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 main() {
  int t;
  cin >> t;
  while (t--) {
    int n, k = 0;
    cin >> n;
    vector<int> a1(n), a2(n), count(n + 1, 0), l1(n), l2(n), m;
    vector<vector<int>> pos(n + 1);
    vector<bool> vis(n + 1, false);
    bool ok = true;
    for (int i = 0; i < n; i++) {
      cin >> a1[i];
      count[a1[i]]++;
      pos[a1[i]].push_back(i);
    }
    for (int i = 0; i < n; i++) {
      cin >> a2[i];
      count[a2[i]]++;
      pos[a2[i]].push_back(i);
    }
    for (int i = 1; i <= n; i++) {
      if (count[i] != 2) {
        ok = false;
        break;
      }
    }
    if (!ok) {
      cout << -1 << endl;
      continue;
    }
    for (int i = 1; i <= n; i++) {
      if (vis[i]) {
        continue;
      }
      int f, s, p = pos[i][0];
      int j = i;
      int k1 = 0, k2 = 0;
      vector<int> m1, m2, d;
      while (!vis[j]) {
        vis[j] = true;
        l1[p] = j;
        d.push_back(p);
        if (a1[p] == j) {
          j = a2[p];
          l2[p] = j;
        } else {
          j = a1[p];
          l2[p] = j;
        }
        f = pos[j][0];
        s = pos[j][1];
        if (p != f) {
          swap(f, s);
        }
        p = s;
      }
      for (int x : d) {
        if (l1[x] != a1[x]) {
          k1++;
          m1.push_back(x + 1);
        }
        if (l2[x] != a1[x]) {
          k2++;
          m2.push_back(x + 1);
        }
      }
      if (k1 < k2) {
        k += k1;
        for (int x : m1) {
          m.push_back(x);
        }
      } else {
        k += k2;
        for (int x : m2) {
          m.push_back(x);
        }
      }
    }
    cout << k << endl;
    for (int i = 0; i < k; i++) {
      cout << m[i] << ' ';
    }
    cout << endl;
  }
  return 0;
}
