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;
const int maxn = 200005;
int n;
vector<pair<int, int> > poz[maxn];
vector<pair<int, int> > g[maxn];
vector<int> rez;
bool sranje = 0;
int res[maxn];
int boja[maxn];
int komp[maxn];
int broj[maxn][2];
int br = 0;
bool visited[maxn];
void dfs(int v, int color) {
  komp[v] = br;
  broj[br][color]++;
  boja[v] = color;
  visited[v] = 1;
  for (auto c : g[v]) {
    int u = c.first;
    int w = c.second;
    int tren = w ^ color;
    if (visited[u]) {
      if (boja[u] != tren) sranje = 1;
    } else
      dfs(u, tren);
  }
}
void Solve() {
  cin >> n;
  rez.clear();
  sranje = 0;
  br = 0;
  for (int i = 1; i <= n; i++) {
    komp[i] = 0;
    broj[i][0] = broj[i][1] = 0;
    boja[i] = 0;
    visited[i] = 0;
    g[i].clear();
    poz[i].clear();
  }
  for (int j = 0; j < 2; j++) {
    for (int i = 1; i <= n; i++) {
      int x;
      cin >> x;
      poz[x].push_back({i, j});
    }
  }
  for (int i = 1; i <= n; i++) {
    if (poz[i].size() != 2) {
      cout << -1 << endl;
      return;
    }
    auto x = poz[i][0];
    auto y = poz[i][1];
    if (x.first == y.first) continue;
    if (x.second == y.second) {
      g[x.first].push_back({y.first, 1});
      g[y.first].push_back({x.first, 1});
    } else {
      g[x.first].push_back({y.first, 0});
      g[y.first].push_back({x.first, 0});
    }
  }
  for (int i = 1; i <= n; i++) {
    if (visited[i]) continue;
    br++;
    dfs(i, 0);
  }
  if (sranje) {
    cout << -1 << endl;
    return;
  }
  for (int i = 1; i <= br; i++) res[i] = (broj[i][0] > broj[i][1] ? 1 : 0);
  for (int i = 1; i <= n; i++) {
    if (boja[i] == res[komp[i]]) rez.push_back(i);
  }
  cout << rez.size() << endl;
  for (auto c : rez) cout << c << " ";
  cout << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t = 1;
  cin >> t;
  while (t--) Solve();
  return 0;
}
