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:
//ΔAGC017F
#include<iostream>
#include<cstdio>
#include<fstream>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<cmath>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef long long LL;
typedef double DB;
const int N = 22;
const int M = 1<<N;
const int mod = 1e9+7;
int n,m,k,b[N],c[N];
int v[M][N],f[M],dp[M];
int main()
{
	int i,j,x,y,z;
	cin>>n>>k>>m;
	n--;
	while(m--){
		cin>>x>>y>>z;
		x--,y--;
		b[x]|=(1<<y);
		if(z)
			c[x]|=(1<<y);
	}
	m=1<<n;
	for(j=0;j<m;j=j+1){
		z=0;
		for(i=0;i<n;i=i+1)
			v[j][i]=-1;
		for(i=1;i<n;i=i+1){
			x=1<<i,y=1<<(i-1);
			if(j&x){
				if(!(j&y))
					v[j][z]=j^x^y;
				z++;
			}
		}
		f[j]=m;
		for(i=0;i<n;i=i+1)
			if(!(j&(1<<i)))
				f[j]=j^(1<<i);
	}
	dp[0]=1;
	for(z=0;z<k;z=z+1){
		for(j=0;j<m;j=j+1)
			dp[f[j]]=(dp[f[j]]+dp[j])%mod;
		for(i=0;i<n;i=i+1)
			for(j=m-1;j>=0;j=j-1)
				if(v[j][i]>=0)
					dp[v[j][i]]=(dp[v[j][i]]+dp[j])%mod;
		for(j=0;j<m;j=j+1)
			if((j&b[z])!=c[z])
				dp[j]=0;
	}
	z=0;
	for(j=0;j<m;j=j+1)
		z=(z+dp[j])%mod;
	cout<<z<<endl;
	return 0;
}