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 = 3e5 + 5;
const int mod = 998244353;
const int INF = 0x3f3f3f3f;
const long long llINF = 0x3f3f3f3f3f3f3f3f;
long long make_compiler_happy() { return INF & maxn & mod & llINF; }
long long fpow(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long inv(long long x) { return fpow(x, mod - 2); }
int n;
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int T;
  cin >> T;
  for (int kase = (1); kase <= (T); kase++) {
    cin >> n;
    vector<int> a(n), b(n);
    vector<vector<int> > e(n);
    for (int i = 0; i < (n); i++) {
      cin >> a[i];
      a[i]--;
      e[a[i]].push_back(i);
    }
    for (int i = 0; i < (n); i++) {
      cin >> b[i];
      b[i]--;
      e[b[i]].push_back(i);
    }
    int ok = 1;
    for (int i = 0; i < (n); i++) {
      if (((int)e[i].size()) != 2) ok = 0;
    }
    if (!ok) {
      cout << "-1\n";
      continue;
    }
    vector<int> ans, vis(n);
    for (int i = 0; i < (n); i++) {
      if (vis[i] || e[i][0] == e[i][1]) continue;
      vector<int> x, y;
      for (int j = i, k = e[i][0]; !vis[j];) {
        vis[j] = 1;
        k ^= e[j][0] ^ e[j][1];
        if (a[k] == j) {
          x.push_back(k);
          j = b[k];
        } else {
          y.push_back(k);
          j = a[k];
        }
      }
      if (((int)x.size()) < ((int)y.size()))
        ans.insert(ans.end(), (x).begin(), (x).end());
      else
        ans.insert(ans.end(), (y).begin(), (y).end());
    }
    cout << ((int)ans.size()) << '\n';
    for (int i : ans) cout << i + 1 << ' ';
    cout << '\n';
  }
}
