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;
using lli = long long int;
const int MOD = 1000000007;
const int MOD1 = 998244353;
const int maxn = 100010;
const int lim = (int)1e9;
vector<int> v[2 * maxn];
vector<int> adj[2 * maxn];
vector<int> parent;
vector<int> vis;
set<int> st;
int FIND(int node) {
  if (parent[node] < 0) return node;
  return parent[node] = FIND(parent[node]);
}
void DSU(int a, int b) {
  a = FIND(a);
  b = FIND(b);
  if (a != b) {
    if (parent[a] > parent[b]) swap(a, b);
    parent[a] += parent[b];
    parent[b] = a;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 0, n = 0;
  cin >> t;
  while (t--) {
    cin >> n;
    parent.resize(n);
    vis.resize(n);
    for (int i = 0; i < n; ++i) {
      v[i].clear();
      adj[i].clear();
      parent[i] = -1;
      vis[i] = 0;
    }
    vector<int> a(n);
    vector<int> b(n);
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
      --a[i];
      v[a[i]].push_back(i);
    }
    for (int i = 0; i < n; ++i) {
      cin >> b[i];
      --b[i];
      v[b[i]].push_back(n + i);
    }
    bool flag = false;
    for (int i = 0; i < n; ++i) {
      if ((int)v[i].size() != 2) {
        flag = true;
        break;
      }
      if (v[i][0] < n && v[i][1] >= n) DSU(v[i][0], v[i][1] - n);
    }
    if (flag) {
      cout << "-1" << endl;
      continue;
    }
    for (int i = 0; i < n; ++i) {
      if (v[i][0] >= n && v[i][1] >= n) {
        if (FIND(v[i][0] - n) == FIND(v[i][1] - n)) {
          flag = true;
          break;
        } else {
          adj[FIND(v[i][0] - n)].push_back(FIND(v[i][1] - n));
          adj[FIND(v[i][1] - n)].push_back(FIND(v[i][0] - n));
        }
      } else if (v[i][0] < n && v[i][1] < n) {
        if (FIND(v[i][0]) == FIND(v[i][1])) {
          flag = true;
          break;
        } else {
          adj[FIND(v[i][0])].push_back(FIND(v[i][1]));
          adj[FIND(v[i][1])].push_back(FIND(v[i][0]));
        }
      }
    }
    if (flag) {
      cout << "-1" << endl;
      continue;
    }
    st.clear();
    queue<int> q;
    for (int i = 0; i < n; ++i) {
      vector<int> parity[2];
      if (FIND(i) == i && vis[i] == 0) {
        q.push(i);
        vis[i] = 2;
        q.push(0);
      }
      int odd = 0, even = 0;
      while (!q.empty()) {
        int f = q.front();
        q.pop();
        int s = q.front();
        q.pop();
        if (s & 1) {
          odd -= parent[f];
          parity[1].push_back(f);
        } else {
          even -= parent[f];
          parity[0].push_back(f);
        }
        for (auto &j : adj[f]) {
          if (vis[j]) {
            if (vis[j] % 2 != (s + 1) % 2) {
              flag = true;
              break;
            }
            continue;
          }
          q.push(j);
          q.push(s + 1);
          if ((s + 1) & 1)
            vis[j] = 1;
          else
            vis[j] = 2;
        }
        if (flag) break;
      }
      if (flag) break;
      if (odd > even) {
        for (auto &j : parity[0]) st.insert(j);
      } else {
        for (auto &j : parity[1]) st.insert(j);
      }
    }
    if (flag) {
      cout << "-1" << endl;
      continue;
    }
    vector<int> res;
    for (int i = 0; i < n; ++i)
      if (st.count(FIND(i))) res.push_back(i + 1);
    cout << (int)res.size() << endl;
    for (auto &i : res) cout << i << " ";
    cout << endl;
  }
}
