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>
#define ad(x,y) (x=((x)+(y))%1000000007)
#define N 21
#define M 1050005
using namespace std;
 
int n,m,cnt,p0[N],p1[N],f[M][N],dp[M];
int main(){
	scanf("%d%d%d",&n,&m,&cnt); n--;
	int i,j,k,x,y,z;
	for (i=1; i<=cnt; i++){
		scanf("%d%d%d",&x,&y,&z);
		y=n-y;
		if (z) p1[x]|=1<<y; else p0[x]|=1<<y;
	}
	int tot=1<<n;
	for (i=0; i<tot; i++) if ((i&p1[1])==p1[1] && (i&p0[1])==0) dp[i]=1;
	for (i=0; i<tot; i++)
		for (j=0; j<n; j++){
			for (k=j-1; k>=0; k--) if (!(i>>k&1)) break;
			f[i][j]=(k<0)?i:(i|1<<k);
		}
	for (i=2; i<=m; i++){
		for (j=0; j<n; j++)
			for (k=0; k<tot; k++) if (k>>j&1) ad(dp[k],dp[f[k^(1<<j)][j]]);
		for (j=0; j<tot; j++) if ((j&p1[i])!=p1[i] || (j&p0[i])!=0) dp[j]=0;
	}
	for (j=1; j<tot; j++) ad(dp[0],dp[j]);
	printf("%d\n",dp[0]);
	return 0;
}