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;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return x * f;
}
int T, n, a[2][200020], cnt[200020], vis[200020];
pair<int, int> pos[2][200020];
vector<pair<int, int> > vec[200020];
int MYHMD(string s, vector<int> &p, vector<int> &myhmd) {
  int tot = 0;
  for (int i = 0; i < (int)s.length(); ++i) {
    if (s[i] == '0') {
      ++tot;
      myhmd.push_back(p[i]);
      if (i + 1 < (int)s.length()) s[i + 1] = ((s[i + 1] - '0') ^ 1) + '0';
    }
  }
  return tot;
}
void calc(string &s, vector<int> &p, vector<int> &tmp) {
  for (int i = 0; i < (int)s.length(); ++i) {
    if (s[i] == '1') {
      if (i & 1) {
        rotate(s.begin(), s.begin() + i, s.end());
        rotate(p.begin(), p.begin() + i, p.end());
        break;
      }
    }
  }
  vector<int> wzp, zyy;
  int ans1 = MYHMD(s, p, wzp);
  int gu = 0;
  for (int i = (int)s.length() - 1; i >= -1; --i) {
    if (i >= 0 && s[i] == '0')
      ++gu;
    else {
      if (gu & 1) {
        rotate(s.begin(), s.begin() + i, s.end());
        rotate(p.begin(), p.begin() + i, p.end());
        break;
      }
      gu = 0;
    }
  }
  int ans2 = MYHMD(s, p, zyy);
  if (ans1 <= ans2) {
    for (auto x : wzp) tmp.push_back(x);
  } else {
    for (auto x : zyy) tmp.push_back(x);
  }
}
void Solve() {
  n = read();
  for (int i = 1; i <= n; ++i) {
    cnt[i] = vis[i] = 0;
    vec[i].clear();
  }
  for (int i = 0; i < 2; ++i) {
    for (int j = 1; j <= n; ++j) {
      a[i][j] = read();
      ++cnt[a[i][j]];
      vec[a[i][j]].emplace_back(i, j);
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (cnt[i] ^ 2) {
      printf("-1\n");
      return;
    }
    pos[vec[i][0].first][vec[i][0].second] = vec[i][1];
    pos[vec[i][1].first][vec[i][1].second] = vec[i][0];
  }
  vector<int> ans;
  vector<int> p;
  for (int i = 1; i <= n; ++i) {
    if (vis[i]) continue;
    string qwq = "";
    int t = 0, k = i;
    p.clear();
    while ("MYH AK IOI!") {
      if (vis[k]) break;
      vis[k] = 1;
      auto nw = pos[t][k];
      int nt = nw.first, nk = nw.second;
      qwq += (t ^ nt) + '0';
      t = nt ^ 1, k = nk;
      p.push_back(nk);
    }
    calc(qwq, p, ans);
  }
  printf("%d\n", (int)ans.size());
  for (auto x : ans) {
    printf("%d ", x);
  }
  printf("\n");
}
int main() {
  T = read();
  while (T--) {
    Solve();
  }
  return 0;
}
