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;
long long MOD = 998244353;
const long long INF = 1e18;
const double E = 1e-18;
const double PI = acos(-1);
mt19937_64 rnd(time(0));
long long p1 = 31, p2 = 37, p3 = 7, p4 = rnd() % 100 + 1, md1 = MOD,
          md2 = 998244353, md3 = 998244357, md4 = rnd() % 1000000000;
long long n, m, T, k, kl[200001], a[2][200001];
long long ans = 0;
vector<long long> w;
vector<pair<long long, long long> > q;
vector<pair<int, int> > c[2][200001], v[200001];
bool tt[2][200001], nm[2][200001], tl[200001];
long long binpow(long long h, long long r, long long md = MOD) {
  long long l = 1;
  while (r) {
    if (r & 1) l *= h, l %= md;
    h *= h;
    h %= md;
    r /= 2;
  }
  return l;
}
long long gcd(long long x, long long y) {
  if (x < y) swap(x, y);
  while (x && y) {
    x %= y;
    swap(x, y);
  }
  return x + y;
}
long long dfs(int x, int h, bool k) {
  q.push_back(make_pair(x, h));
  tt[x][h] = 1;
  tl[h] = 1;
  nm[x][h] = k;
  long long kl = 0;
  if (x && k) kl++, w.push_back(h);
  for (int i = 0; i < c[x][h].size(); i++)
    if (tt[c[x][h][i].first][c[x][h][i].second]) {
      if (nm[x][h] == nm[c[x][h][i].first][c[x][h][i].second]) kl = 1e18;
    } else {
      long long o = dfs(c[x][h][i].first, c[x][h][i].second, !k);
      kl = min(kl + o, (long long)1e18);
    }
  return kl;
}
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    tl[i] = 0;
    v[i].clear();
    for (int j = 0; j < 2; j++) c[j][i].clear();
  }
  for (int i = 1; i <= n; i++) {
    cin >> a[0][i];
    v[a[0][i]].push_back(make_pair(0, i));
    c[0][i].push_back(make_pair(1, i));
  }
  for (int i = 1; i <= n; i++) {
    cin >> a[1][i];
    v[a[1][i]].push_back(make_pair(1, i));
    c[1][i].push_back(make_pair(0, i));
  }
  for (int i = 1; i <= n; i++)
    if (v[i].size() != 2) {
      cout << "-1\n";
      return;
    }
  for (int i = 1; i <= n; i++) {
    c[v[i][0].first][v[i][0].second].push_back(
        make_pair(v[i][1].first, v[i][1].second));
    c[v[i][1].first][v[i][1].second].push_back(
        make_pair(v[i][0].first, v[i][0].second));
  }
  vector<long long> ans;
  for (int i = 1; i <= n; i++)
    if (!tl[i]) {
      q.clear();
      w.clear();
      vector<long long> w1, w2;
      long long p0 = dfs(0, i, 1);
      w1 = w;
      for (int i = 0; i < q.size(); i++)
        tt[q[i].first][q[i].second] = 0, nm[q[i].first][q[i].second] = 0;
      q.clear();
      w.clear();
      long long p1 = dfs(1, i, 1);
      w2 = w;
      for (int i = 0; i < q.size(); i++)
        tt[q[i].first][q[i].second] = 0, nm[q[i].first][q[i].second] = 0;
      long long p = min(p0, p1);
      if (p >= 1e18) {
        cout << "-1\n";
        return;
      }
      if (p0 < p1)
        for (auto i : w1) ans.push_back(i);
      else
        for (auto i : w2) ans.push_back(i);
    }
  cout << ans.size() << "\n";
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
