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 + 10;
const int MAXM = 2e6 + 10;
const long long MOD = 998244353;
const long long INF = 0x3f3f3f3f;
const unsigned long long BASE = 233;
const int base = 5;
char rubbish[110];
long long read() {
  long long x = 0, f = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = -1;
  } while (ch < '0' || ch > '9');
  do {
    x = x * 10 + ch - 48;
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  return x * f;
}
int N, M, K;
int a[MAXN];
int b[MAXN];
vector<int> ans;
bool vis[MAXN];
vector<int> pos[MAXN];
void init() {
  ans.clear();
  for (int i = 1; i <= N; i++) {
    vis[i] = false;
    pos[i].clear();
  }
}
void solve() {
  int now;
  scanf("%d", &N);
  init();
  for (int i = 1; i <= N; i++) {
    a[i] = read();
    pos[a[i]].push_back(i);
  }
  for (int i = 1; i <= N; i++) {
    b[i] = read();
    pos[b[i]].push_back(i);
  }
  for (int i = 1; i <= N; i++) {
    if (pos[i].size() > 2) {
      printf("-1\n");
      return;
    }
  }
  for (int i = 1; i <= N; i++) {
    if (vis[i]) {
      continue;
    }
    now = pos[i][0];
    vector<int> l, r;
    for (int j = i; !vis[j];) {
      vis[j] = true;
      if (pos[j][0] == now) {
        now = pos[j][1];
      } else {
        now = pos[j][0];
      }
      if (a[now] == j) {
        r.push_back(now);
        j = b[now];
      } else {
        l.push_back(now);
        j = a[now];
      }
    }
    if (l.size() > r.size()) {
      swap(l, r);
    }
    for (auto x : l) {
      ans.push_back(x);
    }
  }
  printf("%d\n", ans.size());
  for (auto x : ans) {
    printf("%d ", (x));
  }
  printf("\n");
}
int main() {
  int T;
  T = 1;
  scanf("%d", &T);
  for (int t = 1; t <= T; t++) {
    solve();
  }
  return 0;
}
