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;
class TaskG {
  int n;
  bool ok;
  vector<vector<int>> b, f;
  vector<array<int, 2>> c, r;
  vector<int> x, y, col, res;
  void dfs(int u) {
    if (col[u])
      x.emplace_back(u);
    else
      y.emplace_back(u);
    for (int v : f[u]) {
      if (col[v] == -1) {
        col[v] = col[u];
        dfs(v);
      } else {
        assert(col[v] == col[u]);
      }
    }
    for (int v : b[u]) {
      if (col[v] == -1) {
        col[v] = !col[u];
        dfs(v);
      } else {
        assert(col[v] != col[u]);
      }
    }
  }

 public:
  TaskG(int n) {
    ok = true;
    this->n = n;
    b.resize(n);
    f.resize(n);
    c.assign(n, {-1, -1});
    r.assign(n, {-1, -1});
    col.assign(n, -1);
    for (int z = 0; z < 2; z++) {
      for (int i = 0; i < n; i++) {
        int x;
        cin >> x;
        --x;
        if (c[x][0] == -1)
          c[x][0] = i, r[x][0] = z;
        else if (c[x][1] == -1)
          c[x][1] = i, r[x][1] = z;
        else
          ok = false;
      }
    }
  }
  void solve() {
    if (not ok) {
      cout << -1 << '\n';
      return;
    }
    for (int x = 0; x < n; x++) {
      if (c[x][0] == c[x][1]) {
        assert(r[x][0] != r[x][1]);
        col[c[x][0] | c[x][1]] = 2;
      } else {
        if (r[x][0] != r[x][1]) {
          f[c[x][0]].emplace_back(c[x][1]);
          f[c[x][1]].emplace_back(c[x][0]);
        } else {
          b[c[x][0]].emplace_back(c[x][1]);
          b[c[x][1]].emplace_back(c[x][0]);
        }
      }
    }
    for (int u = 0; u < n; u++) {
      if (col[u] == -1) {
        col[u] = 0;
        dfs(u);
        if (x.size() < y.size())
          res.insert(res.end(), x.begin(), x.end());
        else
          res.insert(res.end(), y.begin(), y.end());
        x.clear(), y.clear();
      }
    }
    cout << res.size() << '\n';
    for (auto &x : res) {
      cout << (x + 1) << ' ';
    }
    cout << '\n';
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    TaskG G(n);
    G.solve();
  }
}
