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 MaxN = 2e5 + 5;
int n;
int a[2][MaxN];
vector<pair<int, int> > g[MaxN];
int color[MaxN], cmp[MaxN];
int cnt0 = 0, cnt1 = 0, cp = 0;
void dfs(int u, int c, int cp) {
  color[u] = c;
  cmp[u] = cp;
  if (color[u] == 0)
    cnt0++;
  else
    cnt1++;
  for (auto p : g[u]) {
    if (color[p.first] == -1) {
      dfs(p.first, c ^ p.second, cp);
    }
  }
}
void solve_cases() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    color[i] = -1;
    cmp[i] = -1;
  }
  vector<vector<int> > pos(n);
  bool bad = false;
  for (int j = 0; j < 2; j++) {
    for (int i = 0; i < n; i++) {
      cin >> a[j][i];
      a[j][i]--;
      pos[a[j][i]].push_back(i);
    }
  }
  for (int i = 0; i < n; i++) {
    if (pos[i].size() != 2) {
      bad = true;
      break;
    }
    int c1 = pos[i][0], c2 = pos[i][1];
    if (c1 == c2) continue;
    int r1 = a[0][c1] != i;
    int r2 = a[0][c2] != i;
    g[c1].push_back({c2, r1 == r2});
    g[c2].push_back({c1, r1 == r2});
  }
  int ans = 0;
  vector<pair<int, int> > component;
  for (int i = 0; i < n; i++) {
    if (color[i] == -1) {
      cnt0 = 0, cnt1 = 0;
      dfs(i, 0, cp);
      cp++;
      ans += min(cnt0, cnt1);
      component.push_back(make_pair(cnt0, cnt1));
    }
  }
  for (int i = 0; i < n; ++i) {
    for (auto p : g[i]) {
      int j = p.first, diff = p.second;
      if ((color[i] ^ color[j]) != diff) {
        bad = true;
      }
    }
  }
  if (bad)
    cout << -1 << "\n";
  else {
    cout << ans << "\n";
    for (int i = 0; i < n; i++) {
      bool changeZero = component[cmp[i]].first < component[cmp[i]].second;
      if (color[i] ^ changeZero) {
        cout << i + 1 << " ";
      }
    }
    cout << "\n";
  }
  for (int i = 0; i < n; i++) g[i].clear();
  cnt0 = 0, cnt1 = 0, cp = 0;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve_cases();
  }
}
