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 MAXN = 400100;
vector<int> grafo[MAXN];
int a[MAXN], b[MAXN];
pair<int, int> p[MAXN];
int q[MAXN];
bool vis[MAXN];
int n;
vector<int> vec;
void dfs(int x) {
  vis[x] = true;
  vec.push_back(x);
  if (x > n)
    vis[x - n] = true;
  else
    vis[x + n] = true;
  for (int i = 0; i < grafo[x].size(); i++) {
    int at = grafo[x][i];
    if (vis[at] == true) continue;
    dfs(at);
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    vec.clear();
    for (int i = 1; i <= 2 * n; i++) {
      p[i] = {0, 0};
      grafo[i].clear();
      q[i] = 0;
      vis[i] = false;
      a[i] = b[i] = 0;
    }
    for (int i = 1; i <= n; i++) {
      int x;
      cin >> x;
      a[x] = i;
      q[x]++;
      if (p[x].first == 0)
        p[x].first = i;
      else
        p[x].second = i;
    }
    for (int i = 1; i <= n; i++) {
      int x;
      cin >> x;
      b[x] = i;
      q[x]++;
      if (p[x].first == 0)
        p[x].first = i;
      else
        p[x].second = i;
    }
    bool f = true;
    for (int i = 1; i <= n; i++)
      if (q[i] != 2) f = false;
    if (f == false) {
      cout << -1 << endl;
      continue;
    }
    for (int i = 1; i <= n; i++) {
      int p1 = p[i].first, p2 = p[i].second;
      if (a[i] > 0 && b[i] > 0) {
        grafo[p1].push_back(p2);
        grafo[p2].push_back(p1);
        grafo[p1 + n].push_back(p2 + n);
        grafo[p2 + n].push_back(p1 + n);
      } else {
        grafo[p1].push_back(p2 + n);
        grafo[p2].push_back(p1 + n);
        grafo[p2 + n].push_back(p1);
        grafo[p1 + n].push_back(p2);
      }
    }
    vector<int> resp;
    for (int i = 1; i <= n; i++) {
      if (vis[i] == true) continue;
      vec.clear();
      dfs(i);
      int h = vec.size();
      int cnt = 0;
      for (int j = 0; j < vec.size(); j++) {
        int at = vec[j];
        if (at > n) cnt++;
      }
      if (cnt < h - cnt) {
        for (int j = 0; j < vec.size(); j++)
          if (vec[j] > n) resp.push_back(vec[j] - n);
      } else {
        for (int j = 0; j < vec.size(); j++)
          if (vec[j] <= n) resp.push_back(vec[j]);
      }
    }
    cout << resp.size() << endl;
    for (int i = 0; i < resp.size(); i++) cout << resp[i] << " ";
    cout << endl;
  }
}
