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>
#pragma GCC optimize("O3")
using namespace std;
const int N = 2e5 + 5;
int a[5][N], c[5][N], col[N], cnt[2], cmp[N], colcmp[N];
vector<vector<pair<int, int> > > g;
bool used[N];
void dfs(int v, int val, int comp) {
  col[v] = val;
  used[v] = true;
  cmp[v] = comp;
  cnt[val]++;
  for (pair<int, int> to : g[v]) {
    if (used[to.first]) continue;
    dfs(to.first, val ^ to.second, comp);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    g.clear();
    g.resize(n + 1);
    bool ok = true;
    for (int i = 1; i <= n; ++i) {
      c[1][i] = c[2][i] = col[i] = cmp[i] = colcmp[i] = 0;
      used[i] = false;
    }
    for (int i = 1; i <= 2; ++i) {
      for (int j = 1; j <= n; ++j) {
        cin >> a[i][j];
        if (!c[1][a[i][j]])
          c[1][a[i][j]] = j;
        else if (!c[2][a[i][j]])
          c[2][a[i][j]] = j;
        else
          ok = false;
      }
    }
    if (!ok) {
      cout << "-1\n";
      continue;
    }
    vector<int> ans;
    for (int i = 1; i <= n; ++i) {
      if (c[1][i] == c[2][i]) continue;
      int r1 = 1 + (a[2][c[1][i]] == i), r2 = 1 + (a[2][c[2][i]] == i);
      g[c[1][i]].push_back({c[2][i], r1 == r2});
      g[c[2][i]].push_back({c[1][i], r1 == r2});
    }
    int cur = 0;
    for (int i = 1; i <= n; ++i) {
      if (used[i]) continue;
      cnt[0] = cnt[1] = 0;
      dfs(i, 0, ++cur);
      if (cnt[0] < cnt[1]) colcmp[cur] = 1;
    }
    for (int i = 1; i <= n; ++i) {
      if (col[i] ^ colcmp[cmp[i]]) ans.push_back(i);
    }
    cout << ans.size() << "\n";
    for (int p : ans) cout << p << " ";
    cout << "\n";
  }
}
