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;
vector<vector<pair<int, int>>> g;
vector<int> color;
vector<bool> vis;
int cnt0, cnt1;
vector<int> ans;
void dfs1(int cur, int c) {
  if (color[cur] != -1) return;
  color[cur] = c;
  if (color[cur] == 0)
    cnt0++;
  else
    cnt1++;
  for (auto nbr : g[cur]) {
    dfs1(nbr.first, c ^ nbr.second);
  }
}
void dfs2(int cur, int c) {
  if (vis[cur]) return;
  if (color[cur] == c) ans.push_back(cur);
  vis[cur] = true;
  for (auto nbr : g[cur]) {
    dfs2(nbr.first, c);
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  auto solve = []() {
    int n;
    cin >> n;
    vector<vector<pair<int, int>>> col(n);
    for (int i = 0; i < 2; i++) {
      for (int j = 0; j < n; j++) {
        int x;
        cin >> x;
        col[--x].emplace_back(i, j);
      }
    }
    g = vector<vector<pair<int, int>>>(n);
    for (int i = 0; i < n; i++) {
      if ((int)col[i].size() != 2) {
        cout << -1 << '\n';
        return;
      }
      int c1 = col[i][0].second, c2 = col[i][1].second;
      int r1 = col[i][0].first, r2 = col[i][1].first;
      if (c1 == c2) continue;
      if (r1 == r2) {
        g[c1].emplace_back(c2, 1);
        g[c2].emplace_back(c1, 1);
      } else {
        g[c1].emplace_back(c2, 0);
        g[c2].emplace_back(c1, 0);
      }
    }
    color = vector<int>(n, -1);
    vis = vector<bool>(n, false);
    ans.clear();
    for (int i = 0; i < n; i++) {
      if (color[i] == -1) {
        cnt0 = cnt1 = 0;
        dfs1(i, 0);
        if (cnt0 <= cnt1) {
          dfs2(i, 0);
        } else {
          dfs2(i, 1);
        }
      }
    }
    for (int i = 0; i < n; i++) {
      for (auto nbr : g[i]) {
        if ((color[i] ^ color[nbr.first]) != nbr.second) {
          cout << -1 << '\n';
          return;
        }
      }
    }
    cout << ans.size() << '\n';
    for (int i = 0; i < (int)ans.size(); i++) {
      if (i > 0) cout << " ";
      cout << ans[i] + 1;
    }
    cout << '\n';
  };
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}
