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 <cstring>
#include <cstdio>

using namespace std;

const int N=19;
const int M=20;
const int S=1<<N;
const int P=1000000007;

int res[M],val[M];
int f[2][N][S];
int n,m,K,ans,all;

inline void update(int &x,int y){(x+=y)%=P;}

inline void trans(int *tar,int *src,int id,int d)
{
	for (int s=0;s<=all;++s)
		if (src[s])
		{
			int bit=(s>>d)&1;
			for (int i=0;i<2;++i)
				if (!((res[id]>>d)&1)||!(((val[id]>>d)&1)^i))
					if (bit<i)
					{
						int rem=(s>>d+1)<<d+1,lb=rem&-rem;
						update(tar[s^lb^(1<<d)],src[s]);
					}
					else if (bit==i) update(tar[s],src[s]);
		}
}

void dp()
{
	for (int s_=all^res[0];s_;s_=(s_-1)&(all^res[0])) update(f[0][n-1][s_|val[0]],1);
	update(f[0][n-1][val[0]],1);
	for (int i=1,src,tar;i<m;++i)
	{
		tar=i&1,src=tar^1,memset(f[tar],0,sizeof f[tar]);
		trans(f[tar][0],f[src][n-1],i,0);
		for (int j=1;j<n;++j) trans(f[tar][j],f[tar][j-1],i,j);
	}
	ans=0;
	for (int s_=all^res[m-1];s_;s_=(s_-1)&(all^res[m-1])) update(ans,f[(m-1)&1][n-1][s_|val[m-1]]);
	update(ans,f[(m-1)&1][n-1][val[m-1]]);
}

int main()
{
	//freopen("zigzag.in","r",stdin),freopen("zigzag.out","w",stdout);
	scanf("%d%d%d",&n,&m,&K),--n,all=(1<<n)-1;
	for (int i=1,x,y,z;i<=K;++i) scanf("%d%d%d",&x,&y,&z),--x,--y,res[x]|=1<<y,val[x]|=(1<<y)*z;
	dp(),printf("%d\n",ans);
	//fclose(stdin),fclose(stdout);
	return 0;
}