// Solution to KenKen (assumes all target values > 0)
import java.util.*;

class Pair {
  public int r,c;
  public Pair(int r, int c) {
    this.r = r; this.c = c;
  }
}

public class BKenKen1 {
  public static int n,m,t,count;
  public static String op;
  public static int[][] grid;
  public static boolean[][] rowused, colused;
  public static Scanner in;
  public static Stack<Pair> sect;
  public static Stack<Pair> hold;

  public static void main(String[] args) {
    in = new Scanner(System.in);
      n = in.nextInt();
      m = in.nextInt();
      t = in.nextInt();
      op = in.next();
      grid = new int[n][n];
      rowused = new boolean[n][n];
      colused = new boolean[n][n];
      for (int i = 0; i < n; i++) {
        Arrays.fill(grid[i],-1);
        Arrays.fill(rowused[i],false);
        Arrays.fill(colused[i],false);
      }
      sect = new Stack<Pair>();
      hold = new Stack<Pair>();
      for (int i = 0; i < m; i++) {
        int r = in.nextInt();
        int c = in.nextInt();
        sect.push(new Pair(r-1,c-1));
        grid[r-1][c-1] = 0;
      }
/*
      System.out.println("Case " + casenum + ":");
      for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
          System.out.printf("%2d",grid[i][j]);
        }
        System.out.println();
      }
*/
      count = 0;
      switch (op.charAt(0)) {
        case '+':
          solveplus();
          break;
        case '-':
          solveminus();
          break;
        case '*':
          solvestar();
          break;
        case '/':
          solvediv();
          break;
        default: System.out.println("****ERROR: op = " + op);
      }
      System.out.println(count);
  }

  public static void solveplus() {
    // Base case--we just placed the last number:
    if (sect.empty()) {
      if (t==0) {
        count++;
/*
// DEBUG:
System.out.println("Solution:");
        for (Pair x:hold) {
            System.out.println(x.r+","+x.c+": "+grid[x.r][x.c]);
        }
*/
      }
      return;
    }
    // Get next unfilled cell and fill it in all possible ways:
    Pair p = sect.pop();
    hold.push(p);
    for (int i = 1; i <= n; i++) {
      // Make sure sum is still achievable:
      if (i > t) break;
      // Make sure row constraint is not violated:
      if (rowused[p.r][i-1]) continue;
      // Make sure col constraint is not violated:
      if (colused[p.c][i-1]) continue;
      t = t-i;
      rowused[p.r][i-1] = true;
      colused[p.c][i-1] = true;
      int temp = grid[p.r][p.c];
      grid[p.r][p.c] = i;
      solveplus();
      t = t+i;
      rowused[p.r][i-1] = false;
      colused[p.c][i-1] = false;
      grid[p.r][p.c] = temp;
    }
    hold.pop();
    sect.push(p);
  }

  public static void solvestar() {
    // Base case--we just placed the last number:
    if (sect.empty()) {
      if (t == 1) {
        count++;
/*
// DEBUG:
System.out.println("Solution:");
for (Pair x:hold) {
    System.out.println(x.r+","+x.c+": "+grid[x.r][x.c]);
}
*/
      }
      return;
    }
    // Get next unfilled cell and fill it in all possible ways:
    Pair p = sect.pop();
    hold.push(p);
    for (int i = 1; i <= n; i++) {
      // Make sure product is still achievable:
      if (i > t) break;
      // Make sure i is a divisor of t:
      if (t%i != 0) continue;
      // Make sure row constraint is not violated:
      if (rowused[p.r][i-1]) continue;
      // Make sure col constraint is not violated:
      if (colused[p.c][i-1]) continue;
      t = t/i;
      rowused[p.r][i-1] = true;
      colused[p.c][i-1] = true;
      int temp = grid[p.r][p.c];
      grid[p.r][p.c] = i;
      solvestar();
      t = t*i;
      rowused[p.r][i-1] = false;
      colused[p.c][i-1] = false;
      grid[p.r][p.c] = temp;
    }
    hold.pop();
    sect.push(p);
  }

  public static void solveminus() {
    // solve cases where first < second, double count. Assume no "zero"
    // target.
    for (int i = 1; i < n; i++) {
      if (t+i <= n) count+=2;
      else break;
    }
    return;
  }

  public static void solvediv() {
    // solve cases where first < second, double count. Assume no "one"
    // target (since values must be distinct).
    for (int i = 1; i < Math.min(t*i,n); i++) {
      if (t*i <= n)
        count+=2;
      else
        break;
    }
    return;
  }
}