You are given a table a of size 2 × n (i.e. two rows and n columns) consisting of integers from 1 to n.

In one move, you can choose some column j (1 ≤ j ≤ n) and swap values a_{1, j} and a_{2, j} in it. Each column can be chosen no more than once.

Your task is to find the minimum number of moves required to obtain permutations of size n in both first and second rows of the table or determine if it is impossible to do that.

You have to answer t independent test cases.

Recall that the permutation of size n is such an array of size n that contains each integer from 1 to n exactly once (the order of elements doesn't matter).

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of columns in the table. The second line of the test case contains n integers a_{1, 1}, a_{1, 2}, ..., a_{1, n} (1 ≤ a_{1, i} ≤ n), where a_{1, i} is the i-th element of the first row of the table. The third line of the test case contains n integers a_{2, 1}, a_{2, 2}, ..., a_{2, n} (1 ≤ a_{2, i} ≤ n), where a_{2, i} is the i-th element of the second row of the table.

It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case print the answer: -1 if it is impossible to obtain permutation of size n in both first and the second rows of the table, or one integer k in the first line, where k is the minimum number of moves required to obtain permutations in both rows, and k distinct integers pos_1, pos_2, ..., pos_k in the second line (1 ≤ pos_i ≤ n) in any order — indices of columns in which you need to swap values to obtain permutations in both rows. If there are several answers, you can print any.

Example

Input


6
4
1 2 3 4
2 3 1 4
5
5 3 5 1 4
1 2 3 2 4
3
1 2 1
3 3 2
4
1 2 2 1
3 4 3 4
4
4 3 1 4
3 2 2 1
3
1 1 2
3 2 2


Output


0

2
2 3 
1
1 
2
3 4 
2
3 4 
-1
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void ckmax(T& x, T y) {
  x = (y > x ? y : x);
}
template <typename T>
inline void ckmin(T& x, T y) {
  x = (y < x ? y : x);
}
namespace Fread {
const int SIZE = 1 << 20;
char buf[SIZE], *S, *T;
inline char getchar() {
  if (S == T) {
    T = (S = buf) + fread(buf, 1, SIZE, stdin);
    if (S == T) return EOF;
  }
  return *S++;
}
}  // namespace Fread
namespace Fwrite {
const int SIZE = 1 << 20;
char buf[SIZE], *S = buf, *T = buf + SIZE;
inline void flush() {
  fwrite(buf, 1, S - buf, stdout);
  S = buf;
}
inline void putchar(char c) {
  *S++ = c;
  if (S == T) flush();
}
struct _ {
  ~_() { flush(); }
} __;
}  // namespace Fwrite
template <typename T>
inline void read(T& x) {
  x = 0;
  int f = 1;
  char c = Fread::getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = Fread::getchar();
  }
  while (isdigit(c)) x = x * 10 + (c - '0'), c = Fread::getchar();
  x *= f;
}
template <typename T>
inline void write(T x, bool _enter = 0, bool _space = 0) {
  if (!x)
    Fwrite::putchar('0');
  else {
    if (x < 0) Fwrite::putchar('-'), x = -x;
    static char dig[41];
    int top = 0;
    while (x) dig[++top] = (x % 10) + '0', x /= 10;
    while (top) Fwrite::putchar(dig[top--]);
  }
  if (_enter) Fwrite::putchar('\n');
  if (_space) Fwrite::putchar(' ');
}
namespace Fastio {
struct reader {
  template <typename T>
  reader& operator>>(T& x) {
    ::read(x);
    return *this;
  }
  reader& operator>>(char& c) {
    c = Fread::getchar();
    while (c == '\n' || c == ' ') c = Fread::getchar();
    return *this;
  }
  reader& operator>>(char* str) {
    int len = 0;
    char c = Fread::getchar();
    while (c == '\n' || c == ' ') c = Fread::getchar();
    while (c != '\n' && c != ' ') str[len++] = c, c = Fread::getchar();
    str[len] = '\0';
    return *this;
  }
} cin;
const char endl = '\n';
struct writer {
  template <typename T>
  writer& operator<<(T x) {
    ::write(x, 0, 0);
    return *this;
  }
  writer& operator<<(char c) {
    Fwrite::putchar(c);
    return *this;
  }
  writer& operator<<(const char* str) {
    int cur = 0;
    while (str[cur]) Fwrite::putchar(str[cur++]);
    return *this;
  }
} cout;
}  // namespace Fastio
const int MAXN = 2e5;
int n, a[MAXN + 5][2];
vector<int> pos[MAXN + 5];
bool vis[MAXN + 5];
void solve_case() {
  Fastio::cin >> n;
  for (int i = 1; i <= n; ++i) {
    pos[i].clear();
    vis[i] = 0;
  }
  for (int row = 0; row <= 1; ++row)
    for (int i = 1; i <= n; ++i) {
      Fastio::cin >> a[i][row];
      pos[a[i][row]].push_back(i);
    }
  for (int i = 1; i <= n; ++i)
    if (((int)(pos[i]).size()) != 2) {
      Fastio::cout << "-1" << Fastio::endl;
      return;
    }
  vector<int> ans;
  for (int i = 1; i <= n; ++i) {
    if (vis[i]) continue;
    vector<int> col;
    int curp = pos[i][0];
    int curv = i;
    vector<int> vec1, vec2;
    int last_row1 = (a[curp][1] == i);
    int last_row2 = (last_row1 ^ 1);
    vec2.push_back(curp);
    while (true) {
      col.push_back(curp);
      vis[curv] = 1;
      int nxtv = a[curp][a[curp][0] == curv];
      int nxtp = pos[nxtv][pos[nxtv][0] == curp];
      if (nxtv == i) break;
      int cur_row = (a[nxtp][1] == nxtv);
      assert(a[nxtp][cur_row] == nxtv);
      if (cur_row != last_row1) {
        vec1.push_back(nxtp);
      }
      if (cur_row != last_row2) {
        vec2.push_back(nxtp);
      }
      curp = nxtp;
      curv = nxtv;
    }
    if (((int)(vec1).size()) < ((int)(vec2).size())) {
      for (int i = 0; i < ((int)(vec1).size()); ++i) ans.push_back(vec1[i]);
    } else {
      for (int i = 0; i < ((int)(vec2).size()); ++i) ans.push_back(vec2[i]);
    }
  }
  Fastio::cout << ((int)(ans).size()) << Fastio::endl;
  for (int i = 0; i < ((int)(ans).size()); ++i) Fastio::cout << ans[i] << " ";
  Fastio::cout << Fastio::endl;
}
int main() {
  int T;
  Fastio::cin >> T;
  while (T--) {
    solve_case();
  }
  return 0;
}
