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 N = 400010;
int n;
int a[N], b[N];
int cnt[N];
int pos1[N], pos2[N];
int p[N], s[N];
bool st[N];
int find(int first) {
  if (p[first] != first) p[first] = find(p[first]);
  return p[first];
}
void merge(int a, int b) {
  a = find(a), b = find(b);
  if (a != b) {
    s[b] += s[a];
    p[a] = b;
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 1; i <= n + n; i++) p[i] = i, st[i] = pos1[i] = pos2[i] = 0;
    for (int i = 1; i <= n; i++) s[i] = 0, cnt[i] = 0;
    for (int i = n + 1; i <= n + n; i++) s[i] = 1;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      cnt[a[i]]++;
      if (!pos1[a[i]])
        pos1[a[i]] = i;
      else {
        merge(pos1[a[i]], i + n);
        merge(pos1[a[i]] + n, i);
      }
    }
    for (int i = 1; i <= n; i++) {
      cin >> b[i];
      cnt[b[i]]++;
      if (!pos2[b[i]]) {
        pos2[b[i]] = i;
        if (pos1[b[i]] && pos1[b[i]] != i) {
          merge(pos1[b[i]], i);
          merge(pos1[b[i]] + n, i + n);
        }
      } else {
        merge(pos2[b[i]], i + n);
        merge(pos2[b[i]] + n, i);
      }
    }
    bool success = true;
    for (int i = 1; i <= n; i++)
      if (find(i) == find(i + n) || cnt[i] != 2) {
        success = false;
        break;
      }
    if (!success) {
      cout << -1 << endl;
      continue;
    }
    vector<int> ans;
    for (int i = 1; i <= n; i++) {
      int a = find(i), b = find(i + n);
      if (st[b]) {
        ans.push_back(i);
        continue;
      }
      if (s[a] > s[b]) {
        ans.push_back(i);
        st[b] = true;
      } else
        st[a] = true;
    }
    cout << ans.size() << endl;
    for (auto c : ans) cout << c << ' ';
    cout << endl;
  }
  return 0;
}
