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<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define MAXN 20
#define MAXM 20
#define MAXS 1049000
#define MO 1000000007
using namespace std;
int opt[MAXN+5][MAXM+5];
int dp[2][MAXN+5][MAXS+5];
int N,M,K;
int main()
{
	memset(opt,-1,sizeof(opt));
	scanf("%d %d %d",&N,&M,&K);
	N--;
	int A,B,C;
	for(int i=1;i<=K;i++)
	{
		scanf("%d %d %d",&A,&B,&C);
		opt[A][B]=C;
	}
	dp[0][N][0]=1;
	int all=(1<<N),id;
	for(int i=1;i<=M;i++)
	{
		id=i%2;
		for(int s=0;s<all;s++)
			dp[id][0][s]=dp[!id][N][s];
		for(int j=1;j<=N;j++)
		{
			for(int s=0;s<all;s++)
			{
				if(((s&(1<<(j-1)))&&opt[i][j]==0)||dp[id][j-1][s]==0)
					continue;
				if(opt[i][j]!=0)
				{
					int ri=s/(1<<(j-1));
					int low=(ri&(-ri));
					int news=s;
					if(ri!=0)
						news-=(low<<(j-1));
					news+=(1<<(j-1));
					dp[id][j][news]=(1LL*dp[id][j][news]+1LL*dp[id][j-1][s])%MO;
				}
				if(opt[i][j]!=1&&((s&(1<<(j-1)))==0))
				{
					int news=s;
					if(s&(1<<(j-1)))
						news-=(1<<(j-1));
					dp[id][j][news]=(1LL*dp[id][j][news]+1LL*dp[id][j-1][s])%MO;
				}
			}
		}
		memset(dp[!id],0,sizeof(dp[!id]));
	}
	int ans=0;
	for(int s=0;s<all;s++)
		ans=(1LL*ans+1LL*dp[M%2][N][s])%MO;
	printf("%d\n",ans);
	return 0;
}