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 MOD = 1e9 + 7;
const int INF = 1e9;
const long long LNF = 1e18;
void setIO(string s) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
const int N = 2e5 + 7;
int col[N];
vector<vector<pair<int, int> > > g;
vector<int> comp;
int cnt0, cnt1;
void dfs(int u, int s, int cmp) {
  col[u] = s;
  comp[u] = cmp;
  if (s == 0)
    ++cnt0;
  else
    ++cnt1;
  for (auto& i : g[u])
    if (col[i.first] == -1) {
      dfs(i.first, s ^ i.second, cmp);
    }
}
void solve() {
  int n;
  cin >> n;
  vector<vector<int> > a(2, vector<int>(n));
  vector<vector<int> > pos(n);
  for (int i = 0; i < 2; ++i) {
    for (int j = 0; j < n; ++j) {
      cin >> a[i][j];
      --a[i][j];
      pos[a[i][j]].emplace_back(j);
    }
  }
  g = vector<vector<pair<int, int> > >(n);
  for (int i = 0; i < n; ++i) {
    if ((int)pos[i].size() != 2) {
      cout << -1 << "\n\n";
      return;
    }
    int x = pos[i][0], y = pos[i][1];
    if (x == y) continue;
    int rx = a[0][x] != i, ry = a[0][y] != i;
    g[x].emplace_back(y, rx == ry);
    g[y].emplace_back(x, rx == ry);
  }
  memset(col, -1, sizeof col);
  comp = vector<int>(n);
  int cnt = 0, ans = 0;
  vector<pair<int, int> > best;
  for (int i = 0; i < n; ++i) {
    if (col[i] == -1) {
      cnt0 = cnt1 = 0;
      dfs(i, 0, cnt);
      ans += min(cnt0, cnt1);
      best.emplace_back(cnt0, cnt1);
      ++cnt;
    }
  }
  cout << ans << '\n';
  for (int i = 0; i < n; ++i) {
    int bestswitch = best[comp[i]].first > best[comp[i]].second;
    if (col[i] == bestswitch) cout << i + 1 << ' ';
  }
  cout << '\n';
}
int main() {
  setIO("input");
  int t;
  cin >> t;
  while (t--) solve();
}
