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 N = 5e5 + 99;
int n, u, T, a[N], vis[N], cnt[N];
vector<pair<int, int> > ans, v[2], g[N];
map<pair<int, int>, int> mark;
void clear() {
  for (int i = 1; i < n + 1; i++) g[i].clear();
  fill(cnt, cnt + 2 + n, 0);
  fill(vis, vis + 2 + n, 0);
  mark.clear();
  ans.clear();
}
void dfs(int x, int rt, int dist) {
  if (vis[x]) return;
  vis[x] = 1;
  for (int i = 0; i < g[x].size(); i++)
    if (!vis[g[x][i].first] || (g[x][i].first == rt && dist > 1)) {
      v[g[x][i].second].push_back(make_pair(x, g[x][i].first));
      dfs(g[x][i].first, rt, dist + 1);
      return;
    }
}
void solve() {
  clear();
  cin >> n;
  for (int i = 1; i < n + 1; i++) scanf("%d", &a[i]);
  ;
  for (int i = 1; i < n + 1; i++) {
    scanf("%d", &u);
    ;
    cnt[u]++, cnt[a[i]]++;
    g[a[i]].push_back(make_pair(u, 1));
    g[u].push_back(make_pair(a[i], 0));
    mark[make_pair(a[i], u)] = mark[make_pair(u, a[i])] = i;
  }
  for (int i = 1; i < n + 1; i++)
    if (cnt[i] != 2) {
      cout << -1 << endl;
      return;
    }
  for (int i = 1; i < n + 1; i++)
    if (!vis[i]) {
      v[0].clear();
      v[1].clear();
      if (g[i][0].first == g[i][1].first) {
        vis[i] = vis[g[i][0].first] = 1;
        if (g[i][0].second == g[i][1].second)
          v[0].push_back(make_pair(i, g[i][0].first)),
              v[1].push_back(make_pair(i, g[i][0].first));
      } else
        dfs(i, i, 0);
      if (v[0].size() < v[1].size())
        for (int i = 0; i < v[0].size(); i++) ans.push_back(v[0][i]);
      else
        for (int i = 0; i < v[1].size(); i++) ans.push_back(v[1][i]);
    }
  cout << ans.size() << '\n';
  for (int i = 0; i < ans.size(); i++) cout << mark[ans[i]] << " ";
  cout << '\n';
}
int main() {
  cin >> T;
  while (T--) solve();
}
