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;
long long n, nn[2], fq[200069], vl[200069][2], av[2][200069], sq[200069], zs;
bitset<200069> xr, vtd, nr;
vector<pair<long long, long long>> al[200069];
void dfs(long long x) {
  long long i, sz = al[x].size(), l, w;
  vtd[x] = 1;
  nn[nr[x]]++;
  av[nr[x]][nn[nr[x]]] = x;
  for (i = 0; i < sz; i++) {
    l = al[x][i].first;
    w = al[x][i].second;
    if (!vtd[l]) {
      nr[l] = nr[x] ^ w;
      dfs(l);
    }
  }
}
int main() {
  long long t, rr, i, j, ii, e, k;
  bool bad;
  scanf("%lld", &t);
  for (rr = 0; rr < t; rr++) {
    scanf("%lld", &n);
    for (i = 1; i <= n; i++) {
      fq[i] = 0;
      xr[i] = 0;
      al[i].clear();
      vtd[i] = 0;
    }
    bad = 0;
    for (ii = 0; ii < 2; ii++) {
      for (i = 1; i <= n; i++) {
        scanf("%lld", &k);
        if (fq[k] < 2) {
          vl[k][fq[k]] = i;
          xr[k] = xr[k] ^ ii;
          fq[k]++;
        } else {
          bad = 1;
        }
      }
    }
    if (bad) {
      printf("-1\n");
      continue;
    }
    for (i = 1; i <= n; i++) {
      for (ii = 0; ii < 2; ii++) {
        al[vl[i][ii]].push_back({vl[i][!ii], !xr[i]});
      }
    }
    zs = 0;
    for (i = 1; i <= n; i++) {
      if (!vtd[i]) {
        for (ii = 0; ii < 2; ii++) {
          nn[ii] = 0;
        }
        dfs(i);
        e = nn[1] < nn[0];
        for (j = 1; j <= nn[e]; j++) {
          zs++;
          sq[zs] = av[e][j];
        }
      }
    }
    printf("%lld\n", zs);
    for (i = 1; i <= zs; i++) {
      printf("%lld%c", sq[i], " \n"[i == zs]);
    }
  }
}
