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;
inline void pisz(int n) { printf("%d\n", n); }
const int fx[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
const int fxx[9][2] = {{0, 0}, {0, 1},  {0, -1}, {1, 0},  {-1, 0},
                       {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
int a[400005], flag[400005], proc[400005];
vector<int> adj[400005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, neg = 0, pos = 0, ans = 0, err = 0;
    vector<int> output, plus, minus;
    cin >> n;
    for (int i = 0; i < 2 * n; i++) {
      cin >> a[i];
      flag[i + 1] = 0;
    }
    for (int i = 0; i < 2 * n; i++) {
      flag[a[i]]++;
    }
    for (int i = 0; i < n; i++) {
      if (flag[i + 1] != 2) {
        err = 1;
        break;
      }
      flag[i + 1] = -1;
    }
    if (err) {
      cout << -1 << "\n";
      continue;
    }
    for (int i = 0; i < n; i++) {
      proc[i] = 0;
      adj[i].clear();
      if (flag[a[i]] == -1 && flag[a[n + i]] == -1) {
        flag[a[i]] = i;
        flag[a[n + i]] = i;
        if (a[n + i] == a[i]) {
          proc[i] = 1;
          flag[a[i]] = -2;
          flag[a[i + n]] = -2;
        }
      } else {
        if (flag[a[i]] != -1) {
          adj[i].push_back(flag[a[i]]);
          adj[flag[a[i]]].push_back(i);
          flag[a[i]] = -2;
        } else {
          flag[a[i]] = i;
        }
        if (flag[a[i + n]] != -1) {
          adj[i].push_back(flag[a[i + n]]);
          adj[flag[a[i + n]]].push_back(i);
          flag[a[i + n]] = -2;
        } else {
          flag[a[i + n]] = i;
        }
      }
    }
    for (int i = 0; i < n; i++) {
      if (proc[i]) continue;
      int current = i;
      proc[i] = 1;
      int nxt = adj[i][0];
      pos = 1;
      neg = 0;
      int parity = 0;
      plus.clear();
      plus.push_back(current);
      minus.clear();
      while (nxt != i) {
        proc[nxt] = 1;
        if (a[current] == a[nxt] || a[current + n] == a[nxt + n]) {
          if (parity) {
            pos++;
          } else
            neg++;
          parity = !parity;
        } else {
          if (parity) {
            neg++;
          } else
            pos++;
        }
        if (parity) {
          minus.push_back(nxt);
        } else
          plus.push_back(nxt);
        if (adj[nxt][0] == current) {
          current = nxt;
          nxt = adj[nxt][1];
        } else if (adj[nxt][1] == current) {
          current = nxt;
          nxt = adj[nxt][0];
        }
      }
      ans += (pos > neg ? neg : pos);
      if (neg <= pos) {
        if (!minus.empty())
          output.insert(output.end(), minus.begin(), minus.end());
      } else {
        if (!plus.empty())
          output.insert(output.end(), plus.begin(), plus.end());
      }
    }
    cout << ans << "\n";
    for (auto x : output) {
      cout << x + 1 << " ";
    }
    cout << "\n";
  }
  return 0;
}
