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 <cstring>
#include <algorithm>

using namespace std;

const int N=20,M=524288,mo=1e9+7;

typedef long long LL;

int n,m,K,lim[N][N],g[M],f[2][M],p,q,nxt[N][M],h[N][M],ans,T;

int main()
{
	scanf("%d%d%d",&n,&m,&K); T=1<<n-1;
	memset(lim,255,sizeof(lim));
	for (int a,b;K--;scanf("%d",&lim[a-1][b-1])) scanf("%d%d",&a,&b);
	for (int i=1,j,k,la;i<M;i++)
	{
		for (j=T>>1,k=n-2,la=0;k>=0;k--,j>>=1)
		{
			nxt[k][i]=la;
			if (j&i) la=j;
		}
		//for (k=0,j=1,la=0;k<n-1;k++,j<<=1) cnt[k][i]=la+=((j&i)>0);
	}
	g[0]=1;
	for (int i=0,j,k,st;i<m;i++)
	{
		p=0; q=1;
		memcpy(f[0],g,sizeof(g));
		memset(h,0,sizeof(h));
		for (j=0,k=1;j<n-1;j++,p^=1,q^=1,k<<=1)
		{
			memset(f[q],0,sizeof(f[q]));
			for (st=0;st<T;st++) if (f[p][st]>0)
			{
				if (lim[i][j]==-1)
				{
					f[q][st]=(f[q][st]+f[p][st])%mo;
					if (!(k&st))
					{
						if (!nxt[j][st]) h[j+1][st|k]=(h[j+1][st|k]+f[p][st])%mo;
						else f[q][(st|k)^nxt[j][st]]=(f[q][(st|k)^nxt[j][st]]+f[p][st])%mo;
					}
				}else if (lim[i][j]==0)
				{
					if ((k&st)) continue;
					f[q][st]=(f[q][st]+f[p][st])%mo;
				}else
				{
					if ((k&st))
					{
						f[q][st]=(f[q][st]+f[p][st])%mo;
					}else
					{
						if (!nxt[j][st]) h[j+1][st|k]=(h[j+1][st|k]+f[p][st])%mo;
						else f[q][(st|k)^nxt[j][st]]=(f[q][(st|k)^nxt[j][st]]+f[p][st])%mo;
					}
				}
			}
		}
		for (j=2,k=2;j<n;j++,k<<=1)
		{
			for (st=0;st<k;st++) if (h[j-1][st]>0)
			{
				if (lim[i][j-1]!=1) h[j][st]=(h[j][st]+h[j-1][st])%mo;
				if (lim[i][j-1]!=0) h[j][st|k]=(h[j][st|k]+h[j-1][st])%mo;
			}
		}
		for (st=0;st<T;st++) g[st]=(f[p][st]+h[n-1][st])%mo;
	}
	ans=0;
	for (int st=0;st<T;st++) ans=(ans+g[st])%mo;
	printf("%d\n",ans);
	return 0;
}