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 t, n;
int a[200200 * 2];
vector<int> g[200200 * 2];
vector<int> app[200200];
int swaps[200200];
int oapp[200200];
int vis[200200 * 2];
vector<int> incolor[200200 * 2];
int dfs(int c, int color) {
  vis[c] = color;
  incolor[color].push_back(c);
  int flips = (c > n);
  for (int i : g[c]) {
    if (vis[i] == 0) {
      flips += dfs(i, color);
    }
  }
  return flips;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      app[i].clear();
      g[i].clear();
      g[i + n].clear();
      swaps[i] = -1;
      vis[i] = 0;
      vis[i + n] = 0;
      incolor[i].clear();
      incolor[i + n].clear();
    }
    for (int i = 0; i < 2; i++)
      for (int j = 1; j <= n; j++) {
        scanf("%d", &a[i * n + j]);
        app[a[i * n + j]].push_back(n * i + j);
      }
    bool ok = true;
    for (int i = 1; i <= n; i++)
      if (app[i].size() != 2) ok = false;
    if (!ok) {
      printf("-1\n");
      continue;
    }
    for (int i = 1; i <= n; i++) {
      oapp[app[i][0]] = app[i][1];
      oapp[app[i][1]] = app[i][0];
    }
    for (int i = 1; i <= n; i++) {
      if (a[i] == a[i + n]) continue;
      if (oapp[i] <= n)
        g[i + n].push_back(oapp[i]);
      else
        g[i + n].push_back(oapp[i]);
      if (oapp[i + n] > n)
        g[i + n].push_back(oapp[i + n] - n);
      else
        g[i + n].push_back(oapp[i + n] + n);
      if (oapp[i] <= n)
        g[i].push_back(oapp[i] + n);
      else
        g[i].push_back(oapp[i] - n);
      if (oapp[i + n] > n)
        g[i].push_back(oapp[i + n]);
      else
        g[i].push_back(oapp[i + n]);
    }
    vector<int> chosen;
    int ccol = 1;
    for (int i = 1; i <= n; i++) {
      if (vis[i] == 0) {
        int ifnot = dfs(i, ccol);
        int ifso = dfs(i + n, ccol + 1);
        if (ifnot < ifso)
          chosen.push_back(ccol);
        else
          chosen.push_back(ccol + 1);
        ccol += 2;
      }
    }
    vector<int> ans;
    for (int i : chosen) {
      for (int j : incolor[i])
        if (j > n) ans.push_back(j - n);
    }
    printf("%d\n", (int)ans.size());
    for (int i : ans) printf("%d ", i);
    printf("\n");
  }
}
