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<cstdio>
#include<iostream>
using namespace std;
const int N=22;
const int mod=1e9+7;
const int S=1<<20;
int go[S][N],put[S];
int f[S],n,m,k,a,b,c;
int need1[S],need2[S],ans;
int main()
{
	scanf("%d%d%d",&n,&m,&k);
	for (int i=1;i<=k;i++)
	{
		scanf("%d%d%d",&a,&b,&c);
		need1[a]|=(1<<b-1);
		need2[a]|=c*(1<<b-1);
	}
	n--;
	for (int T=0;T<(1<<n);T++)
	{
		for (int i=0;i<n;i++) go[T][i]=-1;
		int id=0;
		for (int i=0;i<n;i++)
		{
			if (T&(1<<i))
			{
				if (i>0&&!(T&(1<<i-1))) go[T][id]=T^(1<<i)^(1<<i-1);
				id++;
			}
		}
	}
	for (int T=0;T<(1<<n);T++)
	{
		put[T]=-1;
		for (int i=n-1;i>=0;i--)
			if (!(T&(1<<i))) {put[T]=T^(1<<i);break;}
	}
	f[0]=1;
	for (int i=1;i<=m;i++)
	{
		for (int T=0;T<(1<<n);T++)
			if (put[T]!=-1) f[put[T]]=(f[put[T]]+f[T])%mod;
		for (int id=0;id<n;id++)
			for (int T=(1<<n)-1;T>=0;T--)
				if (go[T][id]!=-1) f[go[T][id]]=(f[go[T][id]]+f[T])%mod;
		for (int T=0;T<(1<<n);T++)
			if ((need1[i]&T)!=need2[i]) f[T]=0; 
	}
	for (int i=0;i<(1<<n);i++) ans=(ans+f[i])%mod;
	printf("%d",ans);
}