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>
#pragma GCC optimize(2)
using namespace std;
inline int read() {
  int x = 0;
  int f = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    f |= (ch == '-') ? 1 : 0;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return f ? -x : x;
}
const int maxn = 2e5 + 10;
int aa[3][maxn];
int vis[maxn], vis1[maxn];
int in[maxn][2], out[maxn][2];
int cnt = 0, cnt1, cnt3;
int st[maxn];
int st1[maxn], st3[maxn];
set<int> temp;
int t, n;
vector<int> vv;
inline void dfs(int u, int fa) {
  vis1[u] = 1;
  st3[++cnt3] = u;
  st[++cnt] = aa[1][u];
  if (cnt1) {
    if (aa[1][u] != st1[cnt1])
      st1[++cnt1] = aa[1][u];
    else
      st1[++cnt1] = aa[2][u];
  } else {
    if (aa[1][out[u][0]] == aa[1][u] || aa[1][out[u][0]] == aa[2][u])
      st1[++cnt1] = aa[1][out[u][0]];
    else
      st1[++cnt1] = aa[2][out[u][0]];
  }
  for (int i = 0; i <= 1; ++i) {
    int v = out[u][i];
    if (v == fa) continue;
    if (!vis1[v]) dfs(v, u);
  }
}
int main() {
  t = read();
  while (t--) {
    n = read();
    for (int i = 1; i <= n; ++i)
      vis[i] = vis1[i] = 0, out[i][0] = out[i][1] = in[i][0] = in[i][1] = -1;
    for (int i = 1; i <= 2; ++i)
      for (int j = 1; j <= n; ++j) aa[i][j] = read(), ++vis[aa[i][j]];
    int flag = 0;
    for (int i = 1; i <= n; ++i) {
      if (aa[1][i] == aa[2][i]) {
        vis1[i] = 1;
      } else {
        if (in[aa[1][i]][0] == -1)
          in[aa[1][i]][0] = i;
        else
          in[aa[1][i]][1] = i;
        if (in[aa[2][i]][0] == -1)
          in[aa[2][i]][0] = i;
        else
          in[aa[2][i]][1] = i;
      }
      if (vis[i] != 2) {
        flag = 1;
        break;
      }
    }
    if (flag) {
      printf("-1\n");
      continue;
    }
    for (int i = 1; i <= n; ++i) {
      if (out[in[i][0]][0] == -1) {
        out[in[i][0]][0] = in[i][1];
      } else {
        out[in[i][0]][1] = in[i][1];
      }
      if (out[in[i][1]][0] == -1) {
        out[in[i][1]][0] = in[i][0];
      } else {
        out[in[i][1]][1] = in[i][0];
      }
    }
    int ans = 0;
    vv.clear();
    for (int i = 1; i <= n; ++i)
      if (!vis1[i]) {
        cnt = cnt1 = cnt3 = 0;
        temp.clear();
        dfs(i, 0);
        int tmp = 0;
        for (int i = 1; i <= cnt; ++i)
          if (st[i] == st1[i]) ++tmp;
        ans += min(tmp, cnt - tmp);
        if (tmp < cnt - tmp) {
          for (int i = 1; i <= cnt; ++i)
            if (st[i] == st1[i]) vv.push_back(st3[i]);
        } else
          for (int i = 1; i <= cnt; ++i)
            if (st[i] != st1[i]) vv.push_back(st3[i]);
      }
    printf("%d\n", ans);
    for (int i = 0; i < vv.size(); ++i) {
      if (i) printf(" ");
      printf("%d", vv[i]);
    }
    printf("\n");
  }
}
