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 n, a[2][200005], cnt, e[200005], z, minim;
bool viz[200005];
vector<pair<int, int> > Muchii[200005];
void dfs(int nod, int i) {
  e[++z] = i;
  viz[i] = true;
  for (auto it : Muchii[nod]) {
    if (viz[it.second] == false) {
      dfs(it.first, it.second);
    }
  }
}
void solve() {
  vector<int> ans;
  cin >> n;
  vector<int> c(n + 3);
  bool ok = true;
  for (int i = 0; i < 2; ++i)
    for (int j = 1; j <= n; ++j) {
      cin >> a[i][j];
      c[a[i][j]]++;
    }
  for (int i = 1; i <= n; ++i) {
    if (c[i] != 2) {
      ok = false;
      break;
    }
  }
  if (ok == false) {
    cout << -1 << "\n";
    return;
  }
  for (int i = 1; i <= n; ++i) {
    Muchii[a[0][i]].push_back({a[1][i], i});
    Muchii[a[1][i]].push_back({a[0][i], i});
  }
  for (int i = 1; i <= n; ++i) {
    if (a[0][i] == a[1][i]) continue;
    if (viz[i] == false) {
      z = 0;
      dfs(a[1][i], i);
      ++cnt;
      int fr1 = 0, fr2 = 1;
      pair<int, int> last = {a[0][e[1]], a[1][e[1]]};
      for (int j = 2; j <= z; ++j) {
        if (last.first == a[0][e[j]] || last.second == a[1][e[j]]) {
          ++fr1;
          last = {a[1][e[j]], a[0][e[j]]};
        } else {
          last = {a[0][e[j]], a[1][e[j]]};
        }
      }
      last = {a[1][e[1]], a[0][e[1]]};
      for (int j = 2; j <= z; ++j) {
        if (last.first == a[0][e[j]] || last.second == a[1][e[j]]) {
          ++fr2;
          last = {a[1][e[j]], a[0][e[j]]};
        } else {
          last = {a[0][e[j]], a[1][e[j]]};
        }
      }
      minim += min(fr1, fr2);
      if (fr1 > fr2) swap(a[0][e[1]], a[1][e[1]]), ans.push_back(e[1]);
      for (int j = 2; j <= z; ++j) {
        if (a[0][e[j - 1]] == a[0][e[j]] || a[1][e[j - 1]] == a[1][e[j]])
          swap(a[0][e[j]], a[1][e[j]]), ans.push_back(e[j]);
      }
    }
  }
  cout << ans.size() << "\n";
  for (int i = 0; i < ans.size(); ++i) {
    cout << ans[i] << " ";
  }
  cout << "\n";
  for (int i = 1; i <= n; ++i) {
    Muchii[i].clear();
    viz[i] = false;
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}
