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;
const int N = 2e5 + 5;
int gdzie[N], tab[N], tab2[N], vis[N], ile[N];
pair<int, int> E[2][N];
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> tab[i];
      ile[tab[i]]++;
    }
    for (int i = 0; i < n; i++) {
      cin >> tab2[i];
      ile[tab2[i]]++;
    }
    bool b = 1;
    for (int i = 1; i <= n; i++) {
      if (ile[i] != 2) {
        b = 0;
        break;
      }
    }
    if (!b) {
      cout << "-1\n";
      for (int i = 1; i <= n; i++) {
        ile[i] = 0;
      }
      continue;
    }
    for (int i = 0; i < n; i++) {
      if (tab[i] == tab2[i]) continue;
      if (!gdzie[tab[i]])
        gdzie[tab[i]] = i + 1;
      else {
        if (tab[gdzie[tab[i]] - 1] == tab[i]) {
          E[0][i] = {gdzie[tab[i]] - 1, 1};
          E[0][gdzie[tab[i]] - 1] = {i, 1};
        } else {
          E[0][i] = {gdzie[tab[i]] - 1, 0};
          E[1][gdzie[tab[i]] - 1] = {i, 0};
        }
      }
      if (!gdzie[tab2[i]])
        gdzie[tab2[i]] = i + 1;
      else {
        if (tab2[gdzie[tab2[i]] - 1] == tab2[i]) {
          E[1][i] = {gdzie[tab2[i]] - 1, 1};
          E[1][gdzie[tab2[i]] - 1] = {i, 1};
        } else {
          E[1][i] = {gdzie[tab2[i]] - 1, 0};
          E[0][gdzie[tab2[i]] - 1] = {i, 0};
        }
      }
    }
    for (int i = 0; i < n; i++) {
    }
    vector<int> V, V1, V2;
    for (int i = 0; i < n; i++) {
      if (tab[i] == tab2[i]) continue;
      if (!vis[i]) {
        vis[i] = 1;
        int id = 0, col = E[id][i].second, wsk = E[id][i].first;
        id ^= E[id][i].second;
        V1.push_back(i);
        while (wsk != i) {
          vis[wsk] = 1;
          if (col == 1)
            V2.push_back(wsk);
          else
            V1.push_back(wsk);
          col ^= E[id][wsk].second;
          bool c = E[id][wsk].second;
          wsk = E[id][wsk].first;
          id ^= c;
        }
        if (V1.size() > V2.size()) {
          for (int j : V2) {
            V.push_back(j);
          }
        } else {
          for (int j : V1) {
            V.push_back(j);
          }
        }
        V1.clear();
        V2.clear();
      }
    }
    cout << V.size() << "\n";
    for (int i : V) {
      cout << i + 1 << " ";
    }
    cout << "\n";
    for (int i = 0; i <= n; i++) {
      ile[i] = 0;
      E[0][i] = {0, 0};
      E[1][i] = {0, 0};
      gdzie[i] = 0;
      vis[i] = 0;
    }
  }
}
