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>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long int M = 1000000007;
long long int inv(long long int a, long long int b) {
  return 1 < a ? b - inv(b % a, a) * b / a : 1;
}
long long int gcd(long long int x, long long int y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
pair<long long int, long long int> neigh[4] = {
    {-1, 0}, {1, 0}, {0, 1}, {0, -1}};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long int _t;
  cin >> _t;
  for (long long int _t1 = 0; _t1 < (_t); _t1++) {
    long long int n;
    cin >> n;
    vector<long long int> a(n);
    for (long long int _i = 0; _i < (n); _i++) cin >> a[_i];
    vector<long long int> b(n);
    for (long long int _i = 0; _i < (n); _i++) cin >> b[_i];
    int visited[n];
    memset(visited, 0, sizeof(visited));
    for (long long int i = 0; i < (n); i++) {
      a[i]--;
      b[i]--;
      visited[a[i]]++;
      visited[b[i]]++;
    }
    bool bre = false;
    for (long long int i = 0; i < (n); i++) {
      if (visited[i] != 2) bre = true;
    }
    if (bre) {
      cout << -1 << endl;
      continue;
    }
    memset(visited, 0, sizeof(visited));
    map<long long int, long long int> pos;
    for (long long int i = 0; i < (n); i++) {
      if (visited[a[i]])
        a[i] = (a[i] << 1) + 1;
      else
        visited[a[i]] = 1, a[i] <<= 1;
      pos[a[i]] = i << 1;
      if (visited[b[i]])
        b[i] = (b[i] << 1) + 1;
      else
        visited[b[i]] = 1, b[i] <<= 1;
      pos[b[i]] = (i << 1) + 1;
    }
    memset(visited, 0, sizeof(visited));
    vector<long long int> tot;
    for (long long int i = 0; i < (n); i++) {
      if (!visited[i]) {
        visited[i] = true;
        if (b[i] == a[i]) continue;
        long long int top = b[i];
        long long int bot = a[i];
        vector<long long int> flip1;
        vector<long long int> flip2;
        flip2.push_back(i);
        while (true) {
          long long int tn = (top & 1) ? top - 1 : top + 1;
          long long int bn = (bot & 1) ? bot - 1 : bot + 1;
          long long int post = pos[tn];
          long long int posb = pos[bn];
          visited[post >> 1] = true;
          visited[posb >> 1] = true;
          if ((post >> 1) == (posb >> 1)) {
            if ((post >> 1) != i)
              if (post & 1)
                flip1.push_back(post >> 1);
              else
                flip2.push_back(post >> 1);
            break;
          }
          if (post & 1)
            flip1.push_back(post >> 1);
          else
            flip2.push_back(post >> 1);
          if (!(posb & 1))
            flip1.push_back(posb >> 1);
          else
            flip2.push_back(posb >> 1);
          top = (post & 1) ? a[post >> 1] : b[post >> 1];
          bot = (posb & 1) ? a[posb >> 1] : b[posb >> 1];
          if ((top >> 1) == (bot >> 1)) break;
        }
        if (flip1.size() > flip2.size())
          tot.insert(tot.end(), flip2.begin(), flip2.end());
        else
          tot.insert(tot.end(), flip1.begin(), flip1.end());
      }
    }
    cout << tot.size() << endl;
    for (long long int e : tot) cout << e + 1 << " ";
    cout << endl;
  }
}
