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>
using namespace std;
const int mod=1e9+7;
int f[2][22][600005],n,m,K,S[600005],st[600005],bin[32],nxt[600005][22][2];
void add(int &a,int b){a=a+b>=mod?a+b-mod:a+b;}
int main(){
	scanf("%d%d%d",&n,&m,&K);
	n--;
	bin[0]=1;
	for(int i=1;i<=30;i++) bin[i]=bin[i-1]*2;
	for(int i=0;i<(1<<n);i++){
		int pos[2]={0,0};
		for(int p=n;~p;p--){
			nxt[i][p][0]=pos[0];
			nxt[i][p][1]=pos[1];
			pos[(i>>p-1)&1]=p;
		}
	}
	for(int i=1,a,b,c;i<=K;i++){
		scanf("%d%d%d",&a,&b,&c);
		S[a]|=(1<<b-1);
		if(c) st[a]|=(1<<b-1);
	}
	int pre=0,now=1;
	f[1][0][0]=1;
	for(int i=1;i<=m;i++){
		memset(f[pre],0,sizeof(f[pre]));
		for(int p=0;p<n;p++){
			for(int j=0;j<(1<<n);j++)if(f[now][p][j]){
				int pos=nxt[j][p][0],pp=nxt[j][pos][1];
				add(f[now][p+1][j],f[now][p][j]);
				if(pos==p+1) add(f[now][pos][j^(1<<pos-1)^(pp?(1<<pp-1):0)],f[now][p][j]);
			}
		}
		for(int j=0;j<(1<<n);j++){
			if((j&S[i])!=st[i]){
				f[now][n][j]=0;continue;
			}
			add(f[pre][0][j],f[now][n][j]);
		}
		swap(now,pre);
	}
	int ans=0;
	for(int i=0;i<(1<<n);i++) add(ans,f[now][0][i]);
	printf("%d",ans);
	return 0;
}