/*

     题意：求邀请顺序使得去爬山的人最多，每个人有去的条件

     贪心+优先队列：首先按照l和r从小到大排序，每一次将当前人数相同的被邀请者入队，那么只要能当前人数比最多人数条件小，该人能                 被邀请，而且不用考虑最少人数的条件。网上的写的比我简洁，学习了。

     详细解释：http://blog.csdn.net/queuelovestack/article/details/47319361

 */

 /************************************************

 * Author        :Running_Time

 * Created Time  :2015-8-6 14:51:47

 * File Name     :H.cpp

  ************************************************/

 #include <cstdio>

 #include <algorithm>

 #include <iostream>

 #include <sstream>

 #include <cstring>

 #include <cmath>

 #include <string>

 #include <vector>

 #include <queue>

 #include <deque>

 #include <stack>

 #include <list>

 #include <map>

 #include <set>

 #include <bitset>

 #include <cstdlib>

 #include <ctime>

 using namespace std;

 #define lson l, mid, rt << 1

 #define rson mid + 1, r, rt << 1 | 1

 typedef long long ll;

 const int MAXN = 1e5 + ;

 const int INF = 0x3f3f3f3f;

 const int MOD = 1e9 + ;

 struct Node {

     int l, r, id;

     bool operator < (const Node &x) const {

         return r > x.r;

     }

 }node[MAXN];

 priority_queue<Node> Q;

 bool vis[MAXN];

 int ans[MAXN];

 int n;

 bool cmp(Node x, Node y)    {

     if (x.l == y.l) return x.r < y.r;

     return x.l < y.l;

 }

 int main(void)    {     //HDOJ 5360 Hiking

     int T;  scanf ("%d", &T);

     while (T--) {

         scanf ("%d", &n);

         for (int i=; i<=n; ++i)    {

             scanf ("%d", &node[i].l);   node[i].id = i;

         }

         for (int i=; i<=n; ++i)    {

             scanf ("%d", &node[i].r);

         }

         sort (node+, node++n, cmp);

         if (node[].l != ) {

             puts ("");

             for (int i=; i<=n; ++i)    printf ("%d%c", i, (i==n) ? '\n' : ' ');

             continue;

         }

         memset (vis, false, sizeof (vis));

         while (!Q.empty ()) Q.pop ();

         int i = , p = ;

         for (; i<=n && p==node[i].l; ++i)   Q.push (node[i]);

         while (!Q.empty ()) {

             while (!Q.empty ()) {

                 if (p <= Q.top ().r)    {

                     ans[++p] = Q.top ().id;

                     vis[ans[p]] = true;    Q.pop ();   break;

                 }

                 else Q.pop ();

             }

             for (; i<=n && p==node[i].l; ++i)   Q.push (node[i]);

         }

         printf ("%d\n", p);

         printf ("%d", ans[]);

         for (int i=; i<=p; ++i)    {

             printf (" %d", ans[i]);

         }

         for (int i=; i<=n; ++i)    {

             if (!vis[i])    printf (" %d", i);

         }

         puts ("");

     }

     return ;

 }