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<bits/stdc++.h>
const int P=1e9+7;
const int N=1<<20|11111;
int n,m,k,mx;
int lim[25][2];
int F[23][N],D[23][N];
inline int fix(int x){
	return x+(x>>31&P);
}
int main(){
	scanf("%d%d%d",&n,&m,&k);//m lines 
	for(int i=1,a,b,c;i<=k;++i){
		scanf("%d%d%d",&a,&b,&c);
		lim[a][c]|=1<<n-1-b;
	}
	--n;
	if(!n)return puts("1"),0;
	mx=1<<n;
	for(int i=0;i<n;++i){
		for(int j=0;j<mx;++j)if(j>>i&1){
			int x=j,y=i;
			x^=1<<i;
			for(--y;y>=0&&(x>>y&1);--y);
			if(y>=0)x^=1<<y;
			D[i][j]=x;
//			printf("%d,%d %d\n",i,j,x);
		}
	}
	for(int i=0;i<mx;++i)F[n][i]=1;
	for(int i=1;i<=m;++i){
//		printf("(%d %d)\n",lim[i][0],lim[i][1]);
		for(int j=0;j<mx;++j){
			F[0][j]=(j&lim[i][0]|~j&lim[i][1])?0:F[n][j];
//			printf("%d,%d %d\n",i,j,F[0][j]);
		}
		for(int j=0;j<n;++j){
			int*F1=F[j+1],*F0=F[j],*D1=D[j];
			for(int k=0;k<mx;++k){
				F1[k]=F0[k];
				if(k>>j&1)F1[k]=fix(F1[k]+F0[D1[k]]-P);
			}
		}
	}
	printf("%d\n",F[n][mx-1]);
	return 0;
}