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 N = 1e6 + 10;
const long long mod = 1e9 + 7;
inline long long read() {
  long long X = 0;
  bool flag = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    X = (X << 1) + (X << 3) + ch - '0';
    ch = getchar();
  }
  if (flag) return X;
  return ~(X - 1);
}
long long n, m, k;
long long ans;
long long ar[N], br[N];
int d[N], fa[N];
int fid(int x) {
  if (x != fa[x]) {
    int t = fa[x];
    fa[x] = fid(fa[x]);
    d[x] += d[t];
    d[x] %= 2;
  }
  return fa[x];
}
vector<int> vt[N];
vector<int> vv[N];
int ct[N][2];
int pd[N];
vector<int> temp[N][2];
vector<int> as;
int main() {
  int T;
  T = read();
  while (T--) {
    n = read();
    for (int i = 0; i <= n + 10; i++) {
      fa[i] = i, pd[i] = ct[i][0] = ct[i][1] = d[i] = 0;
      vv[i].clear();
      vt[i].clear();
      temp[i][0].clear();
      temp[i][1].clear();
    }
    int f = 0;
    for (int i = 1; i <= n; i++) {
      ar[i] = read();
      pd[ar[i]]++;
      vt[ar[i]].push_back(i);
    }
    for (int i = 1; i <= n; i++) {
      br[i] = read();
      pd[br[i]]++;
      vv[br[i]].push_back(i);
    }
    for (int i = 1; i <= n; i++) {
      if (pd[i] != 2) {
        f = 1;
        break;
      }
      int a = ar[i];
      for (int j = 0; j < vt[a].size(); j++) {
        int v = vt[a][j];
        if (i >= v) continue;
        int fx = fid(i);
        int fy = fid(v);
        if (fx == fy) {
          if ((d[i] + d[v]) % 2 == 0) {
            f = 1;
            break;
          }
        } else {
          fa[fx] = fy;
          d[fx] = (d[i] + d[v] + 1) % 2;
        }
      }
      for (int j = 0; j < vv[a].size(); j++) {
        int v = vv[a][j];
        if (i == v) continue;
        int fx = fid(i);
        int fy = fid(v);
        if (fx == fy) {
          if ((d[i] + d[v]) % 2 == 1) {
            f = 1;
            break;
          }
        } else {
          fa[fx] = fy;
          d[fx] = (d[v] + d[i]) % 2;
        }
      }
      a = br[i];
      for (int j = 0; j < vv[a].size(); j++) {
        int v = vv[a][j];
        if (i >= v) continue;
        int fx = fid(i);
        int fy = fid(v);
        if (fx == fy) {
          if ((d[i] + d[v]) % 2 == 0) {
            f = 1;
            break;
          }
        } else {
          fa[fx] = fy;
          d[fx] = (d[v] + d[i] + 1) % 2;
        }
      }
    }
    for (int i = 1; i <= n; i++) {
      int ff = fid(i);
      ct[ff][d[i]]++;
      temp[ff][d[i]].push_back(i);
    }
    ans = 0;
    as.clear();
    for (int i = 1; i <= n; i++) {
      int gg = fid(i);
      if (gg != i) continue;
      if (ct[gg][0] <= ct[gg][1]) {
        ans += ct[gg][0];
        for (int j = 0; j < temp[gg][0].size(); j++)
          as.push_back(temp[gg][0][j]);
      } else {
        ans += ct[gg][1];
        for (int j = 0; j < temp[gg][1].size(); j++)
          as.push_back(temp[gg][1][j]);
      }
    }
    if (f)
      printf("-1\n");
    else {
      printf("%lld\n", ans);
      for (int i = 0; i < as.size(); i++) printf("%d ", as[i]);
      puts("");
    }
  }
  return 0;
}
