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:
#include<iostream>
#include<cstdio>
#include<cstring>
namespace Qiao_Miao_De_DP
{
	typedef long long ll;
	inline int lowbit(int x){return x&(-x);}
	const int N=21,MOD=1000000007;
	int con[N][N];
	ll f[2][N][1<<N];
	int n,m,S,p;
	void initialize()
	{
		int tmp;
		scanf("%d%d%d",&n,&m,&tmp);
		for(int i=1;i<=m;i++)
			for(int j=1;j<n;j++)
				con[i][j]=-1;
		for(int A,B,C;tmp--;con[A][B]=C)
			scanf("%d%d%d",&A,&B,&C);
		n--,S=1<<n;
	}
	void clear()
	{
		for(int j=1;j<=n+1;j++)
			for(int k=0;k<S;k++)
				f[p][j][k]=0;
	}
	void dp()
	{
		int T;p=0;
		f[0][n+1][0]=1;

		for(int i=0;i<=m;i++)
		{
			p^=1;
			clear();
			for(int k=0;k<S;k++)f[p][1][k]=f[p^1][n+1][k];

			for(int j=1;j<=n;j++)
			{
				for(int k=0;k<S;k++)
				{
//					printf("f[%d][%d][%d] = %lld\n",i,j,k,f[p][j][k]);
					if(!f[p][j][k])continue;

					if(con[i+1][j]<0 || con[i+1][j]==0)
					{
						if(!(k&(1<<(j-1))))
							(f[p][j+1][k]+=f[p][j][k])%=MOD;
					}
					if(con[i+1][j]<0 || con[i+1][j]==1)
					{
						T=k^(k&((1<<(j-1))-1));
						if(!T)T=k^(1<<(j-1));
						else T=k^lowbit(T)^(1<<(j-1));
						(f[p][j+1][T]+=f[p][j][k])%=MOD;
					}
				}
			}
		}
	}
	void solve()
	{
		initialize();
		dp();
		ll ans=0;
		for(int i=0;i<S;i++)
			ans=(ans+f[p][1][i])%MOD;
		printf("%lld\n",ans);
	}
}
int main()
{
//	freopen("in","r",stdin);
//	freopen("out","w",stdout);
	Qiao_Miao_De_DP::solve();
	return 0;
}