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
#include<bits/stdc++.h>
#define mo 1000000007
//#define add(a,b) (a+b>=mo)?(a+=b-mo):(a+=b)
#define lowbit(s) ((s)&(-(s))) 
using namespace std;
int f[2][22][1<<20],a[22][22];
inline void add(int &a,int b){(a+b>=mo)?(a+=b-mo):(a+=b);}
int main()
{
	int n,m,q;scanf("%d%d%d",&n,&m,&q);
	memset(a,255,sizeof(a));
	while (q--)
	{
		int u,v,w;scanf("%d%d%d",&u,&v,&w);
		a[u][v]=w;
	}
	f[1][0][0]=1;
	for (int i=1;i<=m;i++)
	{
		memset(f[(i&1)^1],0,sizeof(f[(i&1)^1]));
		for (int j=0;j<n-1;j++)
		{
			for (int s=0;s<(1<<(n-1));s++)
			{
				add(f[i&1][j+1][s],f[i&1][j][s]*((a[i][j+1]!=1)&&((s&(1<<j))==0)));
				add(f[i&1][j+1][s+(!(bool)(s&(1<<j)))*(-(lowbit(s>>j)<<j)+(1<<j))],f[i&1][j][s]*(a[i][j+1]!=0));
			}
		}
		for (int s=0;s<(1<<(n-1));s++) add(f[(i&1)^1][0][s],f[i&1][n-1][s]);
	}
	int ans=0;
	for (int s=0;s<(1<<(n-1));s++) add(ans,f[m&1][n-1][s]);
	printf("%d\n",ans);
}
	
	
	
	
	
	
	
	
	
	
	