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 maxn = 2e5 + 13;
vector<pair<int, int> > p[maxn];
int ans[maxn];
int cal[maxn];
int a[maxn];
int b[maxn];
int c1[maxn], c2[maxn];
int r1[maxn], r2[maxn];
int vis[maxn];
int num = 0;
queue<int> res;
void dfs(int rt, int now) {
  res.push(rt);
  if (now == 1) num++;
  ans[rt] = now;
  vis[rt] = 1;
  int size = p[rt].size();
  pair<int, int> pp;
  for (int j = 0; j < size; j++) {
    pp = p[rt][j];
    if (vis[pp.first]) continue;
    if (pp.second) {
      dfs(pp.first, now ^ 1);
    } else
      dfs(pp.first, now);
  }
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, lose = 0, end = 0;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
      vis[i] = cal[i] = c1[i] = c2[i] = r1[i] = r2[i] = 0, p[i].clear();
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      cal[a[i]]++;
      if (cal[a[i]] > 2) {
        lose = 1;
      }
    }
    for (int i = 1; i <= n; i++) {
      scanf("%d", &b[i]);
      cal[b[i]]++;
      if (cal[b[i]] > 2) {
        lose = 1;
      }
    }
    if (lose) {
      printf("-1\n");
      continue;
    }
    for (int i = 1; i <= n; i++) {
      if (!vis[a[i]])
        vis[a[i]] = 1, c1[a[i]] = i, r1[a[i]] = 1;
      else
        c2[a[i]] = i, r2[a[i]] = 1;
    }
    for (int i = 1; i <= n; i++) {
      if (!vis[b[i]])
        vis[b[i]] = 1, c1[b[i]] = i, r1[b[i]] = 2;
      else
        c2[b[i]] = i, r2[b[i]] = 2;
    }
    for (int i = 1; i <= n; i++) vis[i] = 0;
    for (int i = 1; i <= n; i++) {
      if (c1[i] != c2[i]) {
        int h = 0, u = c1[i], v = c2[i];
        if (r1[i] == r2[i]) h ^= 1;
        p[u].push_back(make_pair(v, h));
        p[v].push_back(make_pair(u, h));
      }
    }
    for (int i = 1; i <= n; i++) {
      if (!vis[i]) {
        num = 0;
        while (!res.empty()) res.pop();
        dfs(i, 0);
        int s = res.size();
        if (s - num < num) {
          while (!res.empty()) ans[res.front()] ^= 1, res.pop();
        }
      }
    }
    for (int i = 1; i <= n; i++) {
      if (ans[i]) end++;
    }
    printf("%d\n", end);
    for (int i = 1; i <= n; i++) {
      if (ans[i]) printf("%d ", i);
    }
    puts("");
  }
}
