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 = 2e5 + 5;
int A[maxn][2];
bool vis[maxn];
vector<int> v[maxn];
int main() {
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      v[i].clear();
    }
    for (int i = 1; i <= n; i++) {
      scanf("%d", &A[i][0]);
      v[A[i][0]].push_back(i);
    }
    for (int i = 1; i <= n; i++) {
      scanf("%d", &A[i][1]);
      v[A[i][1]].push_back(i);
    }
    int bad = 0;
    for (int i = 1; i <= n; i++) {
      vis[i] = 0;
      if (v[i].size() != 2) {
        bad = 1;
        break;
      }
    }
    if (bad) {
      printf("-1\n");
      continue;
    }
    vector<int> ans;
    ans.clear();
    for (int i = 1; i <= n; i++) {
      if (A[i][0] == A[i][1]) {
        vis[i] = 1;
        continue;
      }
      if (vis[i]) {
        continue;
      }
      int now = i;
      vector<int> v1, v2;
      int st = now;
      v1.push_back(now);
      while (!vis[now]) {
        vis[now] = 1;
        int b = A[now][1];
        int nex;
        if (v[b][0] == now)
          nex = v[b][1];
        else
          nex = v[b][0];
        now = nex;
        if (now == st) break;
        if (A[now][0] == b) {
          v1.push_back(now);
        } else {
          v2.push_back(now);
          swap(A[now][0], A[now][1]);
        }
      }
      if (v1.size() < v2.size()) {
        for (auto j : v1) {
          ans.push_back(j);
        }
      } else {
        for (auto j : v2) {
          ans.push_back(j);
        }
      }
    }
    printf("%d\n", ans.size());
    for (auto i : ans) {
      printf("%d ", i);
    }
    puts("");
  }
  return 0;
}
