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<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<set>
using namespace std;
#define ll long long
#define pb(x) push_back(x)
#define pr(x) cerr<<#x<<"="<<(x)<<endl
#define pri(x,lo) {cerr<<#x<<"={";for (int ol=0;ol<=lo;ol++)cerr<<x[ol]<<",";cerr<<"}"<<endl;}
#define N 1000
#define md 1000000007
int n,m,k,dp[2000010],sta,id,o,p,q,go[2000010][23],put[2000010],need1[30],need2[30],i,ans;
void add(int &x,int y){x+=y;if (x>md) x-=md;}
int main()
{
	scanf("%d %d %d",&n,&m,&k);n--;
	for (i=1;i<=k;i++)
	{
		scanf("%d %d %d",&o,&p,&q);o--;p--;
		need1[o]|=(1<<p);
		need2[o]|=(1<<p)*q;
	}
	memset(go,-1,sizeof(go));
	for (sta=0;sta<(1<<n);sta++)
	{
		id=0;
		for (i=0;i<n;i++)
			if (sta&(1<<i))
			{
				if (i>0&&!(sta&(1<<(i-1)))) go[sta][id]=sta^(1<<i)^(1<<(i-1));
				id++;
			}
	}
	memset(put,-1,sizeof(put));
	for (sta=0;sta<(1<<n);sta++)
		for (i=n-1;i>=0;i--)
			if ((sta&(1<<i))==0){put[sta]=sta^(1<<i);break;}
	dp[0]=1;
	for (i=0;i<m;i++)
	{
		for (sta=0;sta<(1<<n)-1;sta++)  add(dp[put[sta]],dp[sta]);
		for (int id=0;id<n;id++)
			for (int sta=(1<<n)-1;sta>=0;sta--)
			if (go[sta][id]!=-1) add(dp[go[sta][id]],dp[sta]);
		for (sta=0;sta<(1<<n);sta++) if ((sta&need1[i])!=need2[i]) dp[sta]=0;
	}
	for (sta=0;sta<(1<<n);sta++) add(ans,dp[sta]);
	printf("%d\n",ans);
	return 0;
}