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 <bits/stdc++.h>
#define lowbit(x)  ((x)&(-(x)))
using namespace std;

const int mo=int(1e9+7);
int n,m,k,a[20][20],f[1<<19],ans;

void upd(int &x,int y){if ((x+=y)>=mo)  x-=mo;}

void work()
{
	scanf("%d %d %d",&n,&m,&k);
	for (int i=0; i<m; i++)
		for (int j=0; j<n-1; j++)
			a[i][j]=3;
	for (int i=1,x,y,z; i<=k; i++)
		scanf("%d %d %d",&x,&y,&z),a[x-1][y-1]=1<<z;
	f[0]=1;
	for (int i=0; i<m; i++)
		for (int j=0; j<n-1; j++)
			for (int k=(1<<(n-1))-1; ~k; k--)
				if (f[k])
					{
						if (k&(1<<j))
							{
								if (!(a[i][j]&2))  f[k]=0;
							}else
							{
								if (a[i][j]&2)  upd(f[k^((lowbit(k>>j)|1)<<j)],f[k]);
								if (!(a[i][j]&1))  f[k]=0;
							}
					}
	for (int i=0; i<(1<<(n-1)); i++)  upd(ans,f[i]);
	printf("%d",ans);
}

int main()
{
	work();
	return 0;
}