Problem: 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).

<image>

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.

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.

Examples

Input

3 2 1
1 2 0


Output

6


Input

3 2 2
1 1 1
2 1 0


Output

0


Input

5 4 2
1 3 1
4 2 0


Output

172


Input

20 20 0


Output

881396682

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

const int mod = 1e9 + 7;

int N, M, K;
int fix[22][22], prec[1 << 19][22], dp[2][1 << 19];

int main() {
    scanf("%d %d %d", &N, &M, &K);
    N--;

    memset(fix, -1, sizeof(fix));
    for(int i = 0; i < K; i++) {
        int a, b, c; scanf("%d %d %d", &a, &b, &c);
        a--; b--;
        fix[a][b] = c;
    }

    for(int i = 0; i < (1 << N); i++) {
        int la = N;
        for(int j = N - 1; j >= 0; j--) {
            if(i & (1 << j)) la = j;
            prec[i][j] = la;
        }
    }

    int c = (N * M) & 1;
    for(int i = 0; i < (1 << N); i++) {
        dp[c][i] = 1;
    }
    for(int i = M - 1; i >= 0; i--) {
        for(int j = N - 1; j >= 0; j--) {
            int c = (i * N + j) & 1;
            int p = c ^ 1;

            for(int k = 0; k < (1 << N); k++) {
                dp[c][k] = 0;

                if(fix[i][j] != 0) {
                    int t = prec[k][j];
                    if(t == N) {
                        dp[c][k] += dp[p][ k ^ (1 << j) ];
                        dp[c][k] %= mod;
                    }
                    else {
                        dp[c][k] += dp[p][ k ^ (1 << t) ^ (1 << j) ];
                        dp[c][k] %= mod;
                    }
                }
                if(fix[i][j] != 1 && !(k & (1 << j))) {
                    dp[c][k] += dp[p][k];
                    dp[c][k] %= mod;
                }

                //cout << i << ' ' << j << ' ' << k << ' ' << dp[c][k] << endl;
            }
        }
    }
    printf("%d", dp[0][0]);
}