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>
#define mod 1000000007
using namespace std;
inline void Add(int& a,int b)
{
	a=(a+b)%mod;return;
}
int need1[25],need2[25],put[1048580],f[1048580];
int go[1048580][25];
int main(void)
{
	int i,j,S,a,b,c,n,m,k,ans=0;
	scanf("%d%d%d",&n,&m,&k);--n;
	for(i=1;i<=k;++i)
	{
		scanf("%d%d%d",&a,&b,&c);--a;--b;
		need1[a]|=(1<<b);
		need2[a]|=(1<<b)*c;
	}
	memset(go,-1,sizeof(go));
	for(S=0;S<(1<<n);++S)
	{
		int id=0;
		for(i=0;i<n;++i)
		{
			if(S&(1<<i))
			{
				if(i>0&&!(S&(1<<(i-1))))go[S][id]=S^(1<<i)^(1<<(i-1));
				++id;
			}
		}
	}
	for(S=0;S<(1<<n);++S)
	{
		put[S]=-1;
		for(i=n-1;i>=0;--i)
		{
			if(S&(1<<i))continue;
			put[S]=S^(1<<i);
			break;
		}
	}
	f[0]=1;
	for(i=0;i<m;++i)
	{
		for(S=0;S<(1<<n);++S)Add(f[put[S]],f[S]);
		for(j=0;j<n;++j)
		{
			for(S=(1<<n)-1;S>=0;--S)Add(f[go[S][j]],f[S]);
		}
		for(S=0;S<(1<<n);++S)if((S&need1[i])!=need2[i])f[S]=0;
	}
	for(S=0;S<(1<<n);++S)Add(ans,f[S]);
	printf("%d\n",ans);
	return 0;
}
