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 long long mod = 1000000007LL;
int t;
int n;
int a[2][200005];
int par[200005];
vector<int> cols[200005];
vector<int> edges[200005];
int col[200005];
int cnt[200005];
bool rev[200005];
bool dfs(int u, int c, vector<int>& v0, vector<int>& v1) {
  if (col[u] >= 0) {
    if (col[u] != c) return false;
    return true;
  }
  col[u] = c;
  if (c == 0) {
    v0.push_back(u);
  } else {
    v1.push_back(u);
  }
  for (int i = 0; i < edges[u].size(); ++i) {
    int v = edges[u][i];
    if (!dfs(v, !c, v0, v1)) return false;
  }
  return true;
}
int find(int u) { return u == par[u] ? u : par[u] = find(par[u]); }
int main() {
  cin >> t;
  for (int cas = 1; cas <= t; ++cas) {
    cin >> n;
    for (int i = 1; i <= n; ++i) {
      cols[i].clear();
      edges[i].clear();
    }
    for (int i = 0; i < 2; ++i) {
      for (int j = 1; j <= n; ++j) {
        scanf("%d", &a[i][j]);
        cols[a[i][j]].push_back(j);
      }
    }
    bool ok = true;
    for (int i = 1; i <= n; ++i) {
      if (cols[i].size() != 2) {
        ok = false;
        break;
      }
    }
    if (!ok) {
      cout << -1 << endl;
      continue;
    }
    for (int i = 1; i <= n; ++i) {
      par[i] = i;
      cnt[i] = 1;
      col[i] = -1;
      rev[i] = false;
    }
    for (int i = 1; i <= n; ++i) {
      int c1 = cols[i][0];
      int c2 = cols[i][1];
      if ((a[0][c1] == i && a[1][c2] == i) ||
          (a[1][c1] == i && a[0][c2] == i)) {
        int p1 = find(c1);
        int p2 = find(c2);
        if (p1 != p2) {
          par[p1] = p2;
          cnt[p2] += cnt[p1];
        }
      }
    }
    for (int i = 1; i <= n; ++i) {
      int c1 = cols[i][0];
      int c2 = cols[i][1];
      if (c1 == c2) continue;
      if ((a[0][c1] == i && a[0][c2] == i) ||
          (a[1][c1] == i && a[1][c2] == i)) {
        int p1 = find(c1);
        int p2 = find(c2);
        if (p1 == p2) {
          ok = false;
          break;
        }
        edges[p1].push_back(p2);
        edges[p2].push_back(p1);
      }
    }
    if (!ok) {
      cout << -1 << endl;
      continue;
    }
    for (int i = 1; i <= n; ++i) {
      par[i] = find(i);
      if (par[i] == i && col[i] < 0) {
        vector<int> v0, v1;
        v0.clear();
        v1.clear();
        if (!dfs(i, 0, v0, v1)) {
          ok = false;
          break;
        }
        int n0 = 0, n1 = 0;
        for (int j = 0; j < v0.size(); ++j) {
          n0 += cnt[v0[j]];
        }
        for (int j = 0; j < v1.size(); ++j) {
          n1 += cnt[v1[j]];
        }
        if (n0 < n1) {
          for (int j = 0; j < v0.size(); ++j) {
            rev[v0[j]] = true;
          }
        } else {
          for (int j = 0; j < v1.size(); ++j) {
            rev[v1[j]] = true;
          }
        }
      }
    }
    if (!ok) {
      cout << -1 << endl;
    } else {
      vector<int> ans;
      for (int i = 1; i <= n; ++i) {
        par[i] = find(i);
        if (rev[par[i]]) {
          ans.push_back(i);
        }
      }
      cout << ans.size() << endl;
      for (int i = 0; i < ans.size(); ++i) {
        printf("%d", ans[i]);
        if (i < ans.size() - 1)
          cout << " ";
        else
          cout << endl;
      }
    }
  }
  return 0;
}
