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;
template <class T>
using v = vector<T>;
template <class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  string sep;
  for (const auto &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
template <class T, class U>
ostream &operator<<(ostream &os, const set<T, U> &v) {
  os << "{";
  string sep;
  for (const auto &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
template <class A, class B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
long long inf;
v<long long> f, iv, primes, spf;
v<bool> isp;
v<pair<int, int> > dir = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
long long exp(long long a, long long b, long long inf) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % inf;
    a = (a * a) % inf;
    b >>= 1;
  }
  return res;
}
long long inv(long long n) { return exp(n, inf - 2, inf); }
void facts(long long N) {
  f.resize(N + 1);
  iv.resize(N + 1);
  f[0] = iv[0] = 1;
  for (long long i = 1; i <= N; i++) {
    f[i] = (f[i - 1] * i) % inf;
    iv[i] = exp(f[i], inf - 2, inf);
  }
}
long long ncr(long long n, long long r) {
  return (((f[n] * iv[r]) % inf) * iv[n - r]) % inf;
}
void seive(long long N) {
  isp.assign(N + 1, 1);
  spf.resize(N + 1);
  isp[0] = isp[1] = 0;
  spf[0] = 2, spf[1] = 0;
  for (long long i = 2; i <= N; i++) {
    if (isp[i]) {
      primes.push_back(i);
      spf[i] = i;
    }
    for (long long j = 0; j < (long long)primes.size() && i * primes[j] <= N &&
                          primes[j] <= spf[i];
         j++) {
      isp[i * primes[j]] = 0;
      spf[i * primes[j]] = primes[j];
    }
  }
}
void init();
void solve();
int main() {
  init();
  int q = 1;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    solve();
  }
  return 0;
}
void init() {
  inf = (long long)1e9 + 7;
  return;
}
v<v<pair<int, int> > > e;
v<int> vis, v1, v0;
void dfs(int u) {
  for (auto i : e[u]) {
    int v = i.first, s = i.second;
    if (vis[v]) {
      continue;
    } else if (s == 1) {
      if (vis[u] == 1) {
        vis[v] = 1;
        v0.push_back(v);
      } else {
        vis[v] = 2;
        v1.push_back(v);
      }
    } else {
      if (vis[u] == 1) {
        vis[v] = 2;
        v1.push_back(v);
      } else {
        vis[v] = 1;
        v0.push_back(v);
      }
    }
    dfs(v);
  }
}
void solve() {
  int n;
  cin >> n;
  v<v<int> > a(2, v<int>(n));
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < n; j++) {
      cin >> a[i][j];
      a[i][j]--;
    }
  }
  v<int> cnt(n, 0);
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < n; j++) {
      cnt[a[i][j]]++;
    }
  }
  for (int i = 0; i < n; i++) {
    if (cnt[i] != 2) {
      cout << -1 << endl;
      return;
    }
  }
  v<v<int> > pos(n);
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < n; j++) {
      pos[a[i][j]].push_back(n * i + j);
    }
  }
  e.assign(n, v<pair<int, int> >(0));
  vis.assign(n, 0);
  for (int i = 0; i < n; i++) {
    if (pos[i][1] - pos[i][0] == n) {
      continue;
    } else if (pos[i][0] >= n || pos[i][1] < n) {
      int c1 = pos[i][0] % n, c2 = pos[i][1] % n;
      e[c1].push_back({c2, 0});
      e[c2].push_back({c1, 0});
    } else {
      int c1 = pos[i][0] % n, c2 = pos[i][1] % n;
      e[c1].push_back({c2, 1});
      e[c2].push_back({c1, 1});
    }
  }
  v<int> ans;
  for (int i = 0; i < n; i++) {
    if (vis[i] == 0) {
      v1.clear();
      v0.clear();
      v0.push_back(i);
      vis[i] = 1;
      dfs(i);
      if ((long long)v1.size() <= (long long)v0.size()) {
        swap(v0, v1);
      }
      for (int u : v0) {
        ans.push_back(u);
      }
    }
  }
  cout << ans.size() << endl;
  sort(ans.begin(), ans.end());
  for (int i : ans) {
    cout << i + 1 << " ";
  }
  cout << endl;
  return;
}
