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<cstring>
#include<algorithm>
using namespace std;
const int N=23,S=1<<20,mo=1000000007;
template<typename TP>
inline void read(TP &kk)
{
	TP ret=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){ret=ret*10+ch-'0';ch=getchar();}
	kk=ret*f;
}
int n,m,k;
int sc[S],ful,l1[N],l0[N];
int to[S][N],ne[S];
int dp[S];
int ans;

int xi,yi,zi;
signed main()
{
//	freopen("zigzag.in","r",stdin);
//	freopen("zigzag.out","w",stdout);
	memset(ne,-1,sizeof(ne));memset(to,-1,sizeof(to));
	read(n),read(m),read(k);n--;
	ful=(1<<n)-1;
	for(int i=1;i<=k;i++)
	{
		read(xi),read(yi),read(zi);
		yi--;
		l0[xi]|=(1<<yi);
		if(zi)l1[xi]|=(1<<yi);
	}
	for(int s=0;s<=ful;s++)
		for(int j=0;j<n;j++)
			if(s&(1<<j))
			{
				if(j&&!(s&(1<<(j-1))))
				{
					to[s][sc[s]]=s^(1<<j)^(1<<(j-1));
				}
				sc[s]++;
			}
	for(int s=0;s<=ful;s++)
		for(int j=n-1;j>=0;j--)
			if(!(s&(1<<j))){ne[s]=s^(1<<j);break;}
	dp[0]=1;
	for(int i=1;i<=m;i++)
	{
		for(int s=0;s<=ful;s++)
			if(ne[s]!=-1)
				(dp[ne[s]]+=dp[s])%=mo;
		for(int b=0;b<n;b++)
			for(int s=ful;s>=0;s--)
				if(to[s][b]!=-1)
					(dp[to[s][b]]+=dp[s])%=mo;
		for(int s=0;s<=ful;s++)
			if((s&l0[i])!=l1[i]) dp[s]=0;
	}
	for(int s=0;s<=ful;s++) (ans+=dp[s])%=mo;
	printf("%d\n",ans);
	return 0;
}