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:
//Heaplax
#include<bits/stdc++.h>
#define LL long long
#define low(x) ((x)&-(x))
#define LOG(x) cerr<<#x<<" = "<<x<<endl
#define add_edge(u,v) nxt[++cnt]=head[u],head[u]=cnt,to[cnt]=v
#define open(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout)
char ch;bool fs;void re(int& x)
{
	while(ch=getchar(),ch<33);
	if(ch=='-')fs=1,x=0;else fs=0,x=ch-48;
	while(ch=getchar(),ch>33)x=x*10+ch-48;
	if(fs)x=-x;
}
using namespace std;
const int mod=1000000007;
int f[2][21][(1<<20)+1];
int n,m,k;
int mark[21][21];
void add(int& a,int b)
{
	a=a+b<mod?a+b:a+b-mod;
}
int main()
{
	re(n),re(m),re(k);
	memset(mark,-1,sizeof mark);
	for(int a,b,c;k--;)
	{
		re(a),re(b),re(c);
		mark[a][b-1]=c;
	}
	f[0][n-1][0]=1;
	for(int i=1;i<=m;++i)
	{
		bool b=i&1;
		for(int j=0;j<(1<<n);++j)f[b][0][j]=f[!b][n-1][j],f[!b][n-1][j]=0;
		for(int j=0;j<n-1;++j)
			for(int k=0;k<(1<<n);++k)if(f[b][j][k])
			{
				if(!~mark[i][j] || mark[i][j]==(k>>j&1))
					add(f[b][j+1][k],f[b][j][k]);
				if(mark[i][j]!=0 && !(k>>j&1))
				{
					int w=k-(low(k>>j+1)<<j+1)+(1<<j);
					add(f[b][j+1][w],f[b][j][k]);
				}
				f[b][j][k]=0;
			}
	}
	int ans=0;
	for(int i=0;i<(1<<n);++i)
		add(ans,f[m&1][n-1][i]);
	printf("%d\n",ans);
}