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 ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template <typename T>
using vec = vector<T>;
template <typename T>
using Prior = priority_queue<T>;
template <typename T>
using prior = priority_queue<T, vec<T>, greater<T>>;
const int INF = 1e9;
const ll LLINF = (ll)4 * 1e18;
const ll MOD = 1e9 + 7;
const double PI = 3.14159265358;
const double EPS = 1e-8;
const int xx[8] = {0, 1, 0, -1, 1, 1, -1, -1};
const int yy[8] = {1, 0, -1, 0, 1, -1, -1, 1};
void pmod(ll &a, ll b) { a = (a + b) % MOD; }
void mmod(ll &a, ll b) { a = (a - b + MOD) % MOD; }
void tmod(ll &a, ll b) { a = (a * b) % MOD; }
ll POW(ll a, ll b) {
  ll res = 1;
  do {
    if (b % 2) tmod(res, a);
    tmod(a, a);
  } while (b >>= 1);
  return res;
}
template <typename T>
void amax(T &a, T b) {
  if (a < b) a = b;
}
template <typename T>
void amin(T &a, T b) {
  if (a > b) a = b;
}
void solve() {
  int n, ipt;
  bool flag = true;
  cin >> n;
  vec<int> v(2 * n + 10), mem(n + 1);
  for (int i = 1; i <= (n); i++) {
    cin >> ipt;
    if (mem[ipt] == INF) {
      flag = false;
    } else if (mem[ipt]) {
      v[i] = mem[ipt];
      v[mem[ipt]] = i;
      mem[ipt] = INF;
    } else {
      mem[ipt] = i;
    }
  }
  for (int i = 1; i <= (n); i++) {
    cin >> ipt;
    if (mem[ipt] == INF) {
      flag = false;
    } else if (mem[ipt]) {
      v[i + n] = mem[ipt];
      v[mem[ipt]] = i + n;
      mem[ipt] = INF;
    } else {
      mem[ipt] = i + n;
    }
  }
  if (flag == false) {
    cout << -1 << '\n';
    return;
  }
  vec<bool> vis(n + 1), ans(n + 1);
  vec<int> stk;
  int sum, cnt, cur, res = 0;
  bool mov;
  for (int i = 1; i <= (n); i++)
    if (!vis[i]) {
      mov = true;
      cur = i;
      sum = cnt = 0;
      while (true) {
        int abc = (cur > n ? cur - n : cur);
        sum++;
        if (mov) {
          cnt++;
          ans[abc] = true;
        }
        vis[abc] = true;
        stk.emplace_back(abc);
        if (v[cur] > n && vis[v[cur] - n]) break;
        if (v[cur] <= n && vis[v[cur]]) break;
        if ((cur <= n) == (v[cur] <= n)) mov = !mov;
        if (v[cur] > n)
          cur = v[cur] - n;
        else
          cur = v[cur] + n;
      }
      if (sum - cnt < cnt)
        for (auto j : stk) ans[j] = !ans[j];
      res += min(sum - cnt, cnt);
      stk.clear();
    }
  cout << res << '\n';
  for (int i = 1; i <= (n); i++)
    if (ans[i]) cout << i << ' ';
  cout << '\n';
}
int main() {
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
}
