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;
vector<vector<pair<int, int>>> g;
int zero, one;
int id;
void dfs(vector<int>& col, vector<int>& comp, int v, int c, int idx) {
  col[v] = c;
  if (col[v] == 0)
    zero++;
  else
    one++;
  comp[v] = idx;
  for (auto it : g[v]) {
    if (col[it.first] == -1) {
      dfs(col, comp, it.first, it.second ^ c, idx);
    }
  }
}
void p1385G() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a(n, 0);
    vector<int> b(n, 0);
    g = vector<vector<pair<int, int>>>(n);
    vector<vector<int>> cnt(n + 1);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      a[i]--;
      cnt[a[i]].push_back(i);
    }
    for (int i = 0; i < n; i++) {
      cin >> b[i];
      b[i]--;
      cnt[b[i]].push_back(i);
    }
    bool ok = true;
    for (int i = 0; i <= n - 1; i++) {
      if (cnt[i].size() != 2) {
        ok = false;
        break;
      }
      if (cnt[i][0] == cnt[i][1]) continue;
      int r1 = a[cnt[i][0]] == i;
      int r2 = a[cnt[i][1]] == i;
      int flag = r1 == r2;
      g[cnt[i][0]].push_back({cnt[i][1], flag});
      g[cnt[i][1]].push_back({cnt[i][0], flag});
    }
    int ans = 0;
    vector<int> col(n, -1), comp(n, -1);
    vector<pair<int, int>> colcnt;
    id = 0;
    for (int i = 0; i < n; i++) {
      if (col[i] == -1) {
        zero = one = 0;
        dfs(col, comp, i, 0, id);
        id++;
        colcnt.push_back({zero, one});
        ans += min(zero, one);
      }
    }
    for (int i = 0; i < n; i++) {
      for (auto jj : g[i]) {
        if (col[i] ^ col[jj.first] != jj.second) {
          ok = false;
          break;
        }
      }
      if (!ok) break;
    }
    if (!ok) {
      cout << -1 << endl;
      continue;
    } else {
      cout << ans << endl;
      for (int i = 0; i < n; i++) {
        bool flip = colcnt[comp[i]].first < colcnt[comp[i]].second;
        if (flip ^ col[i]) {
          cout << i + 1 << " ";
        }
      }
      cout << endl;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  p1385G();
  return 0;
}
