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<bits/stdc++.h>
using namespace std;
long long read()
{
	char ch=getchar();long long x=0,ff=1;
	while(ch<'0'||ch>'9') {if(ch=='-') ff=-ff;ch=getchar();}
	while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
	return x*ff;
}
void write(long long aa)
{
	if(aa<0) putchar('-'),aa=-aa;
	if(aa>9) write(aa/10);
	putchar(aa%10+'0');
	return;
}
const int mod=1e9+7;
int n,m,K,N,ff,ans;
int mp[25][25],F[2][1<<20];
int up(int aa,int bb)
{
	if(aa&bb) return aa;
	int A=aa%bb;aa/=bb;
	if(aa) aa-=(aa&-aa);
	return aa*bb+bb+A;
}
int main()
{
	n=read(),m=read(),K=read(),N=(1<<(n-1))-1;
	for(int i=1;i<=K;++i) 
	{
		int x=read(),y=read(),z=read();
		mp[x][y]=z+1;
	}
	F[1][0]=1;
	for(int a=1;a<=m;++a)
	for(int b=1,wz=1;b<n;++b,wz<<=1,ff^=1)
	for(int i=0;i<=N;++i) if(F[ff^1][i])
	{
		if(!(i&wz)&&mp[a][b]!=2) (F[ff][i]+=F[ff^1][i])%=mod;
		if(mp[a][b]!=1) (F[ff][up(i,wz)]+=F[ff^1][i])%=mod;
		F[ff^1][i]=0;
	}
	for(int i=0;i<=N;++i) ans=(ans+F[ff^1][i])%mod;
	write(ans);
	return 0;
}