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;
vector<int> ans;
struct edge {
  int u, v;
  bool b;
  edge() {}
  edge(int _u, int _v, bool _b) : u(_u), v(_v), b(_b) {}
};
struct node {
  vector<edge> edges;
};
struct graph {
  vector<node> nodes;
  vector<int> color;
  int n;
  graph(int _n) : n(_n) {
    nodes.resize(n);
    color.resize(n, -1);
  }
  void add_edge(int u, int v, bool b) {
    edge e1(u, v, b);
    nodes[u].edges.push_back(e1);
    edge e2(v, u, b);
    nodes[v].edges.push_back(e2);
  }
  pair<vector<int>, vector<int>> graphColoring(int u) {
    stack<int> S;
    S.push(u);
    color[u] = 0;
    vector<int> c0, c1;
    c0.push_back(u);
    while (!S.empty()) {
      int cur = S.top();
      S.pop();
      for (auto &e : nodes[cur].edges) {
        if (color[e.v] != -1) {
          if (e.b && color[e.u] == color[e.v]) {
            return {vector<int>(), vector<int>()};
          }
          if (!e.b && color[e.u] != color[e.v]) {
            return {vector<int>(), vector<int>()};
          }
          continue;
        }
        if (e.b) {
          color[e.v] = 1 ^ color[e.u];
        } else {
          color[e.v] = color[e.u];
        }
        if (color[e.v]) {
          c1.push_back(e.v);
        } else {
          c0.push_back(e.v);
        }
        S.push(e.v);
      }
    }
    return {move(c0), move(c1)};
  }
  bool graphColoring() {
    for (int i = 0; i < n; i++) {
      if (color[i] == -1) {
        auto p = graphColoring(i);
        if (p.first.size() == p.second.size() && p.first.empty()) {
          return 0;
        }
        if (p.first.size() < p.second.size()) {
          for (auto x : p.first) {
            ans.push_back(x);
          }
        } else {
          for (auto &x : p.second) {
            ans.push_back(x);
          }
        }
      }
    }
    return 1;
  }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    ans.clear();
    vector<int> a1(n), a2(n), freq(n, 0);
    vector<pair<int, int>> pos1(n, {-1, -1}), pos2(n, {-1, -1});
    for (int i = 0; i < n; i++) {
      cin >> a1[i];
      a1[i]--;
      freq[a1[i]]++;
      if (pos1[a1[i]].first == -1) {
        pos1[a1[i]] = {0, i};
      } else {
        pos2[a1[i]] = {0, i};
      }
    }
    for (int i = 0; i < n; i++) {
      cin >> a2[i];
      a2[i]--;
      freq[a2[i]]++;
      if (pos1[a2[i]].first == -1) {
        pos1[a2[i]] = {1, i};
      } else {
        pos2[a2[i]] = {1, i};
      }
    }
    bool OK = true;
    for (int i = 0; i < n; i++) {
      if (freq[i] != 2) {
        OK = false;
        break;
      }
    }
    if (!OK) {
      cout << -1 << '\n';
      continue;
    }
    graph g(n);
    for (int i = 0; i < n; i++) {
      int c1 = pos1[i].second, c2 = pos2[i].second;
      if (pos1[i].first == pos2[i].first) {
        g.add_edge(c1, c2, 1);
      } else {
        g.add_edge(c1, c2, 0);
      }
    }
    bool b = g.graphColoring();
    if (!b) {
      cout << -1 << '\n';
    } else {
      cout << ans.size() << '\n';
      for (auto &x : ans) {
        cout << x + 1 << ' ';
      }
      cout << '\n';
    }
  }
}
