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;
using ll = long long;
vector<vector<pair<int, int>>> g;
vector<int> col, comp;
void dfs(int v, bool &valid) {
  comp.push_back(v);
  for (auto [u, t] : g[v]) {
    if (col[u] == -1) {
      col[u] = col[v] ^ t;
      dfs(u, valid);
    } else {
      valid &= (col[u] == col[v] ^ t);
    }
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<vector<array<int, 2>>> where(n);
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      where[x - 1].push_back({i, 0});
    }
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      where[x - 1].push_back({i, 1});
    }
    bool valid = true;
    for (int i = 0; i < n; i++) {
      valid &= (where[i].size() == 2);
    }
    if (!valid) {
      cout << "-1\n";
      continue;
    }
    g.clear();
    g.resize(n);
    for (int i = 0; i < n; i++) {
      int n1 = where[i][0][0], n2 = where[i][1][0];
      if (n1 != n2) {
        g[n1].emplace_back(n2, (where[i][0][1] == where[i][1][1]));
        g[n2].emplace_back(n1, (where[i][0][1] == where[i][1][1]));
      }
    }
    col.clear();
    col.resize(n, -1);
    for (int i = 0; i < n && valid; i++) {
      if (col[i] == -1) {
        comp.clear();
        col[i] = 0;
        dfs(i, valid);
        int sm = 0;
        for (auto v : comp) {
          sm += col[v];
        }
        if (sm * 2 > comp.size()) {
          for (auto v : comp) {
            col[v] ^= 1;
          }
        }
      }
    }
    if (!valid) {
      cout << "-1\n";
    } else {
      cout << accumulate(col.begin(), col.end(), 0) << '\n';
      for (int i = 0; i < n; i++) {
        if (col[i]) {
          cout << i + 1 << ' ';
        }
      }
      cout << '\n';
    }
  }
}
