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 n, r, Test_num;
bool ok;
int rt[400002], siz[400002];
bool vis[400002];
vector<pair<int, int> > vec[200002];
vector<int> ans;
inline void getroot(int x) {
  if (x == rt[x]) return;
  getroot(rt[x]), rt[x] = rt[rt[x]];
}
int main() {
  scanf("%d", &Test_num);
  while (Test_num--) {
    scanf("%d", &n), ok = 1;
    for (int i = 1; i <= n; ++i)
      vec[i].clear(), rt[i] = i, rt[i + n] = i + n, siz[i] = 1, siz[i + n] = 0,
                      vis[i] = 0;
    for (int i = 1, x; i <= n; ++i)
      scanf("%d", &x), vec[x].push_back(pair<int, int>(1, i));
    for (int i = 1, x; i <= n; ++i)
      scanf("%d", &x), vec[x].push_back(pair<int, int>(2, i));
    for (int i = 1; i <= n; ++i)
      if (vec[i].size() != 2) {
        ok = 0;
        break;
      }
    if (!ok) {
      puts("-1");
      continue;
    }
    for (int i = 1, a, b, c, d; i <= n; ++i) {
      a = vec[i][0].first, b = vec[i][0].second, c = vec[i][1].first,
      d = vec[i][1].second;
      if (b == d) continue;
      getroot(b), getroot(d), getroot(b + n), getroot(d + n);
      if (a == c) {
        if (rt[b] == rt[d]) {
          ok = 0;
          break;
        }
        if (rt[b] != rt[d + n])
          siz[rt[d + n]] += siz[rt[b]], siz[rt[d]] += siz[rt[b + n]],
              rt[rt[b]] = rt[d + n], rt[rt[b + n]] = rt[d];
      } else {
        if (rt[b] == rt[d + n]) {
          ok = 0;
          break;
        }
        if (rt[b] != rt[d])
          siz[rt[d]] += siz[rt[b]], siz[rt[d + n]] += siz[rt[b + n]],
              rt[rt[b]] = rt[d], rt[rt[b + n]] = rt[d + n];
      }
    }
    if (!ok) {
      puts("-1");
      continue;
    }
    for (int i = 1; i <= n; ++i) getroot(i), getroot(i + n);
    ans.clear(), r = 0;
    for (int i = 1; i <= n; ++i)
      if (!vis[i]) {
        getroot(i), getroot(i + n);
        if (rt[i] == i) {
          if (!r) {
            if (siz[rt[i]] > siz[rt[i + n]])
              r = rt[i + n];
            else
              r = rt[i];
          } else {
            if (siz[rt[i]] > siz[rt[i + n]])
              siz[r] += siz[rt[i + n]], rt[rt[i + n]] = r;
            else
              siz[r] += siz[rt[i]], rt[rt[i]] = r;
          }
        }
      }
    for (int i = 1; i <= n; ++i) {
      getroot(i);
      if (rt[i] == r) ans.push_back(i);
    }
    printf("%d\n", ans.size());
    for (int i = 0; i < ans.size(); ++i) printf("%d ", ans[i]);
    puts("");
  }
  return 0;
}
