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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

#define fo(i,j,l) for(int i=j;i<=l;++i)
#define fd(i,j,l) for(int i=j;i>=l;--i)

using namespace std;
typedef long long LL;
const LL N=21,M=(1<<19)+3,mo=1e9+7;

int dir[N][N],m2[N<<1],nex[N<<1];
int n,m,k,a,b,c;
LL f[2][N][M],g[N][M];

int main()
{
	cin>>n>>m>>k; --n;
	fo(i,1,k){
		scanf("%d%d%d",&a,&b,&c);
		dir[a][b]=c+1;
	}
	int u=0,v;
	f[0][0][0]=1;
	m2[1]=1;
	fo(i,2,n+1)m2[i]=m2[i-1]<<1;
	int zd=(m2[n]<<1)-1;
	fo(p,1,m){
		v=u^1;
		fo(l,0,n)fo(i,0,zd)f[v][l][i]=0;
		fo(l,0,n)fo(i,0,zd)g[l][i]=0;
		fd(i,zd,0){
			LL ok=f[u][0][i]%mo; nex[n]=n+1;
			fd(l,n,2)if(i&m2[l])nex[l-1]=l;else nex[l-1]=nex[l];
    		if(!i)g[0][0]=ok;
			fo(l,1,n){
				if((i&m2[l])==0&&dir[p][l]!=1&&nex[l]!=n+1)
				f[u][l][(i^m2[nex[l]])^m2[l]]=f[u][l][(i^m2[nex[l]])^m2[l]]+ok;
				ok=(f[u][l][i]+ok)%mo;
				if(dir[p][l]&&(((i&m2[l])>0)-(dir[p][l]==2)))break;
				if(nex[l]==n+1&&(i&m2[l])>0)g[l][i]=ok;
			}
		}
		fo(i,0,n-1)
		fo(l,0,zd)if(g[i][l]){
			g[i][l]=g[i][l]%mo;
			if(dir[p][i+1]!=2)g[i+1][l]=g[i+1][l]+g[i][l];
			if(dir[p][i+1]!=1)g[i+1][l^m2[i+1]]=g[i+1][l^m2[i+1]]+g[i][l];
		}
		fo(i,0,zd)f[v][0][i]=g[n][i];
		u=v;
	}
	LL ans=0;
	fo(i,0,zd)ans=(ans+f[v][0][i])%mo;
	cout<<ans;
}