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 Hash[(int)(2e5 + 10) << 1], pos1[(int)(2e5 + 10) << 1],
    pos2[(int)(2e5 + 10) << 1];
int res[(int)(2e5 + 10) << 1];
int p[(int)(2e5 + 10) << 1];
bool mark[(int)(2e5 + 10) << 1];
int n;
void init() {
  memset(Hash, 0, (n * 2 + 5) * sizeof(int));
  memset(pos1, 0, (n * 2 + 5) * sizeof(int));
  memset(pos2, 0, (n * 2 + 5) * sizeof(int));
  memset(mark, false, (n * 2 + 5) * sizeof(int));
  memset(res, 0, (n * 2 + 5) * sizeof(int));
  for (int i = 1; i <= n * 2 + 5; i++) p[i] = i;
}
int find(int x) { return p[x] == x ? x : p[x] = find(p[x]); }
void unite(int x, int y) {
  x = find(x), y = find(y);
  if (x == y) return;
  p[x] = y;
}
int main() {
  ios::sync_with_stdio(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> n;
    init();
    for (int i = 1; i <= n; i++) {
      int a;
      cin >> a;
      Hash[a]++;
      if (!pos1[a])
        pos1[a] = 1, pos2[a] = i;
      else
        unite(pos2[a], i + n), unite(pos2[a] + n, i);
    }
    for (int i = 1; i <= n; i++) {
      int tmp;
      cin >> tmp;
      Hash[tmp]++;
      if (!pos1[tmp])
        pos1[tmp] = 2, pos2[tmp] = i;
      else {
        if (pos1[tmp] == 2)
          unite(pos2[tmp], i + n), unite(pos2[tmp] + n, i);
        else if (pos1[tmp] == 1) {
          if (pos2[tmp] != i)
            unite(pos2[tmp], i), unite(pos2[tmp] + n, i + n);
          else
            p[i] = -1, p[i + n] = -1;
        }
        pos1[tmp] = -1;
      }
    }
    bool flag = true;
    for (int i = 1; i <= n; i++)
      if (Hash[i] != 2) {
        flag = false;
        break;
      }
    if (!flag) {
      cout << "-1" << endl;
      continue;
    }
    for (int i = 1; i <= 2 * n; i++)
      if (p[i] != -1) p[i] = find(p[i]);
    for (int i = 1; i <= n; i++)
      if (p[i] != -1) res[p[i]]++;
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      if (p[i] == -1) continue;
      if (res[p[i]] > 0 && res[p[i + n]] >= 0) {
        if (res[p[i]] >= res[p[i + n]])
          res[p[i]] = -1;
        else
          res[p[i + n]] = -1;
      }
      if (res[p[i]] > 0) mark[i] = 1, ans++;
    }
    cout << ans << endl;
    for (int i = 1; i <= n; i++)
      if (mark[i]) cout << i << " ";
    cout << endl;
  }
  return 0;
}
