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;
inline long long mod(long long n, long long m) {
  long long ret = n % m;
  if (ret < 0) ret += m;
  return ret;
}
long long gcd(long long a, long long b) {
  return (b == 0LL ? a : gcd(b, a % b));
}
long long exp(long long a, long long b, long long m) {
  if (b == 0LL) return 1LL;
  if (b == 1LL) return mod(a, m);
  long long k = mod(exp(a, b / 2, m), m);
  if (b & 1LL) {
    return mod(a * mod(k * k, m), m);
  } else
    return mod(k * k, m);
}
const int N = 200100;
int a[N][2];
int cnt[N];
vector<pair<int, int> > pos[N];
vector<pair<int, int> > g[N];
void addEdge(int a, int b, int c) {
  g[a].push_back(pair<int, int>(b, c));
  g[b].push_back(pair<int, int>(a, c));
}
vector<int> res;
int vis[N];
int cor[N];
int c[2];
vector<int> comp[2];
void dfs(int v) {
  c[cor[v]]++;
  comp[cor[v]].push_back(v);
  vis[v] = 1;
  for (auto to : g[v]) {
    if (!vis[to.first]) {
      cor[to.first] = cor[v] ^ to.second;
      dfs(to.first);
    }
  }
}
int ok[N];
void QUERY() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cnt[i] = 0;
    pos[i].clear();
    g[i].clear();
    vis[i] = 0;
    ok[i] = 0;
  }
  for (int j = 0; j < (2); ++j) {
    for (int i = 1; i <= n; i++) {
      cin >> a[i][j];
      cnt[a[i][j]]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (cnt[i] != 2) {
      cout << -1 << "\n";
      return;
    }
  }
  for (int j = 0; j < 2; j++) {
    for (int i = 1; i <= n; i++) {
      pos[a[i][j]].push_back(pair<int, int>(i, j));
    }
  }
  for (int i = 1; i < (n + 1); ++i) {
    assert((int)pos[i].size() == 2);
    if (pos[i][0].first == pos[i][1].first) {
      ok[pos[i][0].first] = 1;
      continue;
    }
    if (pos[i][0].second == pos[i][1].second) {
      addEdge(pos[i][0].first, pos[i][1].first, 1);
    } else {
      addEdge(pos[i][0].first, pos[i][1].first, 0);
    }
  }
  res.clear();
  for (int i = 1; i <= n; i++) {
    if (!vis[i] and !ok[i]) {
      c[0] = c[1] = 0;
      comp[0].clear();
      comp[1].clear();
      dfs(i);
      if (c[0] > c[1]) {
        swap(comp[0], comp[1]);
      }
      for (int x : comp[0]) res.push_back(x);
    }
  }
  sort(res.begin(), res.end());
  for (int x : res) {
    swap(a[x][0], a[x][1]);
  }
  for (int j = 0; j < (2); ++j) {
    for (int i = 1; i <= n; i++) {
      cnt[i] = 0;
    }
    for (int i = 1; i <= n; i++) {
      cnt[a[i][j]]++;
    }
    for (int i = 1; i <= n; i++) {
      if (cnt[i] != 1) {
        cout << -1 << "\n";
        return;
      }
    }
  }
  cout << (int)res.size() << "\n";
  for (int x : res) {
    cout << x << " ";
  }
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    QUERY();
  }
}
