Problem:
Score : 1600 points

Problem Statement

There are N(N+1)/2 dots arranged to form an equilateral triangle whose sides consist of N dots, as shown below.
The j-th dot from the left in the i-th row from the top is denoted by (i, j) (1 \leq i \leq N, 1 \leq j \leq i).
Also, we will call (i+1, j) immediately lower-left to (i, j), and (i+1, j+1) immediately lower-right to (i, j).

Takahashi is drawing M polygonal lines L_1, L_2, ..., L_M by connecting these dots.
Each L_i starts at (1, 1), and visits the dot that is immediately lower-left or lower-right to the current dots N-1 times.
More formally, there exist X_{i,1}, ..., X_{i,N} such that:

L_i connects the N points (1, X_{i,1}), (2, X_{i,2}), ..., (N, X_{i,N}), in this order.

For each j=1, 2, ..., N-1, either X_{i,j+1} = X_{i,j} or X_{i,j+1} = X_{i,j}+1 holds.

Takahashi would like to draw these lines so that no part of L_{i+1} is to the left of L_{i}.
That is, for each j=1, 2, ..., N, X_{1,j} \leq X_{2,j} \leq ... \leq X_{M,j} must hold.

Additionally, there are K conditions on the shape of the lines that must be followed.
The i-th condition is denoted by (A_i, B_i, C_i), which means:

If C_i=0, L_{A_i} must visit the immediately lower-left dot for the B_i-th move.

If C_i=1, L_{A_i} must visit the immediately lower-right dot for the B_i-th move.

That is, X_{A_i, {B_i}+1} = X_{A_i, B_i} + C_i must hold.

In how many ways can Takahashi draw M polygonal lines? Find the count modulo 1000000007.

Notes

Before submission, it is strongly recommended to measure the execution time of your code using "Custom Test".

Constraints

1 \leq N \leq 20

1 \leq M \leq 20

0 \leq K \leq (N-1)M

1 \leq A_i \leq M

1 \leq B_i \leq N-1

C_i = 0 or 1

No pair appears more than once as (A_i, B_i).

Input

Input is given from Standard Input in the following format:

N M K
A_1 B_1 C_1
A_2 B_2 C_2
:
A_K B_K C_K

Output

Print the number of ways for Takahashi to draw M polygonal lines, modulo 1000000007.

Sample Input 1

3 2 1
1 2 0

Sample Output 1

6

There are six ways to draw lines, as shown below. Here, red lines represent L_1, and green lines represent L_2.

Sample Input 2

3 2 2
1 1 1
2 1 0

Sample Output 2

0

Sample Input 3

5 4 2
1 3 1
4 2 0

Sample Output 3

172

Sample Input 4

20 20 0

Sample Output 4

881396682

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

/**
 * @author Pavel Mavrin
 */
public class Main {

    private static final int MOD = (int) (1e9 + 7);

    private void solve() throws IOException {
        int n = nextInt() - 1;
        int m = nextInt();
        int k = nextInt();
        int[][] cc = new int[m][n];
        for (int i = 0; i < k; i++) {
            cc[nextInt() - 1][nextInt() - 1] = nextInt() + 1;
        }
        int[][] d = new int[n + 1][1 << n];
        d[0][0] = 1;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                Arrays.fill(d[j + 1], 0);
                for (int x = 0; x < (1 << n); x++) {
                    if (d[j][x] > 0) {
                        int y = (x >> j) & 1;
                        if (cc[i][j] != 2 && y == 0) {
                            d[j + 1][x] += d[j][x];
                            if (d[j + 1][x] >= MOD) {
                                d[j + 1][x] -= MOD;
                            }
                        }
                        if (cc[i][j] != 1) {
                            int xx = y == 1 ? x : (x - (1 << j)) & x + (1 << j);
                            d[j + 1][xx] += d[j][x];
                            if (d[j + 1][xx] >= MOD) {
                                d[j + 1][xx] -= MOD;
                            }
                        }
                    }
                }
            }
            for (int x = 0; x < (1 << n); x++) {
                d[0][x] = d[n][x];
            }
        }
        int res = 0;
        for (int x = 0; x < (1 << n); x++) {
            res += d[0][x];
            if (res >= MOD) {
                res -= MOD;
            }
        }
        out.println(res);

    }

    // ------------------

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st;
    PrintWriter out = new PrintWriter(System.out);

    String next() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(br.readLine());
        }
        return st.nextToken();
    }

    int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    public static void main(String[] args) throws IOException {
        new Main().run();
    }

    private void run() throws IOException {
        solve();
        out.close();
    }

}