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<algorithm>
#include<math.h>
#include<stdio.h>
#include<string.h>
using namespace std;
const int mod=1000000007;
long long n,m,p;
long long bin[23];
int f[2][23][1048580];
long long c[23][23];
inline int modify(int s,int i){
	if(s&bin[i]){
		return s;
	}
	int w=s&(bin[n]-bin[i]);
	if(w){
		s-=w&(-w);
	}
	return s+bin[i];
}
int main(){
	cin>>n>>m>>p;
	n--;
	for(int i=1;i<=m;i++){
		for(int j=1;j<=n;j++){
			c[i][j]=-1;
		}
	}
	for(int i=1;i<=p;i++){
		int x,y,z;
		cin>>x>>y>>z;
		c[x][y]=z;
	}
	bin[0]=1;
	for(int i=1;i<=n;i++){
		bin[i]=bin[i-1]*2;
	}
	int now=0;
	f[0][0][0]=1;
	for(int i=1;i<=m;i++){
		for(int j=0;j<=n-1;j++){
			for(int s=0;s<=bin[n]-1;s++){
				if(f[now][j][s]){
					if(c[i][j+1]<=0&&!(s&bin[j])){
						f[now][j+1][s]=(f[now][j+1][s]+f[now][j][s])%mod;
					}
					if(c[i][j+1]!=0){
						int x=modify(s,j);
						f[now][j+1][x]=(f[now][j+1][x]+f[now][j][s])%mod;
					}
				}
			}
		}
		memset(f[now^1],0,sizeof(f[now^1]));
		for(int s=0;s<=bin[n]-1;s++){
			f[now^1][0][s]=f[now][n][s];
		}
		now^=1;
	}
	long long ans=0;
	for(int s=0;s<=bin[n]-1;s++){
		ans=(ans+f[now][0][s])%mod;
	}
	cout<<ans<<endl;
	return 0;
}
/*

in:
3 2 1
1 2 0

out:
6

*/