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 = 4e5 + 7;
int n, N;
namespace Graph {
int head[maxN], cnt;
struct Eddge {
  int nex, to;
  Eddge(int a = -1, int b = 0) : nex(a), to(b) {}
} edge[maxN << 1];
inline void addEddge(int u, int v) {
  edge[cnt] = Eddge(head[u], v);
  head[u] = cnt++;
}
inline void init() {
  cnt = 0;
  for (int i = 1; i <= N; i++) head[i] = -1;
}
};  // namespace Graph
using namespace Graph;
unsigned long long h = 1e9 + 7;
unsigned long long Bas[maxN];
int a[maxN], b[maxN];
vector<int> pos[maxN];
namespace SCC {
int dfn[maxN], low[maxN], tot, Stap[maxN], Stop, Bcnt, Bsiz[maxN], _ID[maxN];
unsigned long long tp[maxN];
vector<int> Belong[maxN];
bool instack[maxN];
bool cmp(int e1, int e2) {
  return tp[e1] == tp[e2] ? Bsiz[e1] < Bsiz[e2] : tp[e1] < tp[e2];
}
void clear_SCC() {
  tot = Stop = Bcnt = 0;
  for (int i = 1; i <= N; i++) {
    dfn[i] = 0;
    instack[i] = false;
  }
}
void Tarjan(int u) {
  dfn[u] = low[u] = ++tot;
  Stap[++Stop] = u;
  instack[u] = true;
  for (int i = head[u], v; ~i; i = edge[i].nex) {
    v = edge[i].to;
    if (!dfn[v]) {
      Tarjan(v);
      low[u] = min(low[u], low[v]);
    } else if (instack[v])
      low[u] = min(low[u], dfn[v]);
  }
  if (low[u] == dfn[u]) {
    Bcnt++;
    Bsiz[Bcnt] = 0;
    tp[Bcnt] = 0;
    _ID[Bcnt] = Bcnt;
    Belong[Bcnt].clear();
    int v;
    do {
      v = Stap[Stop--];
      if (v > n) {
        Bsiz[Bcnt]++;
        Belong[Bcnt].push_back(v - n);
      }
      tp[Bcnt] = tp[Bcnt] + Bas[v > n ? b[v - n] : a[v]];
      instack[v] = false;
    } while (u ^ v);
  }
}
};  // namespace SCC
using namespace SCC;
int main() {
  Bas[0] = 1;
  for (int i = 1; i < maxN; i++) Bas[i] = Bas[i - 1] * h;
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    N = n << 1;
    init();
    for (int i = 1; i <= n; i++) pos[i].clear();
    bool ok = true;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      pos[a[i]].push_back(i);
      if (pos[a[i]].size() > 2) ok = false;
    }
    for (int i = 1; i <= n; i++) {
      scanf("%d", &b[i]);
      pos[b[i]].push_back(n + i);
      if (pos[b[i]].size() > 2) ok = false;
    }
    if (!ok) {
      printf("-1\n");
      continue;
    }
    for (int i = 1; i <= n; i++) {
      if (a[i] == b[i]) continue;
      addEddge(i, pos[b[i]][0] == n + i ? pos[b[i]][1] : pos[b[i]][0]);
    }
    for (int i = 1; i <= n; i++) {
      if (a[i] == b[i]) continue;
      addEddge(n + i, pos[a[i]][0] == i ? pos[a[i]][1] : pos[a[i]][0]);
    }
    clear_SCC();
    for (int i = 1; i <= N; i++)
      if (!dfn[i]) Tarjan(i);
    sort(_ID + 1, _ID + Bcnt + 1, cmp);
    vector<int> ans;
    ans.clear();
    for (int i = 1, ith; i <= Bcnt; i += 2) {
      ith = _ID[i];
      for (int j : Belong[ith]) {
        ans.push_back(j);
      }
    }
    int len = (int)ans.size();
    printf("%d\n", len);
    for (int i = 0; i < len; i++)
      printf("%d%c", ans[i], i == len - 1 ? '\n' : ' ');
  }
  return 0;
}
