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 + 5;
const int mod = 1e9 + 7;
long long qpow(long long a, long long b) {
  long long res = 1;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
struct graph {
  int head[maxn], nxt[maxn << 1], to[maxn << 1], w[maxn << 1], sz;
  void init() { memset(head, -1, sizeof(head)); }
  graph() { init(); }
  void push(int a, int b, int c) {
    nxt[sz] = head[a], to[sz] = b, w[sz] = c, head[a] = sz++;
  }
  int& operator[](const int a) { return to[a]; }
} g;
int a[maxn], b[maxn], cnt[maxn];
bool prec[maxn];
vector<pair<int, int> > p[maxn];
vector<int> rec;
vector<int> opt;
bool vis[maxn];
int color[maxn];
int tmp;
void dfs(int now) {
  rec.push_back(now);
  vis[now] = 1;
  for (int i = g.head[now]; ~i; i = g.nxt[i]) {
    if (vis[g[i]]) continue;
    if (g.w[i] == 1) {
      if (color[g[i]] == color[now]) {
        tmp++;
        color[g[i]] = !color[now];
      }
      dfs(g[i]);
    } else {
      if (color[g[i]] != color[now]) {
        tmp++;
        color[g[i]] = color[now];
      }
      dfs(g[i]);
    }
  }
}
int solve(int x) {
  tmp = 0;
  color[x] = 0;
  dfs(x);
  int ans = tmp;
  for (int i = 0; i < rec.size(); ++i) {
    prec[rec[i]] = color[rec[i]];
    vis[rec[i]] = 0;
    color[rec[i]] = 0;
  }
  rec.clear();
  tmp = 0;
  color[x] = 1;
  dfs(x);
  if (1 + tmp < ans) {
    for (int i = 0; i < rec.size(); ++i) {
      if (color[rec[i]] == 1) opt.push_back(rec[i]);
    }
  } else {
    for (int i = 0; i < rec.size(); ++i) {
      if (prec[rec[i]] == 1) opt.push_back(rec[i]);
    }
  }
  ans = min(ans, 1 + tmp);
  rec.clear();
  return ans;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    g.sz = 0;
    for (int i = 1; i <= n; ++i) {
      g.head[i] = -1;
      vis[i] = 0;
      color[i] = 0;
      p[i].clear();
      cnt[i] = 0;
    }
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &a[i]);
      cnt[a[i]]++;
      p[a[i]].push_back(make_pair(1, i));
    }
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &b[i]);
      cnt[b[i]]++;
      p[b[i]].push_back(make_pair(2, i));
    }
    bool ok = true;
    for (int i = 1; i <= n; ++i) {
      if (cnt[i] != 2) {
        ok = false;
        break;
      }
      if (p[i][0].first == p[i][1].first) {
        g.push(p[i][0].second, p[i][1].second, 1);
        g.push(p[i][1].second, p[i][0].second, 1);
      } else {
        g.push(p[i][0].second, p[i][1].second, 0);
        g.push(p[i][1].second, p[i][0].second, 0);
      }
    }
    if (!ok) {
      printf("-1\n");
      continue;
    }
    int ans = 0;
    for (int i = 1; i <= n; ++i) {
      if (vis[i]) continue;
      ans += solve(i);
    }
    printf("%d\n", ans);
    for (int i = 0; i < opt.size(); ++i) printf("%d ", opt[i]);
    printf("\n");
    opt.clear();
  }
  return 0;
}
