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;
inline int read() {
  char ch = getchar();
  int nega = 1;
  while (!isdigit(ch)) {
    if (ch == '-') nega = -1;
    ch = getchar();
  }
  int ans = 0;
  while (isdigit(ch)) {
    ans = ans * 10 + ch - 48;
    ch = getchar();
  }
  if (nega == -1) return -ans;
  return ans;
}
int fa[200005 * 2], siz[200005 * 2], n;
int find(int u) { return fa[u] == u ? u : fa[u] = find(fa[u]); }
int a[200005], b[200005];
vector<int> v[200005];
void merge(int u, int v) {
  u = find(u), v = find(v);
  if (u == v) return;
  fa[u] = v;
  siz[v] += siz[u];
}
bool vis[200005 * 2], ch[200005 * 2];
void work() {
  cin >> n;
  for (int i = 1; i <= n; i++) v[i].clear();
  for (int i = 1; i <= n; i++) a[i] = read(), v[a[i]].push_back(i);
  for (int i = 1; i <= n; i++) b[i] = read(), v[b[i]].push_back(i + n);
  for (int i = 1; i <= 2 * n; i++) fa[i] = i;
  for (int i = 1; i <= n; i++) siz[i] = 1;
  for (int i = n + 1; i <= 2 * n; i++) siz[i] = 0;
  for (int i = 1; i <= n * 2; i++) vis[i] = ch[i] = 0;
  for (int i = 1; i <= n; i++) {
    if (v[i].size() != 2) {
      cout << "-1\n";
      return;
    }
    int a = v[i][0], b = v[i][1];
    if (a > b) swap(a, b);
    if (a == b - n) continue;
    if (a > n) {
      a -= n, b -= n;
      merge(a, b + n), merge(a + n, b);
    } else if (b <= n)
      merge(a, b + n), merge(a + n, b);
    else {
      b -= n;
      merge(a, b), merge(a + n, b + n);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (find(i) == find(i + n)) {
      cout << "-1\n";
      return;
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    int u = find(i), v = find(i + n);
    if (vis[u] || vis[v]) continue;
    ans += min(siz[u], siz[v]);
    vis[u] = vis[v] = 1;
    if (siz[u] < siz[v])
      ch[u] = 1;
    else
      ch[v] = 1;
  }
  cout << ans << endl;
  for (int i = 1; i <= n; i++) {
    int u = find(i);
    if (ch[u]) printf("%d ", i);
  }
  cout << "\n";
}
signed main() {
  int T = read();
  while (T--) work();
  return 0;
}
