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;
class Solution {
 public:
  int solve(vector<vector<int>>& A, vector<int>& ans) {
    int n = A.size();
    vector<int> C(n, 0);
    for (auto& V : A) {
      for (auto x : V) {
        ++C[x];
      }
    }
    for (auto x : C) {
      if (x != 2) {
        return -1;
      }
    }
    vector<vector<int>> pos(n);
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < 2; ++j) {
        pos[A[i][j]].push_back(i);
      }
    }
    vector<vector<pair<int, int>>> graph(n);
    for (int i = 0; i < n; ++i) {
      int a = pos[i][0], b = pos[i][1];
      if (a == b) {
        continue;
      }
      if ((A[a][0] == i && A[b][0] == i) || (A[a][1] == i && A[b][1] == i)) {
        graph[a].emplace_back(b, 1);
        graph[b].emplace_back(a, 1);
      } else {
        graph[a].emplace_back(b, 0);
        graph[b].emplace_back(a, 0);
      }
    }
    vector<int> cnts(2, 0);
    vector<vector<int>> comps(2, vector<int>(n, -1));
    vector<int> colors(n, -1);
    vector<int> que(n, -1);
    auto bfs = [&](int s, int clr = 0) {
      colors[s] = clr;
      int sz = 0;
      que[sz++] = s;
      cnts[0] = cnts[1] = 0;
      for (int i = 0; i < sz; ++i) {
        int u = que[i];
        int c = colors[u];
        comps[c][cnts[c]++] = u;
        for (auto& P : graph[u]) {
          int v = P.first, d = P.second;
          int nc = c ^ d;
          if (colors[v] != -1 && colors[v] != nc) {
            return false;
          }
          if (colors[v] == -1) {
            colors[v] = nc;
            que[sz++] = v;
          }
        }
      }
      return true;
    };
    int res = 0;
    for (int i = 0; i < n; ++i) {
      if (colors[i] == -1) {
        if (!bfs(i)) {
          return -1;
        }
        int s = 0;
        if (cnts[0] < cnts[1]) {
          res += cnts[0];
          s = 0;
        } else {
          res += cnts[1];
          s = 1;
        }
        for (int j = 0; j < cnts[s]; ++j) {
          ans.push_back(comps[s][j]);
        }
      }
    }
    return res;
  }
};
int main(int argc, char** argv) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  Solution sol;
  int t;
  cin >> t;
  while (t-- > 0) {
    int n;
    cin >> n;
    vector<vector<int>> A(n, vector<int>(2, 0));
    for (int i = 0; i < 2; ++i) {
      for (int j = 0; j < n; ++j) {
        cin >> A[j][i];
        --A[j][i];
      }
    }
    vector<int> ans;
    auto res = sol.solve(A, ans);
    if (res == ans.size()) {
      cout << res << '\n';
      for (int i = 0; i < res; ++i) {
        cout << ans[i] + 1 << (i == res - 1 ? '\n' : ' ');
      }
    } else {
      cout << -1 << '\n';
    }
  }
  return 0;
}
