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;
int cnt0, cnt1, timer;
void dfs(int v, vector<int> &colour, vector<int> &comp,
         vector<vector<pair<int, int>>> &g) {
  if (colour[v] == 0)
    cnt0++;
  else
    cnt1++;
  comp[v] = timer;
  for (auto &now : g[v])
    if (colour[now.first] == -1) {
      colour[now.first] = (colour[v] ^ now.second);
      dfs(now.first, colour, comp, g);
    }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    timer = 0;
    int n;
    cin >> n;
    vector<int> a(n), b(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      a[i]--;
    }
    for (int i = 0; i < n; i++) {
      cin >> b[i];
      b[i]--;
    }
    vector<int> cnt(n);
    for (int i = 0; i < n; i++) cnt[a[i]]++, cnt[b[i]]++;
    bool tf = true;
    for (int i = 0; i < n; i++)
      if (cnt[i] != 2) tf = false;
    if (!tf) {
      cout << -1 << '\n';
      continue;
    }
    vector<int> pos1(n, -1), pos2(n, -1);
    for (int i = 0; i < n; i++) {
      if (pos1[a[i]] == -1)
        pos1[a[i]] = i;
      else
        pos2[a[i]] = i;
    }
    for (int i = 0; i < n; i++) {
      if (pos1[b[i]] == -1)
        pos1[b[i]] = i;
      else
        pos2[b[i]] = i;
    }
    vector<vector<pair<int, int>>> g(n);
    for (int i = 0; i < n; i++) {
      if (pos1[i] == pos2[i]) continue;
      int ina_1 = (a[pos1[i]] != i), ina_2 = (a[pos2[i]] != i);
      g[pos1[i]].push_back({pos2[i], ina_1 == ina_2});
      g[pos2[i]].push_back({pos1[i], ina_1 == ina_2});
    }
    vector<int> colour(n, -1), comp(n, -1);
    int ans = 0;
    vector<bool> arr;
    for (int i = 0; i < n; i++)
      if (colour[i] == -1) {
        cnt0 = cnt1 = 0;
        colour[i] = 0;
        dfs(i, colour, comp, g);
        timer++;
        ans += min(cnt0, cnt1);
        arr.push_back((bool)(cnt0 < cnt1));
      }
    for (int i = 0; i < n; i++)
      for (auto &now : g[i])
        if (colour[i] != (colour[now.first] ^ now.second)) tf = false;
    if (!tf) {
      cout << -1 << '\n';
      continue;
    }
    cout << ans << '\n';
    for (int i = 0; i < n; i++)
      if (arr[comp[i]] ^ colour[i]) cout << i + 1 << ' ';
    cout << '\n';
  }
  return 0;
}
