/**
 * CERC 2018
 * Problem Solution: The Lord of the Kings
 * Idea: Find Steiner tree of particular graph
 * @author Josef Malik
 */

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <vector>

using namespace std;

typedef long long int ll;
typedef pair<int, int> pii;

#define PB push_back
#define MP make_pair

#define FOR(prom, a, b) for(int prom = (a); prom < (b); prom++)
#define FORD(prom, a, b) for(int prom = (a); prom > (b); prom--)
#define FORDE(prom, a, b) for(int prom = (a); prom >= (b); prom--)
#define R1(a) do{scanf("%d", &(a));}while(0)
#define R2(a, b) do{scanf("%d%d", &(a), &(b));}while(0)
#define R3(a, b, c) do{scanf("%d%d%d", &(a), &(b), &(c));}while(0)
#define SV(vec) do{int s_v_;scanf("%d", &(s_v_));vec.PB(s_v_);}while(0)
#define MM(co, cim) memset((co), (cim), sizeof((co)))
#define DEB(x) cerr << ">>> " << #x << " : " << x << endl;
#define INF 100000007

#define MAXS 16
#define MAXT 11

int km[][2] = {{-2, -1}, {-2, 1}, {-1, -2}, {-1, 2}, {1, -2}, {1, 2}, {2, -1}, {2, 1}};

int n, m, sx, sy, t, ts, vcnt;
char c;
int d[MAXS*MAXS][MAXS*MAXS];
vector<int> terms;
int dp[MAXS*MAXS][1 << MAXT];

int gv (int x, int y)
{
  return x * m + y; 
}

int isv (int x, int y)
{
  return (x >= 0 && x < n && y >= 0 && y < m);
}

int main ()
{
  //load grid structure
  R2(n, m);
  vcnt = n * m;
  FOR(i, 0, vcnt) FOR(j, 0, vcnt) d[i][j] = INF;
  FOR(i, 0, vcnt) d[i][i] = 0;
  //load terminals
  R2(sx, sy);
  --sx;--sy;
  terms.PB(gv(sx, sy));
  scanf(" %c", &c);
  R1(t);
  FOR(i, 0, t)
  {
    int x, y;
    R2(x, y);
    --x;--y;
    terms.PB(gv(x, y));
    if (c == 'B' && ((sx + sy) % 2) != ((x + y) % 2))
    {
      printf("-1\n");
      return 0;
    }
  }
  ts = terms.size();
  //translate grid->graph
  if (c == 'K')
  {
    FOR(i, 0, n) FOR(j, 0, m)
    {
      FOR(di, -1, 2) FOR(dj, -1, 2) if (isv(i + di, j + dj))
      {
        int x = gv(i,j);
        int y = gv(i + di, j + dj);
        if (x != y) d[x][y] = d[y][x] = 1;
      }
    }
  }
  else if (c == 'R')
  {
    FOR(i, 0, n) FOR(j, 0, m)
    {
      FOR(di, -1, 2) FOR(dj, -1, 2) if (di * di + dj * dj == 1)
      {
        int cm = 0;
        while (isv(i + di * cm, j + dj * cm))
        {
          int x = gv(i,j);
          int y = gv(i + di * cm, j + dj * cm);
          if (x != y) d[x][y] = d[y][x] = 1;
          ++cm;
        }
      }
    }
  }
  else if (c == 'Q')
  {
    FOR(i, 0, n) FOR(j, 0, m)
    {
      FOR(di, -1, 2) FOR(dj, -1, 2) if (di * di + dj * dj > 0)
      {
        int cm = 0;
        while (isv(i + di * cm, j + dj * cm))
        {
          int x = gv(i,j);
          int y = gv(i + di * cm, j + dj * cm);
          if (x != y) d[x][y] = d[y][x] = 1;
          ++cm;
        }
      }
    }
  }
  else if (c == 'B')
  {
    FOR(i, 0, n) FOR(j, 0, m)
    {
      FOR(di, -1, 2) FOR(dj, -1, 2) if (di * di + dj * dj > 1)
      {
        int cm = 0;
        while (isv(i + di * cm, j + dj * cm))
        {
          int x = gv(i,j);
          int y = gv(i + di * cm, j + dj * cm);
          if (x != y) d[x][y] = d[y][x] = 1;
          ++cm;
        }
      }
    } 
  }
  else
  {
    FOR(i, 0, n) FOR(j, 0, m)
    {
      FOR(k, 0, 8) if (isv(i + km[k][0], j + km[k][1]))
      {
        int x = gv(i,j);
        int y = gv(i + km[k][0], j + km[k][1]);
        if (x != y) d[x][y] = d[y][x] = 1;
      }
    }
  }
  //fw
  FOR(k, 0, vcnt) FOR(i, 0, vcnt) FOR(j, 0, vcnt) d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
  
  //steiner
  FOR(i, 0, vcnt) FOR(j, 0, 1 << ts) dp[i][j] = INF;
  FOR(i, 0, ts) dp[terms[i]][1 << i] = 0;
  FOR(mask, 0, 1 << ts)
  {
    FOR(i, 0, vcnt)
    {
      for (int submask = mask; submask > 0; submask = (submask - 1) & mask)
      {
        dp[i][mask] = min(dp[i][mask], dp[i][submask] + dp[i][mask - submask]); 
      }
      FOR(j, 0, vcnt) dp[j][mask] = min(dp[j][mask], dp[i][mask] + d[i][j]);
    }
  }
  //get result
  int res = dp[gv(sx, sy)][(1 << ts) - 1];
  printf("%d\n", res < INF ? res : -1);
  return 0;
}