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<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
typedef long long LL;
int gi() {
	int w=0;bool q=1;char c=getchar();
	while ((c<'0'||c>'9') && c!='-') c=getchar();
	if (c=='-') q=0,c=getchar();
	while (c>='0'&&c <= '9') w=w*10+c-'0',c=getchar();
	return q?w:-w;
}
const int S=1<<19|1;
const int N=40;
int f[S],g[S],to[S][N];
int ban[N][N];
int main()
{
	int n=gi()-1,m=gi(),lim=gi(),i,t,s,a,b,c;
	const int mod=1e9+7;
	
	while (lim--)
		a=gi()-1,b=gi()-1,c=gi(),ban[a][b]|=1<<!c;

	for (s=0;s<1<<n;s++)
		for (i=0;i<n;i++)
			if (s>>i&1) to[s][i]=s;
			else {
				to[s][i]=s|1<<i;
				for (t=i+1;t<n;t++)
					if (s>>t&1) {
						to[s][i]^=1<<t;
						break;
					}
			}
	
	f[0]=1;
	for (t=0;t<m;t++) {
		for (i=0;i<n;i++) {
			memset(g,0,4*(1<<n));
			for (s=0;s<1<<n;s++)
				if (f[s]) {
					if (~ban[t][i]&1)
						if (!(s>>i&1))
							(g[s]+=f[s])%=mod;
					if (~ban[t][i]&2)
						(g[to[s][i]]+=f[s])%=mod;
				}
			memcpy(f,g,4*(1<<n));
		}
	}
	int ans=0;
	for (s=0;s<1<<n;s++) (ans+=f[s])%=mod;
	cout<<ans<<endl;
	return 0;
}
