/*

     题意：bob按照指定顺序行走，他的狗可以在他到达下一个点之前到一个景点并及时返回，问狗最多能走多少个景点

     匈牙利算法：按照狗能否顺利到一个景点分为两个集合，套个模板

 */

 #include <cstdio>

 #include <algorithm>

 #include <cstring>

 #include <cmath>

 #include <vector>

 using namespace std;

 const int MAXN = 1e2 + ;

 const int INF = 0x3f3f3f3f;

 struct P

 {

     int x, y;

     double nxt;

 }bob[MAXN], dog[MAXN];

 double d[MAXN][MAXN];

 bool vis[MAXN];

 int lk[MAXN];

 int lk2[MAXN];

 vector<int> G[MAXN];

 double get_dis(int x1, int y1, int x2, int y2)

 {

     return sqrt ((1.0) * ((x1-x2) * (x1-x2) + (y1-y2) * (y1-y2)));

 }

 bool DFS(int u)

 {

     for (int i=; i<G[u].size (); ++i)

     {

         int v = G[u][i];

         if (!vis[v])

         {

             vis[v] = true;

             if (lk[v] == - || DFS (lk[v]))

             {

                 lk[v] = u;    lk2[u] = v;    return true;

             }

         }

     }

     return false;

 }

 void hungary(int n)

 {

     int res = ;    memset (lk, -, sizeof (lk));    memset (lk2, -, sizeof (lk2));

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

     {

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

         if (DFS (i))    res++;

     }

     printf ("%d\n", n + res);

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

     {

         printf ("%d %d ", bob[i].x, bob[i].y);

         if (lk2[i] != -)    printf ("%d %d ", dog[lk2[i]].x, dog[lk2[i]].y);

     }

     printf ("%d %d\n", bob[n].x, bob[n].y);

 }

 int main(void)        //UVA 670 The dog task

 {

 //    freopen ("UVA_670.in", "r", stdin);

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

     while (t--)

     {

         int n, m;    scanf ("%d%d", &n, &m);

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

         {

             scanf ("%d%d", &bob[i].x, &bob[i].y);    G[i].clear ();

         }

         for (int i=; i<=m; ++i)

         {

             scanf ("%d%d", &dog[i].x, &dog[i].y);

         }

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

         {

             for (int j=; j<=m; ++j)

             {

                 d[i][j] = get_dis (bob[i].x, bob[i].y, dog[j].x, dog[j].y);

             }

         }

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

         {

             bob[i].nxt = get_dis (bob[i].x, bob[i].y, bob[i+].x, bob[i+].y);

         }

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

         {

             for (int j=; j<=m; ++j)

             {

                 if (d[i][j] + d[i+][j] <=  * bob[i].nxt)    G[i].push_back (j);

             }

         }

         hungary (n);

         if (t)    puts ("");

     }

     return ;

 }