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>
using namespace std;
const int mo=1e9+7;
int n,m,k,f[20][1<<19],S[1<<19][20],B[2][22][22];
void inc(int &x,int y){ x=x+y>=mo ? x+y-mo : x+y;}
int main()
{
	int i,j,l,x,y,z;
	cin>>n>>m>>k; n--;
	for(i=1;i<=k;i++){
		scanf("%d %d %d",&x,&y,&z),y--;
		B[z][x][y]=1;
	  }
	f[0][0]=1;
	for(i=0;i<(1<<n);i++){
		S[i][n]=(1<<n);
		for(j=n-1;j>=0;j--){
			S[i][j]=S[i][j+1];
			if((i>>j)&1)
				S[i][j]=(1<<j);
		  }
	  }
	f[0][0]=1;
	for(l=1;l<=m;l++){
		for(j=0;j<n;j++)
			for(i=0;i<(1<<n);i++){
				y=((i>>j)&1);
				for(x=y;x<2;x++){
					if(B[x^1][l][j]) continue;
					if(x==y) inc(f[j+1][i],f[j][i]);
					else{
						z=i^(1<<j);
						if(S[i][j]<(1<<n)) z^=S[i][j];
						inc(f[j+1][z],f[j][i]);
					  }
				  }
			  }
		for(i=0;i<(1<<n);i++){
			f[0][i]=f[n][i];
			for(j=1;j<=n;j++) f[j][i]=0;
		  }
	  }
	int ans=0;
	for(i=0;i<(1<<n);i++)
		inc(ans,f[0][i]);
	cout<<ans;
	return 0;
}
