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>
using namespace std;
int n,m,k;
int nd1[21],nd2[21];
int go[1<<21][21];
int nxt[1<<21];
int dp[1<<21];
const int md=1e9+7;
int ans;
int main()
{
	scanf("%d%d%d",&n,&m,&k);
	for(int i=1;i<=k;i++)
	{
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		a--;b--;
		nd1[a]|=(1<<b);
		nd2[a]|=(1<<b)*c;
	}
	n--;
	for(int i=0;i<(1<<n);i++)
	{
		int id=0;
		for(int j=0;j<n;j++)
		{
			go[i][j]=-1;
			if(i>>j&1)
			{
				if(j && !(i>>(j-1)&1))
				{
					go[i][id]=i^(1<<j)^(1<<(j-1));
				}
				id++;
			}
		}
	}
	for(int i=0;i<(1<<n);i++)
	{
		for(int j=n-1;j>=0;j--)
		{
			if(i>>j&1)continue;
			nxt[i]=i|(1<<j);
			break;
		}
	}
	dp[0]=1;
	for(int l=0;l<m;l++)
	{
		for(int i=0;i<(1<<n)-1;i++)
		{
			dp[nxt[i]]+=dp[i];
			if(dp[nxt[i]]>=md)dp[nxt[i]]-=md;
		}
		for(int id=0;id<n;id++)
		{
			for(int i=(1<<n)-1;i>=0;i--)
			{
				if(go[i][id]==-1)continue;
				dp[go[i][id]]+=dp[i];
				if(dp[go[i][id]]>=md)dp[go[i][id]]-=md;
			}
		}
		for(int i=0;i<(1<<n);i++)
		{
			if((i&nd1[l])!=nd2[l])dp[i]=0;
		}
	}
	for(int i=0;i<(1<<n);i++)
	{
		ans+=dp[i];
		if(ans>=md)ans-=md; 
	}
	printf("%d",ans);
	return 0;
}