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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
const int N = 200005;
int n;
int a[N];
int b[N];
int q[N];
vector<pair<int, int> > g[N];
vector<int> gi[N];
bool c[N];
void solv() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) scanf("%d", &b[i]);
  for (int i = 1; i <= n; ++i) q[i] = 0;
  for (int i = 1; i <= n; ++i) {
    ++q[a[i]];
    ++q[b[i]];
  }
  for (int i = 1; i <= n; ++i) {
    if (q[i] != 2) {
      printf("-1\n");
      return;
    }
  }
  for (int x = 1; x <= n; ++x) {
    g[x].clear();
    gi[x].clear();
  }
  for (int i = 1; i <= n; ++i) {
    g[a[i]].push_back(make_pair(b[i], 0));
    g[b[i]].push_back(make_pair(a[i], 1));
    gi[a[i]].push_back(i);
    gi[b[i]].push_back(i);
  }
  for (int x = 1; x <= n; ++x) c[x] = false;
  vector<int> ans;
  for (int x = 1; x <= n; ++x) {
    if (c[x]) continue;
    if (g[x][0].first == g[x][1].first) {
      c[x] = true;
      c[g[x][0].first] = true;
      if (g[x][0].second == g[x][1].second) {
        ans.push_back(gi[x][0]);
      }
      continue;
    }
    int p = -1;
    int y = x;
    vector<int> vi;
    vector<int> vz;
    while (1) {
      c[y] = true;
      int hh = -1;
      int hi = -1;
      int hz = -1;
      for (int i = 0; i < g[y].size(); ++i) {
        int h = g[y][i].first;
        if (h != p) {
          hh = h;
          hz = g[y][i].second;
          hi = gi[y][i];
        }
      }
      assert(hh != -1);
      vi.push_back(hi);
      vz.push_back(hz);
      p = y;
      y = hh;
      if (y == x) break;
    }
    int q[2] = {};
    for (int i = 0; i < vz.size(); ++i) {
      q[vz[i]]++;
    }
    if (q[0] < q[1]) {
      for (int i = 0; i < vz.size(); ++i) {
        if (vz[i] == 0) ans.push_back(vi[i]);
      }
    } else {
      for (int i = 0; i < vz.size(); ++i) {
        if (vz[i] == 1) ans.push_back(vi[i]);
      }
    }
  }
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); ++i) printf("%d ", ans[i]);
  printf("\n");
}
int main() {
  int tt;
  scanf("%d", &tt);
  while (tt--) solv();
  return 0;
}
