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 maxi = 1e6 + 10;
string s;
int n;
int a[5][maxi];
vector<pair<int, int> > v[maxi];
vector<int> ans, g[maxi], g1[maxi];
int u[maxi];
map<pair<int, int>, int> mp;
int comp[maxi];
int sz[maxi];
int res, total;
int ob[maxi];
vector<int> nep, par;
void dfs(int x, int y) {
  ob[x] = 1;
  sz[y]++;
  comp[x] = y;
  for (int i : g[x])
    if (!ob[i]) dfs(i, y);
}
void dfs1(int x, int y) {
  total += sz[x];
  if (y)
    nep.push_back(x);
  else {
    res += sz[x];
    par.push_back(x);
  }
  ob[x] = 1;
  for (int i : g1[x])
    if (!ob[i]) dfs1(i, y ^ 1);
}
void solve() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    comp[i] = 0;
    v[i].clear();
    u[i] = 0;
    ob[i] = 0;
    g[i].clear();
    g1[i].clear();
    sz[i] = 0;
  }
  res = 0;
  mp.clear();
  ans.clear();
  for (int j = 1; j <= 2; j++)
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[j][i]);
      v[a[j][i]].push_back({i, j});
    }
  for (int i = 1; i <= n; i++)
    if (v[i].size() != 2) {
      printf("-1\n");
      return;
    }
  for (int i = 1; i <= n; i++) {
    pair<int, int> node1 = v[i][0];
    pair<int, int> node2 = v[i][1];
    if (node1.second != node2.second && node1.first != node2.first) {
      g[node1.first].push_back(node2.first);
      g[node2.first].push_back(node1.first);
    }
  }
  int cur = 0;
  for (int i = 1; i <= n; i++)
    if (!ob[i]) dfs(i, ++cur);
  for (int i = 1; i <= n; i++) {
    pair<int, int> node1 = v[i][0];
    pair<int, int> node2 = v[i][1];
    if (node1.second == node2.second) {
      if (comp[node1.first] == comp[node2.first]) {
        printf("-1\n");
        return;
      }
      int xi = comp[node1.first];
      int yi = comp[node2.first];
      if (!mp[{xi, yi}]) {
        g1[xi].push_back(yi);
        g1[yi].push_back(xi);
        mp[{xi, yi}] = 1;
        mp[{yi, xi}] = 1;
      }
    }
  }
  for (int i = 1; i <= cur; i++) ob[i] = 0;
  for (int i = 1; i <= cur; i++)
    if (!ob[i]) {
      total = 0;
      res = 0;
      nep.clear();
      par.clear();
      dfs1(i, 0);
      if (res > total - res) swap(nep, par);
      for (int j : par) u[j] = 1;
    }
  for (int i = 1; i <= n; i++)
    if (u[comp[i]]) ans.push_back(i);
  printf("%d\n", (int)ans.size());
  for (int i : ans) printf("%d ", i);
  printf("\n");
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
}
