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() {
  int x = 0;
  bool t = false;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') t = true, ch = getchar();
  while (ch <= '9' && ch >= '0') x = x * 10 + ch - 48, ch = getchar();
  return t ? -x : x;
}
int n, a[3][400400], cnt[400400];
struct P {
  int x, y;
} p[400400][3];
int f[400400], val[400400], cho[400400];
int getf(int x) { return (f[x] == x) ? x : (f[x] = getf(f[x])); }
void Link(int x, int y) {
  if (getf(x) == getf(y)) return;
  val[getf(y)] += val[getf(x)];
  f[getf(x)] = getf(y);
}
int main() {
  int T = read();
  while (T--) {
    n = read();
    bool fl = true;
    for (int i = 1; i <= 2; ++i)
      for (int j = 1; j <= n; ++j) a[i][j] = read();
    for (int i = 1; i <= 2; ++i)
      for (int j = 1; j <= n; ++j)
        if (a[i][j] < 1 || a[i][j] > n) fl = false;
    if (!fl) {
      puts("-1");
      continue;
    }
    for (int i = 1; i <= n; ++i) cnt[i] = 0;
    for (int i = 1; i <= 2; ++i)
      for (int j = 1; j <= n; ++j) p[a[i][j]][++cnt[a[i][j]]] = (P){i, j};
    for (int i = 1; i <= n; ++i)
      if (cnt[i] != 2) fl = false;
    int ans = 0;
    for (int i = 1; i <= n + n; ++i) f[i] = i, val[i] = (i <= n);
    for (int i = 1; i <= n; ++i)
      if (p[i][1].x == p[i][2].x)
        Link(p[i][1].y, p[i][2].y + n), Link(p[i][1].y + n, p[i][2].y);
      else
        Link(p[i][1].y, p[i][2].y), Link(p[i][1].y + n, p[i][2].y + n);
    for (int i = 1; i <= n; ++i)
      if (getf(i) == getf(i + n)) fl = false;
    if (!fl) {
      puts("-1");
      continue;
    }
    for (int i = 1; i <= n + n; ++i) cho[i] = 0;
    for (int i = 1; i <= n; ++i)
      if (getf(i) == i) {
        if (val[getf(i)] < val[getf(i + n)])
          cho[getf(i)] = 1;
        else
          cho[getf(i + n)] = 1;
        ans += min(val[getf(i)], val[getf(i + n)]);
      }
    printf("%d\n", ans);
    for (int i = 1; i <= n; ++i)
      if (cho[getf(i)]) printf("%d ", i);
    if (ans) puts("");
  }
  return 0;
}
