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;
using namespace std;
const int maxn = 200000 + 10;
int col[2][maxn];
int row[2][maxn];
vector<pair<int, int> > G[maxn];
int p[2][maxn];
int co[maxn];
int vis[maxn];
stack<int> S;
int n;
int dfs(int u, int c) {
  co[u] = c;
  S.push(u);
  for (auto e : G[u]) {
    int nc = c;
    if (e.second) nc = c ^ 1;
    if (co[e.first] == -1) {
      if (dfs(e.first, nc) == maxn) return maxn;
    } else {
      if (co[e.first] != nc) return maxn;
    }
  }
  return 0;
}
inline void solve() {
  map<int, int> ff;
  for (auto i = (0); i < (2); ++i) {
    for (auto j = (1); j <= (n); ++j) col[i][j] = -1, row[i][j] = -1;
  }
  for (auto i = (0); i < (2); ++i) {
    for (auto j = (1); j <= (n); ++j) {
      if (ff.count(p[i][j]) == 0) ff[p[i][j]] = 0;
      if (++ff[p[i][j]] == 3) {
        puts("-1");
        return;
      }
      if (col[0][p[i][j]] == -1) {
        col[0][p[i][j]] = j;
        row[0][p[i][j]] = i;
      } else {
        col[1][p[i][j]] = j;
        row[1][p[i][j]] = i;
      }
    }
  }
  for (auto i = (1); i <= (n); ++i) {
    co[i] = -1;
    if (row[0][i] == row[1][i]) {
      G[col[0][i]].push_back({col[1][i], 1});
      G[col[1][i]].push_back({col[0][i], 1});
    } else {
      G[col[0][i]].push_back({col[1][i], 0});
      G[col[1][i]].push_back({col[0][i], 0});
    }
  }
  int ans = 0;
  for (; !S.empty();) S.pop();
  vector<int> ops;
  for (auto i = (1); i <= (n); ++i) {
    if (vis[i]) continue;
    pair<int, vector<int> > tmp2, tmp;
    tmp.first = dfs(i, 0);
    for (; !S.empty(); S.pop()) {
      if (co[S.top()] == 1) tmp.second.push_back(S.top());
      tmp.first += co[S.top()], co[S.top()] = -1;
    }
    tmp2.first = dfs(i, 1);
    for (; !S.empty(); S.pop()) {
      if (co[S.top()] == 1) tmp2.second.push_back(S.top());
      tmp2.first += co[S.top()], vis[S.top()] = 1;
    }
    int cc = min(tmp.first, tmp2.first);
    if (cc >= maxn) {
      puts("-1");
      return;
    }
    if (tmp.first == cc) {
      for (auto e : tmp.second) ops.push_back(e);
    } else {
      for (auto e : tmp2.second) ops.push_back(e);
    }
    ans += cc;
  }
  printf("%d\n", ans);
  for (auto e : ops) printf("%d ", e);
  puts("");
}
int main(void) {
  int T;
  scanf("%d", &(T));
  for (; T--;) {
    scanf("%d", &(n));
    for (auto i = (1); i <= (n); ++i) G[i].clear(), vis[i] = 0;
    for (auto i = (0); i < (2); ++i)
      for (auto j = (1); j <= (n); ++j) scanf("%d", &(p[i][j]));
    solve();
  }
  return 0;
}
