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>;
using ld = long double;
const int mod = 1e9 + 7;
const ll inf = 1e9;
const ll INF = 2e18;
const int N = 2e6 + 66;
const int M = 5005;
vector<pii> pos[N], g[N];
int col[N], cnt0, cnt1, ans[N], a[2][N];
bool ok;
void dfs(int v, int c = 0) {
  if (c)
    cnt1++;
  else
    cnt0++;
  col[v] = c;
  for (auto [to, cost] : g[v]) {
    if (col[to] == -1) dfs(to, c ^ cost);
    if ((col[to] ^ col[v]) != cost) {
      ok = false;
    }
  }
}
void calc(int v, int c = 0) {
  if (col[v] ^ c)
    ans[v] = 1;
  else
    ans[v] = 0;
  for (auto [to, cost] : g[v]) {
    if (ans[to] == -1) calc(to, c);
  }
}
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    pos[i].clear();
    g[i].clear();
  }
  for (int j = 0; j <= 1; ++j) {
    for (int i = 1; i <= n; ++i) {
      cin >> a[j][i];
      pos[a[j][i]].push_back({j, i});
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (pos[i].size() != 2) {
      return void(cout << -1);
    }
  }
  for (int i = 1; i <= n; ++i) {
    int v = pos[i][0].second, u = pos[i][1].second;
    int cost = (pos[i][0].first == pos[i][1].first);
    if (v == u) continue;
    g[v].push_back({u, cost});
    g[u].push_back({v, cost});
  }
  ok = true;
  fill(col + 1, col + 1 + n, -1);
  fill(ans + 1, ans + 1 + n, -1);
  for (int i = 1; i <= n; ++i) {
    if (col[i] == -1) {
      cnt0 = cnt1 = 0;
      dfs(i);
      calc(i, cnt0 < cnt1);
    }
  }
  if (!ok) {
    return void(cout << -1);
  }
  int cnt = 0;
  for (int i = 1; i <= n; ++i) {
    assert(ans[i] != -1);
    cnt += ans[i];
  }
  cout << cnt << "\n";
  for (int i = 1; i <= n; ++i) {
    if (ans[i]) cout << i << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  int tt = 1;
  cin >> tt;
  while (tt--) {
    solve();
    cout << "\n";
  }
}
