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 <iostream>
#include <algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<iomanip>
#define sqr(x) (x)*(x)
using namespace std;
int n,m,k,i,j,l,mod=1e9+7,ans,dp[2][1<<20],ned1[25],ned2[25],put[1<<20],x,y,z;
int bi[1<<20][22],cnt;
int main()
{
	cin>>n>>m>>k;
	for (i=1;i<=k;i++)
	{
		cin>>x>>y>>z;
		y--;
		ned1[x]|=(1<<y);
		ned2[x]|=(1<<y)*z;
	}
	n--;
	for (i=0;i<(1<<n);i++)
	{
		for (j=0;j<n;j++) bi[i][j]=(1<<n);
		cnt=0;
		for (j=1;j<n;j++)
		{
			if (((1<<j)&i)) cnt++;
			if (((1<<j)&i)&&(!((1<<(j-1))&i)))
			{
				bi[i][cnt]=(i^(1<<j)^(1<<(j-1)));
			}
		}
	}
	/*for (i=0;i<(1<<n);i++)
	{
		cerr<<i<<':';
		for (j=0;j<n;j++)
		{
			cerr<<bi[i][j]<<' ';
		}
		cerr<<endl;
	}*/
	for (i=0;i<(1<<n);i++)
	{
		//put[i]=(~((~i)^((~i)&(-(~i)))));
		put[i]=(1<<n);
		for (j=n-1;j>=0;j--)
		{
			if (!(i&(1<<j)))
			{
				put[i]=i^(1<<j);
				break;
			}
		}
	}
	/*for (i=0;i<(1<<n);i++) cerr<<put[i]<<' ';
	cerr<<endl;*/
	dp[0][0]=1;
	for (i=1;i<=m;i++)
	{
		memcpy(dp[i&1],dp[(i-1)&1],sizeof(dp[i&1]));
		for (j=0;j<(1<<n);j++)
		{
			dp[i&1][put[j]]+=dp[i&1][j];
			dp[i&1][put[j]]%=mod;
		}
		/*for (j=0;j<(1<<n);j++)
		{
			//if ((ned1[i]&j)!=ned2[i]) dp[i&1][j]=0;
			cerr<<dp[i&1][j]<<' ';
		}*/
		//cerr<<endl;
		for (l=0;l<n;l++)
		{
			for (j=(1<<n)-1;j>=0;j--)
			{
				dp[i&1][bi[j][l]]+=dp[i&1][j];
				dp[i&1][bi[j][l]]%=mod;
			}
		}
		for (j=0;j<(1<<n);j++)
		{
			if ((ned1[i]&j)!=ned2[i]) dp[i&1][j]=0;
			//cerr<<dp[i&1][j]<<' ';
		}
		//cerr<<endl;
	}
	for (i=0;i<(1<<n);i++)
	{
		ans+=dp[m&1][i];
		ans%=mod;
	}
	cout<<ans;
	return 0;
}