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 <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <iostream>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fod(i,a,b) for(int i=a;i>=b;i--)
#define LL long long
#define mo 1000000007
#define M 1048580
#define N 21
using namespace std;
int f[N][M];
int n,m,l,a[N][N],cf[N],qs[M][N][2];
int main()
{
	cin>>n>>m>>l;
	memset(a,255,sizeof(a));
	cf[0]=1;
	fo(i,1,n) cf[i]=cf[i-1]*2;
	fo(i,1,l)
	{
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		a[x][y]=z;
	}
	n--;
	fo(c,0,cf[n]-1)
	{
		int w1=0,w2=0;
		fod(i,n,0)
		{
			qs[c][i][0]=w1,qs[c][i][1]=w2;
			if(i>0)
			{
				if(c&cf[i-1]) w2=i;
				else w1=i; 
			}
		}
	}
	f[0][0]=1;
	LL ans;
	fo(i,1,m)
	{
		fo(j,0,n-1)
		{
			fo(c,0,cf[n]-1)
			{
				if(!f[j][c]) continue;
				int p=qs[c][j][0];
				if(p&&p<=n)
				{
					fo(k,j+1,p-1) 
					{
						if(a[i][k]==0) 
						{
							f[j][c]=0;
							break;
						} 
					}
					if(a[i][p]!=1) 
					{
						(f[p][c]+=f[j][c]);
						if(f[p][c]>=mo) f[p][c]-=mo;
					}
					if(a[i][p]!=0) 
					{
						int q=qs[c][p][1];
						if(q&&q<=n) 
						{
							int v=(c|cf[p-1])-cf[q-1];
							(f[p][v]+=f[j][c]);
							if(f[p][v]>=mo) f[p][v]-=mo;
						}
						else 
						{
							int v=c|cf[p-1];
							(f[p][v]+=f[j][c]);
							if(f[p][v]>=mo) f[p][v]-=mo;
						}
					}
				}
				else 
				{
					fo(k,j+1,n) 
					{
						if(a[i][k]==0) 
						{
							f[j][c]=0;
							break;
						} 
					}
					(f[n][c]+=f[j][c]);
					if(f[n][c]>=mo) f[n][c]-=mo;
				}
				f[j][c]=0;
			}
		}
		if(i==m)
		{
			ans=0;
			fo(c,0,cf[n]-1) (ans+=f[n][c])%=mo;
		}
		else
		fo(c,0,cf[n]-1)
		{
			f[0][c]=f[n][c];
			f[n][c]=0;
		}
	}
	printf("%lld\n",ans);
}