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;
int cnt[200005];
int pos[200005][2];
int node[200005][2];
int vis[200005];
bool change[200005];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int cases;
  cin >> cases;
  while (cases--) {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
      cnt[i] = 0;
      vis[i] = 0;
    }
    for (int j = 0; j < 2; j++)
      for (int i = 0; i < n; i++) {
        int x;
        cin >> x;
        x--;
        cnt[x]++;
        if (cnt[x] == 1)
          pos[x][0] = i;
        else
          pos[x][1] = i;
        node[i][j] = x;
      }
    bool ok = 1;
    for (int i = 0; i < n; i++) {
      if (cnt[i] != 2) {
        ok = 0;
        break;
      }
    }
    if (!ok) {
      cout << "-1\n";
      continue;
    }
    int ans = 0;
    for (int i = 0; i < n; i++) {
      if (vis[i]) continue;
      vis[i] = 1;
      change[i] = 0;
      if (node[i][0] == node[i][1]) continue;
      int val = node[i][0];
      int pr = i;
      int nex = pos[val][0];
      if (nex == pr) nex = pos[val][1];
      int cnt = 1, cnt1 = 0;
      vector<int> cycle = {i};
      while (!vis[nex]) {
        cycle.push_back(nex);
        vis[nex] = 1;
        cnt++;
        if (node[pr][0] == val && node[nex][0] == val ||
            node[pr][1] == val && node[nex][1] == val)
          change[nex] = !change[pr];
        else
          change[nex] = change[pr];
        cnt1 += change[nex];
        if (node[nex][0] == val)
          val = node[nex][1];
        else
          val = node[nex][0];
        pr = nex;
        nex = pos[val][0];
        if (nex == pr) nex = pos[val][1];
      }
      if (node[pr][0] == val && node[nex][0] == val ||
          node[pr][1] == val && node[nex][1] == val) {
        if (change[nex] == change[pr]) {
          ok = 0;
          break;
        }
      } else {
        if (change[nex] != change[pr]) {
          ok = 0;
          break;
        }
      }
      if (cnt - cnt1 < cnt1) {
        cnt1 = cnt - cnt1;
        for (int p : cycle) change[p] = !change[p];
      }
      ans += cnt1;
    }
    if (!ok) {
      cout << "-1\n";
      continue;
    }
    cout << ans << "\n";
    for (int i = 0; i < n; i++)
      if (change[i]) cout << i + 1 << " ";
    cout << "\n";
  }
  return 0;
}
