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;
void solve() {
  cerr << "BEGIN" << endl;
  int n;
  cin >> n;
  vector<int> a(n), b(n);
  for (int i = 0; (i) < (int)(n); (i)++) {
    cin >> a[i];
  }
  for (int i = 0; (i) < (int)(n); (i)++) {
    cin >> b[i];
  }
  map<int, int> cnt;
  for (int i = 0; (i) < (int)(n); (i)++) {
    cnt[a[i]]++;
    cnt[b[i]]++;
  }
  for (auto it : cnt) {
    if (it.second != 2) {
      cout << -1 << '\n';
      return;
    }
  }
  map<int, int> pos;
  vector<vector<pair<int, int>>> g(n);
  for (int i = 0; (i) < (int)(n); (i)++) {
    if (pos.count(a[i]) == 0) {
      pos[a[i]] = i * 2;
    } else {
      int p = pos[a[i]];
      int idx = p / 2;
      if (p % 2 == 0) {
        g[idx].emplace_back(i, 1);
        g[i].emplace_back(idx, 1);
      } else {
        g[idx].emplace_back(i, 0);
        g[i].emplace_back(idx, 0);
      }
    }
    if (pos.count(b[i]) == 0) {
      pos[b[i]] = i * 2 + 1;
    } else {
      int p = pos[b[i]];
      int idx = p / 2;
      if (p % 2 == 0) {
        g[idx].emplace_back(i, 0);
        g[i].emplace_back(idx, 0);
      } else {
        g[idx].emplace_back(i, 1);
        g[i].emplace_back(idx, 1);
      }
    }
  }
  vector<int> color(n, -1);
  int c0 = 0, c1 = 0;
  vector<int> c0v, c1v;
  function<void(int, int)> dfs = [&](int u, int c) {
    color[u] = c;
    if (c) {
      c0++;
      c0v.push_back(u);
    } else {
      c1++;
      c1v.push_back(u);
    }
    for (pair<int, int> e : g[u]) {
      int v = e.first;
      int rev = e.second;
      if (color[v] != -1) {
        assert(color[v] == (c ^ rev));
        continue;
      }
      dfs(v, c ^ rev);
    }
  };
  int ans = 0;
  vector<int> res;
  for (int i = 0; (i) < (int)(n); (i)++) {
    if (color[i] == -1) {
      dfs(i, 0);
    }
    ans += min(c0, c1);
    if (c0 < c1) {
      for (auto it : c0v) res.push_back(it);
    } else {
      for (auto it : c1v) res.push_back(it);
    }
    c0 = 0, c1 = 0;
    c0v.clear(), c1v.clear();
  }
  cout << ans << '\n';
  for (int i = 0; (i) < (int)(res.size()); (i)++) {
    cout << res[i] + 1 << " ";
  }
  cout << '\n';
}
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
