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 L long long
#define vi vector<int>
#define pb push_back
#define pi pair<int,int>
#define pii pair<pi,int>
#define aa first
#define bb second
#define xx aa.aa
#define yy aa.bb
#define zz bb
#define mp make_pair
#define mpp(a,b,c) mp(mp(a,b),c)
using namespace std;
const int q=1000000007;
int n,m,p,l[30][30],r[30][30],f[1050000],g[1050000],h[1050000][25];
inline void add(int &a,int b)
{
	a=(a+b)%q;
}
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int i,j,k,u;
	scanf("%d%d%d",&n,&m,&p);
	n--;
	for(i=1;i<=m;i++)
	  for(j=0;j<n;j++)
	    r[i][j]=1;
	while(p--)
	  {
	   scanf("%d%d%d",&i,&j,&k);
	   l[i][j-1]=r[i][j-1]=k;
	  }
	for(i=0;i<(1<<n);i++)
	  for(j=0;j<n;j++)
	    {
		 for(k=j+1;k<n;k++)
		   if(i&(1<<k))
		     break;
		 h[i][j]=(k==n?i:(i^(1<<k)))^(1<<j);
		}
	f[0]=1;
	for(i=1;i<=m;i++)
	  for(j=0;j<n;j++)
	    {
		 memset(g,0,sizeof(g));
		 for(k=0;k<(1<<n);k++)
		   for(u=l[i][j];u<=r[i][j];u++)
		     if(u>=((k>>j)&1))
		       add(g[(u==((k>>j)&1))?k:h[k][j]],f[k]);
		 memcpy(f,g,sizeof(f));
		}
	for(i=0,k=0;i<(1<<n);i++)
	  add(k,f[i]);
	printf("%d\n",k);
	return 0;
}