#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int l, int r) { return uniform_int_distribution<int>(l, r)(rng); }
char gc(void) {
  const int S = 3e7;
  static char buf[S], *s = buf, *t = buf;
  if (s == t) t = buf + fread(s = buf, 1, S, stdin);
  if (s == t) return EOF;
  return *s++;
}
void read(char &c) {
  while (c = gc(), c == ' ' || c == '\n')
    ;
}
void read(string &s) {
  char c;
  while (c = gc(), c != ' ' && c != '\n' && c != EOF) s += c;
}
void read(char *s) {
  char c;
  while (c = gc(), c != ' ' && c != '\n' && c != EOF) *s = c, s++;
}
template <class T>
void read(T &ret) {
  char c;
  if (c = gc(), c == EOF) return;
  while (c != '-' && c != '.' && (c < '0' || c > '9')) c = gc();
  int sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = gc(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  if (c == ' ' || c == '\n') {
    ret *= sgn;
    return;
  }
  T bit = 0.1;
  while (c = gc(), c >= '0' && c <= '9') ret += (c - '0') * bit, bit /= 10;
  ret *= sgn;
}
void I(){};
template <class T, class... A>
void I(T &a, A &...x) {
  read(a);
  I(x...);
}
void write(char x) { putchar(x); }
void write(string x) { cout << x; }
void write(const char *x) { printf(x); }
void write(double x) { printf("%.10lf", x); }
void write(long double x) { printf("%.10Lf", x); }
template <class T>
void write(T x) {
  if (x < 0) putchar('-'), x *= -1;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
void O(){};
template <class T, class... A>
void O(T a, A... x) {
  write(a);
  O(x...);
}
template <class T>
void Min(T &a, const T b) {
  if (a > b) a = b;
}
template <class T>
void Max(T &a, const T b) {
  if (a < b) a = b;
}
const double PI = acos(-1);
const int inf = 0x3f3f3f3f;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const int N = 2030 + 10;
const int mod = 998244353;
int n, ar[N];
void solve() {
  I(n);
  for (int i = 1; i <= n; i++) I(ar[i]);
  for (int i = 1; i <= n; i++) {
    if (ar[i] % 2 != i % 2) {
      O(-1);
      return;
    }
  }
  vector<int> v;
  for (int i = n; i >= 3; i -= 2) {
    int p = 1;
    while (ar[p] != i) p += 2;
    v.push_back(p);
    reverse(ar + 1, ar + 1 + p);
    p = 2;
    while (ar[p] != i - 1) p += 2;
    p--;
    v.push_back(p);
    reverse(ar + 1, ar + 1 + p);
    p += 2;
    v.push_back(p);
    reverse(ar + 1, ar + 1 + p);
    p = 3;
    v.push_back(p);
    reverse(ar + 1, ar + 1 + p);
    p = i;
    v.push_back(p);
    reverse(ar + 1, ar + 1 + p);
  }
  for (int i = 1; i <= n; i++)
    ;
  O(v.size(), '\n');
  for (auto i : v) O(i, ' ');
}
int main() {
  ;
  int _;
  I(_);
  while (_--) {
    solve();
    O('\n');
  }
}

### Prompt

Generate 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int l, int r) { return uniform_int_distribution<int>(l, r)(rng); }
char gc(void) {
  const int S = 3e7;
  static char buf[S], *s = buf, *t = buf;
  if (s == t) t = buf + fread(s = buf, 1, S, stdin);
  if (s == t) return EOF;
  return *s++;
}
void read(char &c) {
  while (c = gc(), c == ' ' || c == '\n')
    ;
}
void read(string &s) {
  char c;
  while (c = gc(), c != ' ' && c != '\n' && c != EOF) s += c;
}
void read(char *s) {
  char c;
  while (c = gc(), c != ' ' && c != '\n' && c != EOF) *s = c, s++;
}
template <class T>
void read(T &ret) {
  char c;
  if (c = gc(), c == EOF) return;
  while (c != '-' && c != '.' && (c < '0' || c > '9')) c = gc();
  int sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = gc(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  if (c == ' ' || c == '\n') {
    ret *= sgn;
    return;
  }
  T bit = 0.1;
  while (c = gc(), c >= '0' && c <= '9') ret += (c - '0') * bit, bit /= 10;
  ret *= sgn;
}
void I(){};
template <class T, class... A>
void I(T &a, A &...x) {
  read(a);
  I(x...);
}
void write(char x) { putchar(x); }
void write(string x) { cout << x; }
void write(const char *x) { printf(x); }
void write(double x) { printf("%.10lf", x); }
void write(long double x) { printf("%.10Lf", x); }
template <class T>
void write(T x) {
  if (x < 0) putchar('-'), x *= -1;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
void O(){};
template <class T, class... A>
void O(T a, A... x) {
  write(a);
  O(x...);
}
template <class T>
void Min(T &a, const T b) {
  if (a > b) a = b;
}
template <class T>
void Max(T &a, const T b) {
  if (a < b) a = b;
}
const double PI = acos(-1);
const int inf = 0x3f3f3f3f;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const int N = 2030 + 10;
const int mod = 998244353;
int n, ar[N];
void solve() {
  I(n);
  for (int i = 1; i <= n; i++) I(ar[i]);
  for (int i = 1; i <= n; i++) {
    if (ar[i] % 2 != i % 2) {
      O(-1);
      return;
    }
  }
  vector<int> v;
  for (int i = n; i >= 3; i -= 2) {
    int p = 1;
    while (ar[p] != i) p += 2;
    v.push_back(p);
    reverse(ar + 1, ar + 1 + p);
    p = 2;
    while (ar[p] != i - 1) p += 2;
    p--;
    v.push_back(p);
    reverse(ar + 1, ar + 1 + p);
    p += 2;
    v.push_back(p);
    reverse(ar + 1, ar + 1 + p);
    p = 3;
    v.push_back(p);
    reverse(ar + 1, ar + 1 + p);
    p = i;
    v.push_back(p);
    reverse(ar + 1, ar + 1 + p);
  }
  for (int i = 1; i <= n; i++)
    ;
  O(v.size(), '\n');
  for (auto i : v) O(i, ' ');
}
int main() {
  ;
  int _;
  I(_);
  while (_--) {
    solve();
    O('\n');
  }
}

```