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<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define MOD 1000000007
void add(int &x,int y){x+=y;if(x>=MOD)x-=MOD;}
inline int read()
{
	int x=0;bool t=false;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')t=true,ch=getchar();
	while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
	return t?-x:x;
}
int n,m,K,ans,Lim[22][22];
int f[2][22][1<<20],nxt[1<<20][22];
int main()
{
	n=read()-1;m=read();K=read();
	memset(Lim,-1,sizeof(Lim));
	for(int i=1;i<=K;++i)
	{
		int a=read(),b=read();
		Lim[a][b-1]=read();
	}
	int S=1<<n;
	for(int i=0;i<S;++i)
	{
		int v=-1;
		for(int j=n-1;~j;--j)
		{
			if(i&(1<<j))v=j;
			nxt[i][j]=v;
		}
	}
	f[0][n][0]=1;
	for(int i=1,nw=1,pw=0;i<=m;++i,nw^=1,pw^=1)
	{
		for(int j=0;j<S;++j)f[nw][0][j]=f[pw][n][j];
		memset(f[pw],0,sizeof(f[pw]));
		for(int j=0;j<n;++j)
			for(int k=0;k<S;++k)
				if(f[nw][j][k])
					for(int l=0;l<=1;++l)
					{
						if((~Lim[i][j])&&l!=Lim[i][j])continue;
						int t=(k>>j)&1;
						if(l==0&&t==1)continue;
						if(l==t)add(f[nw][j+1][k],f[nw][j][k]);
						else
						{
							int p=nxt[k][j];
							if(p==-1)add(f[nw][j+1][k+(1<<j)],f[nw][j][k]);
							else add(f[nw][j+1][k+(1<<j)-(1<<p)],f[nw][j][k]);
						}
					}
	}
	for(int i=0;i<S;++i)add(ans,f[m&1][n][i]);
	printf("%d\n",ans);
	return 0;
}