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
#include <bits/stdc++.h>
#define P 1000000007

using namespace std;

long long F[22][1 << 19];

int K[22][1 << 19], n, m, k, S[22][22];

int main()
{
	memset(S, -1, sizeof S), scanf("%d%d%d", &n, &m, &k);
	for(int i = 1; i <= k; i++) {
		int x, y, z;
		scanf("%d%d%d", &x, &y, &z), S[--x][--y] = z;
	}
	for(int i = 0; i < (1 << n - 1); i++)
		for(int lst = -1, j = n - 2; j >= 0; j--) {
			if((i >> j) & 1) lst = j;
			K[j][i] = i + (1 << j) - (lst != -1 ? 1 << lst : 0);
		}
	F[0][0] = 1;
	for(int O = 0; O < m; O++, memcpy(F[0], F[n - 1], sizeof F[0])) {
		for(int T = 0; T < n - 1; T++) {
			memset(F[T + 1], 0, sizeof F[T + 1]);
			for(int i = 0; i < (1 << n - 1); i++) {
				if(S[O][T] != 1 && !((i >> T) & 1)) F[T + 1][i] += F[T][i];
				if(S[O][T] != 0) F[T + 1][K[T][i]] += F[T][i];
			}
		}
		for(int i = 0; i < (1 << n - 1); i++) F[n - 1][i] %= P;
	}
	long long ans = 0;
	for(int i = 0; i < (1 << n - 1); i++) ans = (ans + F[n - 1][i]) % P;
	return printf("%lld\n", ans), 0;
}