/*

     题意：在坐标轴上一群蚂蚁向左或向右爬，问经过ts后，蚂蚁的位置和状态

     思维题：本题的关键1：蚂蚁相撞看作是对穿过去，那么只要判断谁是谁就可以了

                 关键2：蚂蚁的相对位置不变    关键3：order数组记录顺序

 */

 #include <cstdio>

 #include <algorithm>

 #include <iostream>

 #include <cstring>

 #include <string>

 #include <cmath>

 using namespace std;

 const int MAXN = 1e4 + ;

 const int INF = 0x3f3f3f3f;

 const char dir_name[][] = {"L", "Turning", "R"};

 struct Ant

 {

     int pos, dir, id;

     bool operator < (const Ant &a)    const

     {

         return pos < a.pos;

     }

 }pre[MAXN], now[MAXN];

 int order[MAXN];

 int main(void)        //UVA 10881 Piotr's Ants

 {

 //    freopen ("UVA_10881.in", "r", stdin);

     int T;    int cas = ;    int len, t, n;

     scanf ("%d", &T);

     while (T--)

     {

         scanf ("%d%d%d", &len, &t, &n);

         char ch;

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

         {

             int p, d;    char ch;

             scanf ("%d %c", &p, &ch);

             d = ((ch=='L') ? - : );

             pre[i] = (Ant) {p, d, i};

             now[i] = (Ant) {p+t*d, d, };

         }

         sort (pre+, pre++n);        //计算相对位置

         for (int i=; i<=n; ++i)    order[pre[i].id] = i;    //输入(输出)的顺序

         sort (now+, now++n);

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

         {

             if (now[i].pos == now[i+].pos)

                 now[i].dir = now[i+].dir = ;

         }

         printf ("Case #%d:\n", ++cas);

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

         {

             int x = order[i];

             if (now[x].pos <  || now[x].pos > len)    puts ("Fell off");

             else

             {

                 printf ("%d %s\n", now[x].pos, dir_name[now[x].dir+]);

             }

         }

         puts ("");

     }

     return ;

 }

 /*

 Case #1:

 2 Turning

 6 R

 2 Turning

 Fell off

 Case #2:

 3 L

 6 R

 10 R

 */