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;
template <class T1, class T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p);
template <class T>
ostream &operator<<(ostream &os, vector<T> &v);
template <class T>
ostream &operator<<(ostream &os, set<T> &v);
template <class T1, class T2>
ostream &operator<<(ostream &os, map<T1, T2> &v);
const long long mod = 1e9 + 7;
const int N = 2e5 + 5;
vector<vector<pair<int, int> > > adj;
int n;
vector<int> a, b;
vector<bool> vis;
void dfs(int s, vector<int> &idx, int &last) {
  last = s;
  vis[s] = true;
  for (pair<int, int> i : adj[s]) {
    if (vis[i.first]) {
      continue;
    }
    idx.push_back(i.second);
    dfs(i.first, idx, last);
  }
}
void TEST_CASES(int cas) {
  scanf("%d", &n);
  a = vector<int>(n);
  b = vector<int>(n);
  vis = vector<bool>(n);
  adj.clear();
  adj.resize(n);
  vector<int> cnt(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i]--;
    cnt[a[i]]++;
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &b[i]);
    b[i]--;
    cnt[b[i]]++;
  }
  for (int i = 0; i < n; i++) {
    if (cnt[i] != 2) {
      printf("-1\n");
      return;
    }
  }
  for (int i = 0; i < n; i++) {
    adj[a[i]].push_back({b[i], i + 1});
    adj[b[i]].push_back({a[i], -i - 1});
  }
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    if (vis[i]) continue;
    vector<int> idx;
    int src = i;
    int last;
    dfs(i, idx, last);
    for (pair<int, int> i : adj[last]) {
      if (i.first == src) {
        bool flag = true;
        for (int x : idx) {
          x = abs(x);
          if (x == abs(i.second)) {
            flag = false;
            break;
          }
        }
        if (!flag) continue;
        ;
        idx.push_back(i.second);
        break;
      }
    };
    ;
    ;
    int pos = 0, neg = 0;
    for (int i : idx) {
      if (i > 0)
        pos++;
      else
        neg++;
    }
    if (neg < pos) {
      for (int &i : idx) i = -i;
    }
    for (int i : idx) {
      if (i > 0) ans.push_back(i);
    }
  }
  printf("%d\n", ans.size());
  for (int &i : ans) {
    printf("%d ", i);
  }
  puts("");
}
int main() {
  int t = 1, cas = 0;
  scanf("%d", &t);
  while (t--) {
    TEST_CASES(++cas);
  }
  return 0;
}
template <class T1, class T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p) {
  os << "{" << p.first << ", " << p.second << "} ";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, vector<T> &v) {
  os << "[ ";
  for (int i = 0; i < v.size(); i++) {
    os << v[i] << " ";
  }
  os << " ]";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, set<T> &v) {
  os << "[ ";
  for (T i : v) {
    os << i << " ";
  }
  os << " ]";
  return os;
}
template <class T1, class T2>
ostream &operator<<(ostream &os, map<T1, T2> &v) {
  for (auto i : v) {
    os << "Key : " << i.first << " , Value : " << i.second << endl;
  }
  return os;
}
