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;
struct poz {
  int c1, c2, r1, r2, nr;
};
struct drum {
  int y, t;
  drum(int y, int t) {
    this->y = y;
    this->t = t;
  }
};
struct nodS {
  int cul;
  bool viz;
  vector<drum> vecini;
};
int t, n, i, nr0, nr1, ans;
bool ok;
int a[200005];
int b[200005];
poz fr[200005];
nodS nod[200005];
void add(int x, int y, int val) {
  nod[x].vecini.push_back(drum(y, val));
  nod[y].vecini.push_back(drum(x, val));
}
void dfs(int poz, int val) {
  nod[poz].viz = true;
  nod[poz].cul = val;
  if (val)
    nr1++;
  else
    nr0++;
  for (auto it : nod[poz].vecini) {
    if (nod[it.y].viz) continue;
    dfs(it.y, val ^ it.t);
  }
}
void dev(int poz) {
  nod[poz].viz = false;
  for (auto it : nod[poz].vecini) {
    if (nod[it.y].viz) dev(it.y);
  }
}
void norm(int poz) {
  nod[poz].viz = true;
  nod[poz].cul ^= 1;
  for (auto it : nod[poz].vecini) {
    if (nod[it.y].viz) continue;
    norm(it.y);
  }
}
int main() {
  cin >> t;
  while (t) {
    t--;
    cin >> n;
    ok = true;
    ans = 0;
    for (i = 1; i <= n; i++) {
      fr[i].r1 = fr[i].c1 = fr[i].r2 = fr[i].c2 = fr[i].nr = 0;
      nod[i].viz = false;
      nod[i].vecini.clear();
    }
    for (i = 1; i <= n; i++) {
      cin >> a[i];
      fr[a[i]].nr++;
      if (fr[a[i]].r1) {
        fr[a[i]].r2 = 1;
        fr[a[i]].c2 = i;
      } else {
        fr[a[i]].r1 = 1;
        fr[a[i]].c1 = i;
      }
    }
    for (i = 1; i <= n; i++) {
      cin >> b[i];
      fr[b[i]].nr++;
      if (fr[b[i]].r1) {
        fr[b[i]].r2 = 2;
        fr[b[i]].c2 = i;
      } else {
        fr[b[i]].r1 = 2;
        fr[b[i]].c1 = i;
      }
    }
    for (i = 1; i <= n; i++) {
      if (fr[i].nr != 2) ok = false;
      if (fr[i].c1 == fr[i].c2) continue;
      if (fr[i].r1 == fr[i].r2)
        add(fr[i].c1, fr[i].c2, 1);
      else
        add(fr[i].c1, fr[i].c2, 0);
    }
    if (!ok) {
      cout << "-1\n";
      continue;
    }
    for (i = 1; i <= n; i++) {
      if (!nod[i].viz) {
        nr0 = nr1 = 0;
        dfs(i, 0);
        if (nr1 > nr0) {
          dev(i);
          norm(i);
        }
        ans += min(nr0, nr1);
      }
    }
    cout << ans << '\n';
    for (i = 1; i <= n; i++) {
      if (nod[i].cul) cout << i << ' ';
    }
    cout << '\n';
  }
  return 0;
}
