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>>> indWhere;
vector<vector<pair<int, int>>> G;
vector<bool> Used;
vector<vector<int>> comp;
vector<int> Assign;
vector<int> Ans;
void dfs1(int u, int cC) {
  Used[u] = true;
  for (pair<int, int> item : G[u]) {
    int v = item.first;
    if (!Used[v]) dfs1(v, cC);
  }
  comp[cC].push_back(u);
}
void dfs2(int u, bool flag) {
  Assign[u] = flag;
  for (auto item : G[u]) {
    if (Assign[item.first] == -1) {
      dfs2(item.first, flag ^ item.second);
    }
  }
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int ans = 0;
    int n;
    scanf("%d", &n);
    indWhere.assign(n, vector<pair<int, int>>());
    for (int i = 0; i < 2; ++i) {
      for (int j = 0; j < n; ++j) {
        int tmp;
        scanf("%d", &tmp);
        tmp--;
        indWhere[tmp].push_back(make_pair(i, j));
        if (indWhere[tmp].size() > 2) {
          ans = -1;
        }
      }
    }
    if (ans) {
      puts("-1");
      continue;
    }
    G.assign(n, vector<pair<int, int>>());
    for (int i = 0; i < n; ++i) {
      int r0 = indWhere[i][0].first, c0 = indWhere[i][0].second;
      int r1 = indWhere[i][1].first, c1 = indWhere[i][1].second;
      if (c0 == c1) continue;
      G[c0].push_back(make_pair(c1, r0 == r1));
      G[c1].push_back(make_pair(c0, r0 == r1));
    }
    Used.assign(n, false);
    comp.clear();
    for (int i = 0; i < n; ++i) {
      int curComp = comp.size();
      comp.push_back(vector<int>());
      if (!Used[i]) dfs1(i, curComp);
    }
    Assign.assign(n, -1);
    Ans.clear();
    for (int i = 0; i < comp.size(); ++i) {
      int tot = comp[i].size();
      if (tot <= 1) continue;
      int cntTrue = 0;
      dfs2(comp[i][0], true);
      for (int j = 0; j < comp[i].size(); ++j) {
        if (Assign[comp[i][j]]) cntTrue++;
      }
      bool flag = cntTrue > tot - cntTrue;
      for (int j = 0; j < comp[i].size(); ++j) {
        if (flag ^ Assign[comp[i][j]]) Ans.push_back(comp[i][j]);
      }
    }
    if (Ans.empty()) {
      printf("0\n\n");
    } else {
      printf("%d\n", Ans.size());
      for (int i = 0; i < Ans.size(); ++i) {
        printf("%d%c", Ans[i] + 1, i == Ans.size() - 1 ? '\n' : ' ');
      }
    }
  }
  return 0;
}
