#include <bits/stdc++.h>
using namespace std;
inline char gc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long read() {
  char c = getchar();
  long long su = 0, f = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) su = su * 10 + c - '0';
  return su * f;
}
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    write(-x);
    return;
  }
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writesp(long long x) { write(x), putchar(' '); }
inline void writeln(long long x) {
  write(x);
  putchar('\n');
}
int T, n, a[400005], bl;
vector<int> ans;
signed main() {
  T = read();
  while (T--) {
    n = read();
    bl = 0;
    for (int i = 1; i <= n; i++) {
      a[i] = read();
      if (a[i] % 2 != i % 2) bl = 1;
    }
    if (bl)
      puts("-1");
    else {
      for (int i = n; i >= 3; i -= 2) {
        int x = 0, y = 0;
        for (int j = 1; j <= i; j++)
          if (a[j] == i) {
            x = j;
          }
        for (int j = 1; j <= i; j++)
          if (a[j] == i - 1) {
            y = j;
          }
        ans.push_back(x);
        reverse(a + 1, a + x + 1);
        for (int j = 1; j <= i; j++)
          if (a[j] == i - 1) {
            y = j;
          }
        ans.push_back(y - 1);
        reverse(a + 1, a + y - 1 + 1);
        ans.push_back(i);
        reverse(a + 1, a + i + 1);
        for (int j = 1; j <= i; j++)
          if (a[j] == i) {
            x = j;
          }
        ans.push_back(x);
        reverse(a + 1, a + x + 1);
        ans.push_back(i);
        reverse(a + 1, a + i + 1);
      }
      for (int i = 1; i <= n; i++) {
        if (a[i] != i) bl = 1;
      }
      if (bl)
        puts("-1");
      else {
        writeln(ans.size());
        for (auto u : ans) writesp(u);
        puts("");
      }
    }
    ans.clear();
  }
}

### Prompt

Your challenge is to write a cpp solution to 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;
inline char gc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long read() {
  char c = getchar();
  long long su = 0, f = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) su = su * 10 + c - '0';
  return su * f;
}
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    write(-x);
    return;
  }
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writesp(long long x) { write(x), putchar(' '); }
inline void writeln(long long x) {
  write(x);
  putchar('\n');
}
int T, n, a[400005], bl;
vector<int> ans;
signed main() {
  T = read();
  while (T--) {
    n = read();
    bl = 0;
    for (int i = 1; i <= n; i++) {
      a[i] = read();
      if (a[i] % 2 != i % 2) bl = 1;
    }
    if (bl)
      puts("-1");
    else {
      for (int i = n; i >= 3; i -= 2) {
        int x = 0, y = 0;
        for (int j = 1; j <= i; j++)
          if (a[j] == i) {
            x = j;
          }
        for (int j = 1; j <= i; j++)
          if (a[j] == i - 1) {
            y = j;
          }
        ans.push_back(x);
        reverse(a + 1, a + x + 1);
        for (int j = 1; j <= i; j++)
          if (a[j] == i - 1) {
            y = j;
          }
        ans.push_back(y - 1);
        reverse(a + 1, a + y - 1 + 1);
        ans.push_back(i);
        reverse(a + 1, a + i + 1);
        for (int j = 1; j <= i; j++)
          if (a[j] == i) {
            x = j;
          }
        ans.push_back(x);
        reverse(a + 1, a + x + 1);
        ans.push_back(i);
        reverse(a + 1, a + i + 1);
      }
      for (int i = 1; i <= n; i++) {
        if (a[i] != i) bl = 1;
      }
      if (bl)
        puts("-1");
      else {
        writeln(ans.size());
        for (auto u : ans) writesp(u);
        puts("");
      }
    }
    ans.clear();
  }
}

```