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).

<image>

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<iostream>
#include<cstdio>
#define MN 20
#define mod 1000000007
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
int f[1<<MN],g[1<<MN],n,m,K,d[MN+5][MN+5],C[1<<MN],ans=0;
inline void R(int&x,int y){x+=y;x>=mod?x-=mod:0;}
int main()
{
	n=read();m=read();K=read();f[0]=1;
	for(int i=1;i<=K;++i) 
	{
		int x=read(),y=read();
		d[x][y]=read()+1;	
	}
	for(int i=1;i<1<<n;++i)
		for(int j=0;j<n;++j) if(i&(1<<j)) {C[i]=i^(1<<j);break;}	
	for(int i=1;i<=m;++i) for(int j=1;j<n;++j) 
	{
		for(int k=0;k<1<<n;++k) if(f[k])
		{
			int pre=k&((1<<j-1)-1),aft=k-pre;
			if(d[i][j]!=2&&!(aft&(1<<j-1))) R(g[k],f[k]);
			if(d[i][j]!=1) R(g[pre|(1<<j-1)|C[aft]],f[k]);
		}
		for(int k=0;k<1<<n;++k) f[k]=g[k],g[k]=0;
	}
	for(int i=0;i<1<<n;++i) R(ans,f[i]);
	cout<<ans;
	return 0;
}