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;
const int maxn = 200010;
int c[maxn], id[maxn];
int a[2][maxn];
vector<vector<int>> cnt;
vector<pair<int, int>> g[maxn];
vector<pair<int, int>> pos[maxn];
void dfs(int u, int color, int num) {
  c[u] = color;
  id[u] = num;
  ++cnt[num][color];
  for (pair<int, int>& p : g[u]) {
    if (c[p.first] == -1) {
      dfs(p.first, color ^ p.second, num);
    }
  }
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    fill(pos, pos + n, vector<pair<int, int>>());
    for (int i = 0; i < 2; ++i) {
      for (int j = 0; j < n; ++j) {
        scanf("%d", &a[i][j]);
        pos[a[i][j] - 1].emplace_back(i, j);
      }
    }
    bool flag = true;
    fill(g, g + n, vector<pair<int, int>>());
    for (int i = 0; i < n; ++i) {
      if (pos[i].size() != 2) {
        flag = false;
        break;
      }
      int c1 = pos[i][0].second, c2 = pos[i][1].second;
      if (c1 == c2) continue;
      int r1 = pos[i][0].first, r2 = pos[i][1].first;
      g[c1].emplace_back(c2, r1 == r2);
      g[c2].emplace_back(c1, r1 == r2);
    }
    if (flag) {
      cnt.clear();
      fill(c, c + n, -1);
      fill(id, id + n, 0);
      int ans = 0, num = 0;
      for (int i = 0; i < n; ++i) {
        if (c[i] == -1) {
          cnt.emplace_back(2);
          dfs(i, 0, num);
          ans += min(cnt[num][0], cnt[num][1]);
          ++num;
        }
      }
      printf("%d\n", ans);
      for (int i = 0; i < n; ++i) {
        if (c[i] ^ (cnt[id[i]][0] < cnt[id[i]][1])) {
          printf("%d ", i + 1);
        }
      }
      puts("");
    } else {
      puts("-1");
    }
  }
  return 0;
}
