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 <cstdio>
#include <cstring>
#include <cassert>
#include <cctype>
using namespace std;
typedef long long lint;
#define cout cerr
#define ni (next_num<int>())
template<class T>inline T next_num(){
	T i=0;char c;
	while(!isdigit(c=getchar())&&c!='-');
	bool flag=c=='-';
	flag?c=getchar():0;
	while(i=i*10-'0'+c,isdigit(c=getchar()));
	return flag?-i:i;
}
template<class T1,class T2>inline void apmax(T1 &a,const T2 &b){if(a<b)a=b;}
template<class T1,class T2>inline void apmin(T1 &a,const T2 &b){if(b<a)a=b;}
const int N=19,O=1000000007;
int f[1<<N];
int a[N+2][N];
int main(){
	int n=ni-1,m=ni;
	memset(a,-1,sizeof(a));
	for(int tot=ni,u,v;tot--;u=ni,v=ni-1,a[u][v]=ni);
	memset(f,0,sizeof(f));
	f[0]=1;
	for(int i=1;i<=m;i++){
		for(int j=0;j<n;j++){
			for(int s=(1<<n)-1;s>=0;s--){
				int F=f[s];
				if(F){
					f[s]=((s>>j)&1)==0&&a[i][j]!=1?F:0;//go left
					if(a[i][j]!=0){//go right
						if((s>>j)&1){
							f[s]=F;
						}else{
							int lo=s&((1<<j)-1),t=s^lo;
							(f[(t^(t&-t))|(1<<j)|lo]+=F)%=O;
						}
					}
				}
			}
		}
	}
	lint ans=0;
	for(int s=0,ts=1<<n;s<ts;s++){
		ans+=f[s];
	}
	printf("%lld\n",ans%O);
	return 0;
}