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>
#pragma GCC optimize("O2")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int MAX = 2e5 + 5;
const long long MAX2 = 11;
const long long MOD = 1000000007;
const long long MOD2 = 1000005329;
const long long INF = 2e18;
const int dr[] = {1, 0, -1, 0, 1, 1, -1, -1, 0};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1, 0};
const double pi = acos(-1);
const double EPS = 1e-9;
const int block = 555;
int tc, n, x[MAX], y[MAX], nx, nw, hit;
bool vis[MAX], vs[MAX], st[MAX], ava;
vector<int> v[MAX], ans, lp;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> tc;
  for (int ntc = 1; ntc <= tc; ++ntc) {
    cin >> n;
    for (int i = 1; i <= n; ++i) v[i].clear(), vis[i] = vs[i] = st[i] = 0;
    for (int i = 1; i <= n; ++i) cin >> x[i], v[x[i]].push_back(i);
    for (int i = 1; i <= n; ++i) cin >> y[i], v[y[i]].push_back(i);
    ava = 1;
    for (int i = 1; i <= n; ++i)
      if (v[i].size() != 2) ava = 0;
    if (!ava) {
      cout << "-1\n";
      continue;
    }
    ans.clear();
    for (int i = 1; i <= n; ++i)
      if (!vis[i]) {
        lp.clear(), nw = i, hit = 0;
        while (!vis[nw]) {
          vis[nw] = 1;
          lp.push_back(nw);
          if (vs[x[nw]]) {
            st[nw] = vs[y[nw]] = 1, ++hit;
            nw = nw == v[y[nw]][0] ? v[y[nw]][1] : v[y[nw]][0];
          } else
            vs[x[nw]] = 1, nw = nw == v[x[nw]][0] ? v[x[nw]][1] : v[x[nw]][0];
        }
        if (hit > lp.size() - hit) {
          for (int j : lp)
            if (!st[j]) ans.push_back(j);
        } else {
          for (int j : lp)
            if (st[j]) ans.push_back(j);
        }
      }
    cout << ans.size() << '\n';
    for (int i : ans) cout << i << ' ';
    cout << '\n';
  }
  return 0;
}
