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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define N 1000010
#define M 21
#define mod 1000000007
using namespace std;
int all,n,p[N],g[N][M];
inline int updv(int x,int p)
{
	return p=n-p+1,x|(1<<(p-1));
}
inline int delv(int x,int p)
{
	return p=n-p+1,x&(all^(1<<(p-1)));
}
inline int getv(int x,int p)
{
	return p=n-p+1,(x&(1<<(p-1)))>>(p-1);
}
inline int show(int x)
{
	for(int i=1;i<=n;i++) cout<<getv(x,i);
	cout<<endl;return 0;
}
int x[M],y[M],f[N];
int main()
{
	int m,k;scanf("%d%d%d",&n,&m,&k);
	n--,all=(1<<n)-1;
	for(int i=1;i<=k;i++)
	{
		int a,b,c;scanf("%d%d%d",&a,&b,&c);
		x[a]=updv(x[a],b),(c?y[a]=updv(y[a],b):0);
	}
	for(int i=0;i<=all;i++)
	{
		for(int j=1,k=0;j<=n;j++)
			if(getv(i,j))
			{
				k++;
				if(!getv(i,j-1))
					g[i][k]=updv(delv(i,j),j-1);
			}
		for(int j=n;j>=1;j--)
			if(!getv(i,j))
			{	p[i]=updv(i,j);break;	}
	}
	f[0]=1;
	for(int i=1;i<=m;i++)
	{
		for(int j=0;j<=all;j++)
			if(f[j]&&p[j]) (f[p[j]]+=f[j])%=mod;
		for(int t=1;t<=n;t++)
			for(int j=0;j<=all;j++)
				if(f[j]&&g[j][t]) (f[g[j][t]]+=f[j])%=mod;
		for(int j=0;j<=all;j++)
			if((x[i]&j)^y[i]) f[j]=0;
	}
	int ans=0;
	for(int i=0;i<=all;i++) (ans+=f[i])%=mod;
	return !printf("%d\n",ans);
}