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 a[200005], b[200005], CNT[200005], posa[200005], posb[200005], bel[200005],
    ans[200005], sz[200005], cnt[200005];
vector<int> v;
int main(int argc, char** argv) {
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++)
      cnt[i] = 0, posa[i] = 0, posb[i] = 0, bel[i] = 0, sz[i] = 0, CNT[i] = 0,
      ans[i] = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      ++CNT[a[i]];
      if (!posa[a[i]])
        posa[a[i]] = i;
      else
        posb[a[i]] = i;
    }
    for (int i = 1; i <= n; i++) {
      cin >> b[i];
      ++CNT[b[i]];
      if (!posa[b[i]])
        posa[b[i]] = i;
      else
        posb[b[i]] = i;
    }
    int flag = 1;
    for (int i = 1; i <= n; i++) flag &= CNT[i] == 2;
    if (!flag) {
      puts("-1");
      continue;
    }
    for (int i = 1; i <= n; i++) {
      if (!bel[i]) {
        ans[i] = a[i];
        bel[i] = i;
        int now = i;
        while (1) {
          int x, p;
          if (ans[now] == a[now]) {
            x = b[now], p = posa[x];
            if (p == now) p = posb[x];
          } else {
            x = a[now], p = posa[x];
            if (p == now) p = posb[x];
          }
          if (bel[p]) break;
          bel[p] = i, ans[p] = x, now = p;
        }
      }
    }
    for (int i = 1; i <= n; i++) ++sz[bel[i]], cnt[bel[i]] += ans[i] == a[i];
    for (int i = 1; i <= n; i++) {
      if (sz[bel[i]] - cnt[bel[i]] > cnt[bel[i]])
        if (ans[i] == a[i])
          ans[i] = b[i];
        else
          ans[i] = a[i];
    }
    for (int i = 1; i <= n; i++) {
      if (ans[i] != a[i]) v.push_back(i);
    }
    cout << v.size() << "\n";
    for (auto x : v) cout << x << " ";
    puts("");
    v.clear();
  }
  return 0;
}
