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<cmath>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n,mm,K;
int lim[30],w[30],dp[1<<20],two[1<<19][12];
const int mo=1e9+7;
int main(){
	scanf("%d%d%d",&n,&mm,&K);
	for (int i=1;i<=K;i++){
		int k1,k2,k3; scanf("%d%d%d",&k1,&k2,&k3);
		lim[k1]|=(1<<k2-1); w[k1]|=(k3<<k2-1);
	}
	memset(two,0xff,sizeof two);
	memset(dp,0x00,sizeof dp); dp[0]=1;
	int N=n-1; int m=0;
	for (int i=0;i<(1<<N);i++){
		int len=0;
		for (int j=1;j<N;j++)
			if ((i&(1<<j))&&(i&(1<<j-1))==0) two[i][len++]=i-(1<<j)+(1<<j-1);
		m=max(m,len);
	}
	for (int now=1;now<=mm;now++){
		for (int i=0;i<m;i++)
			for (int j=(1<<N)-1;j>=0;j--)
				if (two[j][i]>=0) dp[two[j][i]]=(dp[two[j][i]]+dp[j])%mo;
		for (int i=0;i<(1<<N);i++)
			for (int j=N-1;j>=0&&(i&(1<<j))==0;j--)
				dp[i+(1<<j)]=(dp[i+(1<<j)]+dp[i])%mo;
		for (int i=0;i<(1<<N);i++)
			if ((i&lim[now])!=w[now]) dp[i]=0;
	//	cout<<lim[now]<<" "<<w[now]<<endl;
	//	for (int i=0;i<(1<<N);i++) cout<<dp[i]<<" "; cout<<endl;
	}
	int ans=0;
	for (int i=0;i<(1<<N);i++) ans=(ans+dp[i])%mo;
	printf("%d\n",ans);
	return 0;
}