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<pair<int, int>> g[200011];
pair<pair<int, int>, pair<int, int>> coord[200011];
int occ[200011];
int visited[200011];
int n;
bool valid = false;
vector<int> ret[2];
vector<int> ans;
void init() {
  int i;
  valid = true;
  for (i = 1; i <= n; ++i) {
    g[i].clear();
    coord[i].first.first = coord[i].first.second = 0;
    coord[i].second.first = coord[i].second.second = 0;
    occ[i] = 0;
    visited[i] = -1;
  }
  ret[0].clear();
  ret[1].clear();
  ans.clear();
}
void dfs(int v) {
  for (auto ne : g[v]) {
    int nv = ne.first;
    if (visited[nv] == -1) {
      visited[nv] = visited[v] ^ ne.second;
      ret[visited[nv]].push_back(nv);
      dfs(nv);
    } else if ((visited[nv] ^ visited[v]) != ne.second) {
      valid = false;
    }
  }
}
void solve() {
  int i;
  for (i = 1; i <= n; ++i) {
    if (coord[i].first.first != coord[i].second.first) {
      if (coord[i].first.second != coord[i].second.second) {
        g[coord[i].first.second].emplace_back(coord[i].second.second, 0);
        g[coord[i].second.second].emplace_back(coord[i].first.second, 0);
      }
    } else {
      g[coord[i].first.second].emplace_back(coord[i].second.second, 1);
      g[coord[i].second.second].emplace_back(coord[i].first.second, 1);
    }
  }
  for (i = 1; i <= n; ++i) {
    if (visited[i] == -1) {
      visited[i] = 1;
      ret[1].push_back(i);
      dfs(i);
      if (ret[0].size() < ret[1].size()) {
        for (auto v : ret[0]) ans.push_back(v);
      } else {
        for (auto v : ret[1]) ans.push_back(v);
      }
      ret[0].clear();
      ret[1].clear();
    }
  }
  if (!valid) {
    printf("-1\n");
  } else {
    printf("%d\n", (int)ans.size());
    for (auto v : ans) {
      printf("%d ", v);
    }
    printf("\n");
  }
}
int main() {
  int T, i;
  scanf("%d", &T);
  while (T-- > 0) {
    scanf("%d", &n);
    init();
    for (i = 1; i <= n; ++i) {
      int v;
      scanf("%d", &v);
      if (coord[v].first.first == 0) {
        coord[v].first.first = 1;
        coord[v].first.second = i;
      } else {
        coord[v].second.first = 1;
        coord[v].second.second = i;
      }
      occ[v]++;
    }
    for (i = 1; i <= n; ++i) {
      int v;
      scanf("%d", &v);
      if (coord[v].first.first == 0) {
        coord[v].first.first = 2;
        coord[v].first.second = i;
      } else {
        coord[v].second.first = 2;
        coord[v].second.second = i;
      }
      occ[v]++;
    }
    for (i = 1; i <= n; ++i) {
      if (occ[i] != 2) {
        valid = false;
      }
    }
    if (!valid) {
      printf("-1\n");
      continue;
    }
    solve();
  }
  return 0;
}
