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;
int n, ct[400020], a[400020], b[400020], fa[400020], vis[400020];
vector<pair<int, int> > loc[400020];
unordered_map<int, vector<int> > mp;
vector<int> ans;
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
inline void solve() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) loc[i].clear(), ct[i] = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]), ++ct[a[i]];
    loc[a[i]].push_back({1, i});
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &b[i]), ++ct[b[i]];
    loc[b[i]].push_back({2, i});
  }
  for (int i = 1; i <= n; ++i)
    if (ct[i] != 2) {
      puts("-1");
      return;
    }
  for (int i = 1; i <= n + n; ++i) fa[i] = i;
  for (int i = 1; i <= n; ++i) {
    auto a = loc[i][0], b = loc[i][1];
    if (a.first != b.first) {
      int p = find(a.second), q = find(b.second);
      if (p != q) fa[q] = p;
      p = find(a.second + n), q = find(b.second + n);
      if (p != q) fa[q] = p;
    } else {
      int p = find(a.second), q = find(b.second + n);
      if (p != q) fa[q] = p;
      p = find(a.second + n), q = find(b.second);
      if (p != q) fa[q] = p;
    }
  }
  mp.clear();
  for (int i = 1; i <= n; ++i)
    if (find(i) == find(i + n)) {
      puts("-1");
      return;
    } else {
      mp[find(i)].push_back(i);
      mp[find(i + n)].push_back(i + n);
    }
  ans.clear();
  for (int i = 1; i <= n; ++i) vis[i] = 0;
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) {
      int cl = 0, cr = 0;
      for (auto v : mp[find(i)])
        if (v > n) ++cl;
      for (auto v : mp[find(i + n)])
        if (v > n) ++cr;
      if (cl < cr) {
        for (auto v : mp[find(i)]) {
          if (v > n) ans.push_back(v - n);
          vis[v > n ? v - n : v] = 1;
        }
      } else {
        for (auto v : mp[find(i + n)]) {
          if (v > n) ans.push_back(v - n);
          vis[v > n ? v - n : v] = 1;
        }
      }
    }
  printf("%d\n", ans.size());
  for (auto v : ans) printf("%d ", v);
  puts("");
}
int main() {
  int T = 1;
  scanf("%d", &T);
  while (T--) solve();
}
