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<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<map>
#include<vector>
using namespace std;
int n,m,k,rw=1000000007;
int yq[30][30],nx[(1<<20)+5][25],dp[2][21][(1<<20)+5];
int main()
{
	scanf("%d%d%d",&n,&m,&k);--n;
	for(int i=1;i<=m;i++)
		for(int j=0;j<n;j++)
			yq[i][j]=-1;
	for(int i=1,a,b,c;i<=k;i++)
	{
		scanf("%d%d%d",&a,&b,&c);
		yq[a][b-1]=c;
	}
	for(int i=0;i<(1<<n);i++)
	{
		int la=-1;
		for(int j=n-1;j>=0;j--)
		{
			if((i&(1<<j)))la=j;
			nx[i][j]=la;
		}
        //for(int j=0;j<n;j++)cout<<nx[i][j];
	}
	dp[0][n][0]=1;
	int now=1;
	for(int i=1;i<=m;i++)
	{
		for(int s=0;s<(1<<n);s++)
			dp[now][0][s]=dp[now^1][n][s];
		for(int j=0;j<=n;j++)
			for(int s=0;s<(1<<n);s++)
				dp[now^1][j][s]=0;
		for(int j=0;j<n;j++)
		{
			for(int s=0;s<(1<<n);s++)
			{
				if(dp[now][j][s])
				{
					for(int k=0;k<=1;k++)
					{
						if(yq[i][j]!=-1&&yq[i][j]!=k)continue;
						int t=((s>>j)&1);
						if(k==0&&t==1)continue;
						if(k==t)(dp[now][j+1][s]+=dp[now][j][s])%=rw;
						if(k==1&&t==0)
						{
							int la=nx[s][j];
							if(la==-1)(dp[now][j+1][s|(1<<j)]+=dp[now][j][s])%=rw;
							else (dp[now][j+1][(s|(1<<j))^(1<<la)]+=dp[now][j][s])%=rw;
						}
					}
				}
			}
		}
		now^=1;
	}
	int ans=0;
	for(int s=0;s<(1<<n);s++)
	{
		(ans+=dp[now^1][n][s])%=rw;
	}
	printf("%d\n",(ans+rw)%rw);
}