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>
int ds[200000 * 2], sz[200000 * 2];
int find(int i) { return ds[i] < 0 ? i : (ds[i] = find(ds[i])); }
void join(int i, int j) {
  i = find(i);
  j = find(j);
  if (i == j) return;
  if (ds[i] > ds[j]) {
    ds[i] = j;
    sz[j] += sz[i];
  } else {
    if (ds[i] == ds[j]) ds[i]--;
    ds[j] = i;
    sz[i] += sz[j];
  }
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    static int ii[200000], jj[200000], qu[200000];
    int n, i, j, a, cnt, bad;
    scanf("%d", &n);
    memset(ii, -1, n * sizeof *ii);
    memset(jj, -1, n * sizeof *jj);
    bad = 0;
    for (i = 0; i < n; i++) {
      scanf("%d", &a), a--;
      if (ii[a] == -1)
        ii[a] = i << 1 | 0;
      else if (jj[a] == -1)
        jj[a] = i << 1 | 0;
      else
        bad = 1;
    }
    for (i = 0; i < n; i++) {
      scanf("%d", &a), a--;
      if (ii[a] == -1)
        ii[a] = i << 1 | 1;
      else if (jj[a] == -1)
        jj[a] = i << 1 | 1;
      else
        bad = 1;
    }
    if (bad) {
      printf("-1\n");
      continue;
    }
    memset(ds, -1, n * 2 * sizeof *ds);
    for (i = 0; i < n * 2; i++) sz[i] = i & 1;
    for (a = 0; a < n; a++) {
      i = ii[a], j = jj[a];
      join(i, j ^ 1), join(i ^ 1, j);
    }
    for (i = 0; i < n; i++)
      if (find(i << 1 | 0) == find(i << 1 | 1)) {
        bad = 1;
        break;
      }
    cnt = 0;
    for (i = 0; i < n; i++) {
      int r0 = find(i << 1 | 0), r1 = find(i << 1 | 1);
      if (sz[r0] > sz[r1] || sz[r0] == sz[r1] && r0 > r1) qu[cnt++] = i;
    }
    if (bad) {
      printf("-1\n");
      continue;
    }
    printf("%d\n", cnt);
    while (cnt--) printf("%d ", qu[cnt] + 1);
    printf("\n");
  }
  return 0;
}
