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).

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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 <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int Mod = 1e9+7;
int f[21][1<<19], g[21][1<<19], nxt[21][1<<19];
int n, m, K;
int s0[25], s1[25];
void up(int &x, int y) { x = (x + y) % Mod; }
int main() {
	int i, j, k;
	scanf("%d%d%d", &n, &m, &K);
	for(i = 1; i <= K; i++){
		int x, y, z;
		scanf("%d%d%d", &x, &y, &z);
		if(z == 0) s0[x] |= (1<<y-1);
		else s1[x] |= (1<<y-1);
	}
	n--;
	for(i = 0; i < n; i++){
		for(j = 0; j < 1<<n; j++){
			if(j&(1<<i)){
				nxt[i][j] = j^(1<<i);
				for(k = i+1; k < n; k++){
					if(!(j&(1<<k))){
						nxt[i][j] ^= 1<<k;
						break;
					}
				}
			}
		}
	}
	f[0][0] = 1;
	int mx = (1<<n)-1;
	for(i = 1; i <= m; i++){
		for(j = 0; j < 1<<n; j++) g[n][j] = f[i-1][j];
		for(k = n-1; k >= 0; k--){
			for(j = 0; j < 1<<n; j++){
				g[k][j] = g[k+1][j];
				if(j&(1<<k)) up(g[k][j], g[k+1][nxt[k][j]]);
			}
		}
		for(j = 0; j < 1<<n; j++){
			if(((j^mx)&s0[i]) == s0[i] && (j&s1[i]) == s1[i]) f[i][j] = g[0][j];
		}
	}
	int ans = 0;
	for(i = 0; i < 1<<n; i++) up(ans, f[m][i]);
	printf("%d\n", ans);
	return 0;
}