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 a[200005], b[200005], cnt[200005];
vector<int> pos[200005], co[2];
vector<pair<int, int> > g[200005];
int color[200005];
bool dfs(int u, int c2) {
  color[u] = c2;
  co[c2].emplace_back(u);
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i].first, c = g[u][i].second;
    if (color[v] == -1) {
      if (c == 0) {
        if (!dfs(v, color[u])) return 0;
      } else {
        if (!dfs(v, 1 - color[u])) return 0;
      }
    } else {
      if (c == 0 && color[v] != color[u])
        return 0;
      else if (c == 1 && color[v] == color[u])
        return 0;
    }
  }
  return 1;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    vector<int> ans;
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      cnt[i] = 0;
      pos[i].clear();
      color[i] = -1;
      g[i].clear();
    }
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      cnt[a[i]]++;
      pos[a[i]].emplace_back(i);
    }
    for (int i = 1; i <= n; i++) {
      scanf("%d", &b[i]);
      cnt[b[i]]++;
      pos[b[i]].emplace_back(i);
    }
    bool ok = 1;
    for (int i = 1; i <= n; i++) {
      if (cnt[i] != 2) {
        ok = 0;
        break;
      }
    }
    if (!ok) {
      printf("-1\n");
      continue;
    }
    for (int i = 1; i <= n; i++) {
      if (pos[i][0] == pos[i][1]) continue;
      if (a[pos[i][0]] == a[pos[i][1]] || b[pos[i][0]] == b[pos[i][1]]) {
        g[pos[i][0]].emplace_back(make_pair(pos[i][1], 1));
        g[pos[i][1]].emplace_back(make_pair(pos[i][0], 1));
      } else {
        g[pos[i][0]].emplace_back(make_pair(pos[i][1], 0));
        g[pos[i][1]].emplace_back(make_pair(pos[i][0], 0));
      }
    }
    for (int i = 1; i <= n; i++) {
      if (color[i] != -1) continue;
      co[0].clear();
      co[1].clear();
      if (!dfs(i, 0)) {
        ok = 0;
        break;
      }
      if (co[0].size() < co[1].size()) {
        for (auto u : co[0]) ans.emplace_back(u);
      } else {
        for (auto u : co[1]) ans.emplace_back(u);
      }
    }
    if (!ok) {
      printf("-1\n");
      continue;
    }
    printf("%d\n", ans.size());
    for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
    printf("\n");
  }
}
