#include <bits/stdc++.h>
using namespace std;
template <class T>
bool chkmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
int read() {
  int num = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    num *= 10;
    num += ch - '0';
    ch = getchar();
  }
  return num * f;
}
long long read1() {
  long long num = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    num *= 10;
    num += ch - '0';
    ch = getchar();
  }
  return num * f;
}
int idx(char x) { return x - 'a'; }
int a[1000010];
int ans[4000010];
long long n, m;
int t, tt;
int main() {
  t = read();
  while (t--) {
    n = read();
    for (int i = 1; i <= n; i++) {
      a[i] = read();
    }
    int tot = 0;
    bool flag = true;
    int cnt = 0;
    for (int k = n; k > 1; k -= 2) {
      int x = 1;
      while (a[x] != k) x++;
      int y = 1;
      while (a[y] != k - 1) y++;
      if ((y & 1) || (x % 2 == 0)) {
        flag = false;
        break;
      }
      reverse(a + 1, a + x + 1);
      ans[++tot] = x;
      if (y < x) {
        tt = x;
        x = tt + 1 - x;
        y = tt + 1 - y;
      }
      reverse(a + 1, a + y);
      ans[++tot] = y - 1;
      reverse(a + 1, a + y + 2);
      ans[++tot] = y + 1;
      reverse(a + 1, a + 4);
      ans[++tot] = 3;
      reverse(a + 1, a + k + 1);
      ans[++tot] = k;
    }
    if (a[1] != 1 || flag == false)
      cout << -1 << endl;
    else {
      cout << tot << endl;
      for (int i = 1; i <= tot; i++) cout << ans[i] << ' ';
      if (tot || cnt) cout << endl;
    }
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve the following problem:

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;
template <class T>
bool chkmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
int read() {
  int num = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    num *= 10;
    num += ch - '0';
    ch = getchar();
  }
  return num * f;
}
long long read1() {
  long long num = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    num *= 10;
    num += ch - '0';
    ch = getchar();
  }
  return num * f;
}
int idx(char x) { return x - 'a'; }
int a[1000010];
int ans[4000010];
long long n, m;
int t, tt;
int main() {
  t = read();
  while (t--) {
    n = read();
    for (int i = 1; i <= n; i++) {
      a[i] = read();
    }
    int tot = 0;
    bool flag = true;
    int cnt = 0;
    for (int k = n; k > 1; k -= 2) {
      int x = 1;
      while (a[x] != k) x++;
      int y = 1;
      while (a[y] != k - 1) y++;
      if ((y & 1) || (x % 2 == 0)) {
        flag = false;
        break;
      }
      reverse(a + 1, a + x + 1);
      ans[++tot] = x;
      if (y < x) {
        tt = x;
        x = tt + 1 - x;
        y = tt + 1 - y;
      }
      reverse(a + 1, a + y);
      ans[++tot] = y - 1;
      reverse(a + 1, a + y + 2);
      ans[++tot] = y + 1;
      reverse(a + 1, a + 4);
      ans[++tot] = 3;
      reverse(a + 1, a + k + 1);
      ans[++tot] = k;
    }
    if (a[1] != 1 || flag == false)
      cout << -1 << endl;
    else {
      cout << tot << endl;
      for (int i = 1; i <= tot; i++) cout << ans[i] << ' ';
      if (tot || cnt) cout << endl;
    }
  }
  return 0;
}

```