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;
std::mt19937 rnd(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 200000;
int n;
int a[2][MAXN];
vector<int> ans;
vector<pair<int, int>> pos[MAXN];
int ufpar[MAXN], ufsz[MAXN], ufdiff[MAXN];
int cnt[MAXN][2];
pair<int, int> uffind(int x) {
  if (ufpar[x] == x) return make_pair(x, 0);
  pair<int, int> res = uffind(ufpar[x]);
  res.second ^= ufdiff[x];
  ufpar[x] = res.first;
  ufdiff[x] = res.second;
  return res;
}
bool unite(int a, int b, int c) {
  int adiff;
  tie(a, adiff) = uffind(a);
  int bdiff;
  tie(b, bdiff) = uffind(b);
  if (a == b) return (adiff ^ bdiff ^ c) == 0;
  if (ufsz[a] < ufsz[b]) swap(a, b), swap(adiff, bdiff);
  ufpar[b] = a, ufsz[a] += ufsz[b], ufdiff[b] = adiff ^ bdiff ^ c;
  return true;
}
bool solve() {
  for (int i = (0); i < (n); ++i) pos[i].clear();
  for (int i = (0); i < (2); ++i)
    for (int j = (0); j < (n); ++j) pos[a[i][j]].push_back(make_pair(i, j));
  for (int i = (0); i < (n); ++i)
    if (((int)(pos[i]).size()) != 2) return false;
  for (int i = (0); i < (n); ++i) ufpar[i] = i, ufsz[i] = 1, ufdiff[i] = 0;
  for (int i = (0); i < (n); ++i) {
    int ax, ay, bx, by;
    tie(ax, ay) = pos[i][0];
    tie(bx, by) = pos[i][1];
    if (!unite(ay, by, ax == bx ? 1 : 0)) return false;
  }
  for (int i = (0); i < (n); ++i)
    for (int j = (0); j < (2); ++j) cnt[i][j] = 0;
  for (int i = (0); i < (n); ++i) {
    int root, diff;
    tie(root, diff) = uffind(i);
    ++cnt[root][diff];
  }
  ans.clear();
  for (int i = (0); i < (n); ++i) {
    int root, diff;
    tie(root, diff) = uffind(i);
    if (cnt[root][diff] < cnt[root][1 - diff] ||
        cnt[root][diff] == cnt[root][1 - diff] && diff == 1)
      ans.push_back(i);
  }
  return true;
}
void run() {
  scanf("%d", &n);
  for (int i = (0); i < (2); ++i)
    for (int j = (0); j < (n); ++j) scanf("%d", &a[i][j]), --a[i][j];
  if (!solve()) {
    printf("-1\n");
    return;
  }
  printf("%d\n", ((int)(ans).size()));
  for (int i = (0); i < (((int)(ans).size())); ++i) {
    if (i != 0) printf(" ");
    printf("%d", ans[i] + 1);
  }
  puts("");
}
int main() {
  int ncase;
  scanf("%d", &ncase);
  for (int i = (1); i <= (ncase); ++i) run();
  return 0;
}
