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>
const int bufSize = 1e6;
inline char nc() {
  return getchar();
  static char buf[bufSize], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, bufSize, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline void read(char *s) {
  static char c;
  for (; !isalpha(c); c = nc())
    ;
  for (; isalpha(c); c = nc()) *s++ = c;
  *s = '\0';
}
template <typename T>
inline T read(T &r) {
  static char c;
  static int flag;
  flag = 1, r = 0;
  for (c = nc(); !isdigit(c); c = nc())
    if (c == '-') flag = -1;
  for (; isdigit(c); c = nc()) r = r * 10 + c - 48;
  return r *= flag;
}
const int maxn = 4e5 + 100;
const int maxm = maxn * 3;
int T, n;
int a[2][maxn];
int vis[maxn];
struct node {
  int to, next, val;
} E[maxm];
int head[maxn], tot;
inline void add(const int &x, const int &y, const int &val) {
  E[++tot].next = head[x], E[tot].to = y, E[tot].val = val, head[x] = tot;
}
int col[maxn];
bool flag;
int s[2][maxn], stop[2];
void dfs(int u) {
  s[col[u]][++stop[col[u]]] = u;
  if (flag) return;
  for (int p = head[u]; p; p = E[p].next) {
    int v = E[p].to;
    if (E[p].val) {
      if (col[v] == -1)
        col[v] = col[u] ^ 1, dfs(v);
      else if (col[v] == col[u])
        flag = 1;
    } else if (E[p].val == 0) {
      if (col[v] == -1)
        col[v] = col[u], dfs(v);
      else if (col[v] != col[u])
        flag = 1;
    }
    if (flag) return;
  }
}
int visall[maxn], ans;
int anss[maxn], atop;
int main() {
  read(T);
  while (T--) {
    read(n);
    for (int i = 1; i <= n * 2; ++i)
      col[i] = -1, head[i] = 0, vis[i] = 0, visall[i] = 0;
    atop = tot = flag = 0;
    for (int i = 1; i <= n; ++i) read(a[0][i]), visall[a[0][i]]++;
    for (int i = 1; i <= n; ++i) read(a[1][i]), visall[a[1][i]]++;
    for (int i = 1; i <= n; ++i)
      if (visall[i] != 2) {
        puts("-1");
        goto end;
      }
    for (int i = 1; i <= n; ++i) {
      if (vis[a[0][i]])
        add(vis[a[0][i]], i, 1), add(i, vis[a[0][i]], 1);
      else
        vis[a[0][i]] = i;
    }
    for (int i = 1; i <= n; ++i) {
      if (vis[a[1][i]] == 0)
        vis[a[1][i]] = i + n;
      else if (vis[a[1][i]] <= n)
        add(vis[a[1][i]], i, 0), add(i, vis[a[1][i]], 0);
      else if (vis[a[1][i]] > n)
        add(vis[a[1][i]] - n, i, 1), add(i, vis[a[1][i]] - n, 1);
    }
    for (int i = 1; i <= n; ++i)
      if (col[i] == -1) {
        stop[0] = stop[1] = 0;
        col[i] = 0;
        dfs(i);
        if (flag) {
          printf("-1\n");
          goto end;
        }
        if (stop[0] < stop[1])
          while (stop[0]) anss[++atop] = s[0][stop[0]--];
        else
          while (stop[1]) anss[++atop] = s[1][stop[1]--];
      }
    printf("%d\n", atop);
    while (atop) printf("%d ", anss[atop--]);
    putchar('\n');
  end:;
  }
  return 0;
}
