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 T;
int n;
int a[200005];
int b[200005];
pair<int, int> p[200005];
bool f[200005];
set<int> v;
void solve() {
  scanf("%d", &n);
  v.clear();
  for (int i = 1; i <= n; i++) {
    f[i] = 0;
    p[i].first = 0;
    p[i].second = 0;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", b + i);
  }
  for (int i = 1; i <= n; i++) {
    if (p[a[i]].first == 0) {
      p[a[i]].first = i;
    } else if (p[a[i]].second == 0) {
      p[a[i]].second = i;
    } else {
      printf("-1\n");
      return;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (p[b[i]].first == 0) {
      p[b[i]].first = i;
    } else if (p[b[i]].second == 0) {
      p[b[i]].second = i;
    } else {
      printf("-1\n");
      return;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (p[i].first == 0 || p[i].second == 0) {
      printf("-1\n");
      return;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (f[i]) {
      continue;
    }
    if (p[i].first == p[i].second) {
      continue;
    }
    set<int> s, e;
    int t = i;
    int x = p[t].first;
    while (!f[t]) {
      f[t] = 1;
      if (x == p[t].first) {
        x = p[t].second;
      } else {
        x = p[t].first;
      }
      if (a[x] == t) {
        s.insert(x);
        t = b[x];
      } else {
        e.insert(x);
        t = a[x];
      }
    }
    if (s.empty() || e.empty()) {
      continue;
    } else if (s.size() < e.size()) {
      v.insert(s.begin(), s.end());
    } else {
      v.insert(e.begin(), e.end());
    }
  }
  if (v.empty()) {
    printf("0\n\n");
    return;
  }
  printf("%d\n", v.size());
  for (auto it : v) {
    printf("%d ", it);
  }
  printf("\n");
  return;
}
int main() {
  scanf("%d", &T);
  while (T--) {
    solve();
  }
  return 0;
}
