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<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
ll mod = 1000000007;
ll dp[2][20][1 << 19];
int to[1 << 19][19];
int d[20][20];
int main()
{
	int num, kai, res;
	scanf("%d%d%d", &num, &kai, &res);
	for (int i = 0; i < res; i++)
	{
		int za, zb, zc;
		scanf("%d%d%d", &za, &zb, &zc);
		za--, zb--;
		if (zc == 0)d[za][zb] = -1;
		else d[za][zb] = 1;
	}
	num--;
	for (int p = 0; p < (1 << num); p++)
	{
		for (int i = 0; i < num; i++)
		{
			if (p&(1 << i))continue;
			int t = -1;
			for (int j = i + 1; j < num; j++)
			{
				if (p&(1 << j))
				{
					t = j;
					break;
				}
			}
			if (t == -1)to[p][i] = p + (1 << i);
			else to[p][i] = p + (1 << i) - (1 << t);
			//printf("%d %d %d\n", p, i, to[p][i]);
		}
	}
	dp[0][num][0] = 1;
	int cur = 0;
	for (int i = 0; i < kai; i++)
	{
		for (int j = 0; j < (1 << num); j++)dp[1 - cur][0][j] = dp[cur][num][j];
		for (int j = 0; j <= num; j++)for (int k = 0; k < (1 << num); k++)dp[cur][j][k] = 0;
		cur = 1 - cur;
		for (int j = 0; j < num; j++)
		{
			for (int k = 0; k < (1 << num); k++)
			{
				dp[cur][j + 1][k] += dp[cur][j][k];
				if (dp[cur][j + 1][k] >= mod)dp[cur][j + 1][k] -= mod;
				if ((k&(1 << j)) == 0)
				{
					dp[cur][j + 1][to[k][j]] += dp[cur][j][k];
					if (dp[cur][j + 1][to[k][j]] >= mod)dp[cur][j + 1][to[k][j]] -= mod;
				}
			}
		}
		for (int j = 0; j < num; j++)
		{
			for (int k = 0; k < (1 << num); k++)
			{
				if (d[i][j] == -1 && (k&(1 << j)) != 0)dp[cur][num][k] = 0;
				if (d[i][j] == 1 && (k&(1 << j)) == 0)dp[cur][num][k] = 0;
			}
		}
		//for (int j = 0; j < (1 << num); j++)printf("%lld ", dp[cur][num][j]); printf("\n");
	}
	ll ans = 0;
	for (int i = 0; i < (1 << num); i++)ans = (ans + dp[cur][num][i]) % mod;
	printf("%lld\n", ans);
}