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 A[3][222222];
int B[3][222222];
bool done[222222];
bool done_val[222222];
bool done_solve[222222];
vector<int> pos[222222];
int GetCountsInFirstRow(int x) {
  int p1 = pos[x][0];
  int p2 = pos[x][1];
  return (A[1][p1] == x) + (A[1][p2] == x);
}
void dfs(int val, vector<int>* ids) {
  if (done_val[val]) return;
  int p1 = pos[val][0];
  int p2 = pos[val][1];
  for (int p : {p1, p2}) {
    if (!done[p]) {
      done[p] = true;
      ids->push_back(p);
      int v1 = A[1][p];
      int v2 = A[2][p];
      dfs(v1, ids);
      dfs(v2, ids);
    }
  }
  done_val[val] = true;
}
void Solve(int id, int val, vector<int>* ans) {
  int v1 = A[1][id];
  int v2 = A[2][id];
  int oval = v1 + v2 - val;
  int opos = pos[oval][0] + pos[oval][1] - id;
  if (done_solve[opos]) return;
  if (A[1][opos] != oval) {
    ans->push_back(opos);
  }
  done_solve[opos] = true;
  Solve(opos, oval, ans);
}
void Solve(int n) {
  for (int i = 1; i <= n; i++) {
    pos[i].clear();
    done[i] = false;
    done_val[i] = false;
  }
  for (int i = 1; i <= n; i++) {
    pos[A[1][i]].push_back(i);
    pos[A[2][i]].push_back(i);
  }
  vector<int> ret;
  for (int v = 1; v <= n; v++) {
    if (done_val[v]) continue;
    vector<int> ids;
    dfs(v, &ids);
    if (ids.size() == 1) continue;
    vector<int> ans1;
    for (int id : ids) {
      done_solve[id] = false;
    }
    done_solve[ids[0]] = true;
    Solve(ids[0], A[1][ids[0]], &ans1);
    vector<int> ans2;
    ans2.push_back(ids[0]);
    for (int id : ids) {
      done_solve[id] = false;
    }
    done_solve[ids[0]] = true;
    Solve(ids[0], A[2][ids[0]], &ans2);
    vector<int> ans;
    if (ans1.size() < ans2.size()) {
      ans = ans1;
    } else {
      ans = ans2;
    }
    for (int x : ans) {
      ret.push_back(x);
    }
  }
  cout << ret.size() << endl;
  for (int x : ret) {
    printf("%d ", x);
  }
  cout << endl;
}
int main() {
  int T;
  cin >> T;
  for (int t = 1; t <= T; t++) {
    int n;
    cin >> n;
    map<int, int> counts;
    for (int i = 1; i <= n; i++) {
      cin >> A[1][i];
      counts[A[1][i]]++;
    }
    for (int i = 1; i <= n; i++) {
      cin >> A[2][i];
      counts[A[2][i]]++;
    }
    bool ok = (counts.size() == n);
    for (int i = 1; i <= n; i++) ok &= (counts[i] == 2);
    if (!ok) {
      cout << -1 << endl;
      continue;
    }
    Solve(n);
  }
}
