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;
const int maxm = 4e5 + 5;
vector<int> g[2][maxm];
vector<int> temp[maxm];
vector<int> ans;
int head[maxm], nt[maxm], tail[maxm];
int a[2][maxm];
int mark[maxm];
int pre[maxm];
int cnt[maxm];
int n;
int ffind(int x) { return pre[x] == x ? x : pre[x] = ffind(pre[x]); }
void add(int x, int y) {
  x = ffind(x), y = ffind(y);
  if (x == y) return;
  pre[y] = x;
  cnt[x] += cnt[y];
  nt[tail[x]] = head[y];
  tail[x] = tail[y];
}
void cal(int x) {
  for (int i = head[x]; i; i = nt[i]) {
    if (i > n) {
      ans.push_back(i - n);
      mark[i - n] = 1;
    } else {
      mark[i] = 1;
    }
  }
}
signed main() {
  ios::sync_with_stdio(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[0][i];
    for (int i = 1; i <= n; i++) cin >> a[1][i];
    map<int, int> tot;
    for (int i = 1; i <= n; i++) {
      tot[a[0][i]]++;
      tot[a[1][i]]++;
    }
    int ok = 1;
    for (auto i : tot) {
      if (i.second != 2) {
        ok = 0;
        break;
      }
    }
    if (!ok) {
      cout << -1 << endl;
      continue;
    }
    for (int i = 1; i <= n; i++) {
      g[0][i].clear();
      g[1][i].clear();
    }
    for (int i = 1; i <= n; i++) {
      g[0][a[0][i]].push_back(i);
      g[1][a[1][i]].push_back(i);
    }
    for (int i = 1; i <= n; i++) {
      pre[i] = i;
      cnt[i] = 0;
      head[i] = i;
      tail[i] = i;
      nt[i] = 0;
      pre[i + n] = i + n;
      cnt[i + n] = 1;
      head[i + n] = i + n;
      tail[i + n] = i + n;
      nt[i + n] = 0;
    }
    for (int i = 1; i <= n; i++) {
      if (g[0][i].size() == 2) {
        int x = g[0][i][0], y = g[0][i][1];
        add(x, y + n);
        add(x + n, y);
      } else if (g[1][i].size() == 2) {
        int x = g[1][i][0], y = g[1][i][1];
        add(x, y + n);
        add(x + n, y);
      } else {
        int x = g[0][i][0], y = g[1][i][0];
        add(x, y);
        add(x + n, y + n);
      }
    }
    for (int i = 1; i <= n; i++) {
      if (ffind(i) == ffind(i + n)) {
        ok = 0;
        break;
      }
    }
    if (!ok) {
      cout << -1 << endl;
      continue;
    }
    ans.clear();
    for (int i = 1; i <= n; i++) {
      mark[i] = 0;
    }
    for (int i = 1; i <= n; i++) {
      if (mark[i]) continue;
      int x = ffind(i), y = ffind(i + n);
      if (cnt[x] < cnt[y]) {
        cal(x);
      } else {
        cal(y);
      }
    }
    cout << ans.size() << endl;
    for (auto i : ans) {
      cout << i << ' ';
    }
    cout << endl;
  }
  return 0;
}
