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).

<image>

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 <algorithm>
#include <memory.h>
#include <vector>

using namespace std;
typedef long long LL;

const int maxn = 20;
const int maxm = 1<<19;
const int mod = 1e9+7;

int dp[2][maxm],op;
int nex[maxn][maxm];
int lim[maxn][maxn];
int n,m,K,S;

void inc(int &x,int y) {
	x+=y;x>=mod?x-=mod:0;
}
int main()
{
	#ifdef Amberframe
		freopen("agc017f.in","r",stdin);
		freopen("agc017f.out","w",stdout);
	#endif
	scanf("%d %d %d",&n,&m,&K);S=1<<n-1; 
	for (int i=0;i<n-1;i++)
	for (int j=0;j<S;j++)
	if (j>>i&1) nex[i][j]=j^(1<<i);
	else {
		nex[i][j]=j^(1<<i);
		int k=i+1;while (k<n-1&&!(j>>k&1)) ++k;
		if (k<n-1) nex[i][j]^=1<<k;
	}
	
	for (int i=0;i<m;i++)
	for (int j=0;j<n-1;j++) lim[i][j]=-1;
	for (int a,b,c,i=1;i<=K;i++)
		scanf("%d %d %d",&a,&b,&c),lim[a-1][b-1]=c;
		
	for (int j=0;j<S;j++)
	{
		bool tag=true;
		for (int k=0;k<n-1;k++)
		if (j>>k&1) {
			if (lim[0][k]==0) tag=false;
		}
		else {
			if (lim[0][k]==1) tag=false;
		}
		if (tag) dp[0][j]=1;
	}
	for (int i=1;i<m;i++)
	for (int k=0;k<n-1;k++) {
		op^=1;
		for (int j=0;j<S;j++)
		if (int w=dp[op^1][j]) {
			dp[op^1][j]=0;
			if (j>>k&1) {
				if (lim[i][k]!=0)
					inc(dp[op][j],w);
			}
			else {
				if (lim[i][k]!=1)
					inc(dp[op][j],w);
				if (lim[i][k]!=0)
					inc(dp[op][nex[k][j]],w);
			}
		}
	}
	int ans=0;
	for (int j=0;j<S;j++)
		inc(ans,dp[op][j]);
	printf("%d",ans);
	return 0;
}