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<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<iostream>
#define mmod 1000000007
#define N 22
#define M (1<<20)+10
#define lowbit(x) (x&(-x))
using namespace std;
int n,m,lim[N][N],f[M],g[M],ans;
void init()
{
	int k;scanf("%d%d%d",&n,&m,&k);
	n--;
	memset(lim,-1,sizeof(lim));
	while(k--)
	{
		int a,b,c;scanf("%d%d%d",&a,&b,&c);
		lim[a][b]=c;
	}
}
void upd(int &x,int y)
{
	x=(x+y)%mmod;
}
void solve()
{
	int ss=(1<<n)-1;
	f[0]=1;
	for(int i=1;i<=m;i++)
		for(int j=1;j<=n;j++)
		{
			for(int k=0;k<=ss;k++) g[k]=0;
			for(int k=0;k<=ss;k++)
			{
				if(f[k]==0) continue;
				if(lim[i][j]==0 && (k&(1<<(j-1)))) continue;
				if(lim[i][j]!=1 && (k&(1<<(j-1)))==0) {int s=k;upd(g[s],f[k]);}
				if(lim[i][j]!=0)
				{
					int s=k;
					if((k&(1<<(j-1)))==0) 
					{
						int t=s&((1<<j)-1);
						t|=1<<(j-1);
						s>>=j;
						s-=lowbit(s);
						s<<=j;
						s+=t;
					}
					upd(g[s],f[k]);
				}
			}
			for(int k=0;k<=ss;k++) f[k]=g[k];
		}
	for(int i=0;i<=ss;i++) upd(ans,f[i]);
	printf("%d\n",ans);
}
int main()
{
	init();
	solve();
	return 0;
}