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>
const int maxn = 4e5 + 5;
using namespace std;
vector<int> G[maxn], Gp[2];
int arr[maxn][2], Val[maxn];
vector<pair<int, int>> site[maxn];
bool vis[maxn];
void dfs(int now, int tp) {
  vis[now] = 1;
  Gp[tp].push_back(now);
  for (int k : G[now]) {
    if (!vis[k]) dfs(k, tp);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n, x;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      Val[i] = 0;
      G[i].clear();
      site[i].clear();
      vis[i] = 0;
    }
    for (int i = n + 1; i <= n * 2; i++) {
      Val[i] = 1;
      G[i].clear();
      vis[i] = 0;
    }
    for (int i = 1; i <= n; i++) {
      cin >> x;
      site[x].push_back({i, 1});
    }
    for (int i = 1; i <= n; i++) {
      cin >> x;
      site[x].push_back({i, 2});
    }
    bool fg = 0;
    for (int i = 1; i <= n; i++) {
      fg |= ((int)(site[i].size()) != 2);
    }
    if (fg) {
      cout << -1 << "\n";
      continue;
    }
    for (int i = 1; i <= n; i++) {
      if (site[i][0].first == site[i][1].first) {
        vis[site[i][0].first] = vis[site[i][1].first + n] = 1;
        continue;
      }
      int a = site[i][0].first, b = site[i][1].first;
      if (site[i][0].second == site[i][1].second) {
        G[a].push_back(b + n);
        G[b + n].push_back(a);
        G[a + n].push_back(b);
        G[b].push_back(a + n);
      } else {
        G[a].push_back(b);
        G[b].push_back(a);
        G[a + n].push_back(b + n);
        G[b + n].push_back(a + n);
      }
    }
    vector<int> ans;
    for (int i = 1; i <= n; i++) {
      if (vis[i]) continue;
      Gp[0].clear();
      Gp[1].clear();
      dfs(i, 0);
      dfs(i + n, 1);
      int sum1 = 0, sum2 = 0;
      for (int k : Gp[0]) {
        sum1 += Val[k];
      }
      for (int k : Gp[1]) {
        sum2 += Val[k];
      }
      if (sum2 < sum1) swap(Gp[0], Gp[1]);
      for (int k : Gp[0]) {
        if (k > n) ans.push_back(k - n);
      }
    }
    sort(ans.begin(), ans.end());
    cout << (int)(ans.size()) << "\n";
    for (int i = 0; i < (int)(ans.size()); i++) {
      cout << ans[i] << " ";
    }
    cout << "\n";
  }
}
