>= 1;
  }
  return ret;
}
ll modpow(ll a, ll p, ll mod) {
  ll ret = 1;
  while (p) {
    if (p & 1) {
      ret = ret * a % mod;
    }
    a = a * a % mod;
    p >>= 1;
  }
  return ret;
}
template <class T, class K>
bool chmax(T &a, const K b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T, class K>
bool chmin(T &a, const K b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
const int inf = 1001001001;
const ll INF = 1001001001001001001ll;
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    V<int> a(n);
    for (ll i = 0; i < n; ++i) cin >> a[i];
    bool ok = true;
    for (ll i = 0; i < n; ++i) {
      if (a[i] % 2 == 1 && i % 2 == 1 || a[i] % 2 == 0 && i % 2 == 0) {
        cout << -1 << endl;
        ok = false;
        break;
      }
    }
    if (!ok) continue;
    V<int> ans;
    for (int i = 1; i < n; i += 2) {
      int p, q;
      for (ll j = 0; j < n; ++j) {
        if (a[j] == i) p = j;
      }
      if (p != 0) ans.push_back(p + 1);
      reverse(a.begin(), a.begin() + p + 1);
      for (ll j = 0; j < n; ++j) {
        if (a[j] == i) p = j;
        if (a[j] == i + 1) q = j;
      }
      ans.push_back(q);
      reverse(a.begin(), a.begin() + q);
      ans.push_back(q + 2);
      reverse(a.begin(), a.begin() + q + 2);
      ans.push_back(3);
      reverse(a.begin(), a.begin() + 3);
      ans.push_back(n - i + 1);
      reverse(a.begin(), a.end() - i + 1);
    }
    ans.push_back(n);
    cout << ((int)(ans).size()) << endl;
    for (ll i = 0; i < ((int)(ans).size()); ++i) cout << ans[i] << " ";
    cout << endl;
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
template <class T>
using V = vector<T>;
template <class T>
using VV = V<V<T>>;
template <typename T>
void view(T e) {
  std::cerr << e << std::endl;
}
template <typename T>
void view(const std::vector<T> &v) {
  for (const auto &e : v) {
    std::cerr << e << " ";
  }
  std::cerr << std::endl;
}
template <typename T>
void view(const std::vector<std::vector<T>> &vv) {
  cerr << endl;
  int cnt = 0;
  for (const auto &v : vv) {
    cerr << cnt << "th : ";
    view(v);
    cnt++;
  }
  cerr << endl;
}
std::ostream &operator<<(std::ostream &dest, __int128_t value) {
  std::ostream::sentry s(dest);
  if (s) {
    __uint128_t tmp = value < 0 ? -value : value;
    char buffer[128];
    char *d = std::end(buffer);
    do {
      --d;
      *d = "0123456789"[tmp % 10];
      tmp /= 10;
    } while (tmp != 0);
    if (value < 0) {
      --d;
      *d = '-';
    }
    int len = std::end(buffer) - d;
    if (dest.rdbuf()->sputn(d, len) != len) {
      dest.setstate(std::ios_base::badbit);
    }
  }
  return dest;
}
ll power(ll a, ll p) {
  ll ret = 1;
  while (p) {
    if (p & 1) {
      ret = ret * a;
    }
    a = a * a;
    p >>= 1;
  }
  return ret;
}
ll modpow(ll a, ll p, ll mod) {
  ll ret = 1;
  while (p) {
    if (p & 1) {
      ret = ret * a % mod;
    }
    a = a * a % mod;
    p >>= 1;
  }
  return ret;
}
template <class T, class K>
bool chmax(T &a, const K b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T, class K>
bool chmin(T &a, const K b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
const int inf = 1001001001;
const ll INF = 1001001001001001001ll;
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    V<int> a(n);
    for (ll i = 0; i < n; ++i) cin >> a[i];
    bool ok = true;
    for (ll i = 0; i < n; ++i) {
      if (a[i] % 2 == 1 && i % 2 == 1 || a[i] % 2 == 0 && i % 2 == 0) {
        cout << -1 << endl;
        ok = false;
        break;
      }
    }
    if (!ok) continue;
    V<int> ans;
    for (int i = 1; i < n; i += 2) {
      int p, q;
      for (ll j = 0; j < n; ++j) {
        if (a[j] == i) p = j;
      }
      if (p != 0) ans.push_back(p + 1);
      reverse(a.begin(), a.begin() + p + 1);
      for (ll j = 0; j < n; ++j) {
        if (a[j] == i) p = j;
        if (a[j] == i + 1) q = j;
      }
      ans.push_back(q);
      reverse(a.begin(), a.begin() + q);
      ans.push_back(q + 2);
      reverse(a.begin(), a.begin() + q + 2);
      ans.push_back(3);
      reverse(a.begin(), a.begin() + 3);
      ans.push_back(n - i + 1);
      reverse(a.begin(), a.end() - i + 1);
    }
    ans.push_back(n);
    cout << ((int)(ans).size()) << endl;
    for (ll i = 0; i < ((int)(ans).size()); ++i) cout << ans[i] << " ";
    cout << endl;
  }
}

```