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<bits/stdc++.h>
using namespace std;

const int Size=1<<20;
const int Mod=1e9+7;

int N,M,K,Ans;
int Nxt[Size][20],Tag[25][25];
int F[2][25][Size];

int main(){
	int i,j,k,t,a,b,c,lst=0,now=1;
	scanf("%d%d%d",&N,&M,&K),N--;
	memset(Tag,-1,sizeof(Tag));
	while(K--){
		scanf("%d%d%d",&a,&b,&c);
		Tag[a][b]=c^1;
	}
	for(i=0;i<1<<N;i++)
		for(k=-1,j=N-1;~j;j--){
			Nxt[i][j]=k;
			if(i>>j&1)
				k=j;
		}
	F[0][N][0]=1;
	for(i=1;i<=M;i++){
		memset(F[now],0,sizeof(F[now]));
		memmove(F[now][0],F[lst][N],sizeof(F[now][0]));
		for(j=0;j<N;j++)
			for(k=0;k<1<<N;k++)
				if(F[now][j][k])
					for(t=k>>j&1;t<2;t++)
						if(Tag[i][j+1]^t){
							if((k>>j&1)==t)
								(F[now][j+1][k]+=F[now][j][k])%=Mod;
							else if(~Nxt[k][j])
								(F[now][j+1][k+(1<<j)-(1<<Nxt[k][j])]+=F[now][j][k])%=Mod;
							else (F[now][j+1][k+(1<<j)]+=F[now][j][k])%=Mod;
						}
		swap(lst,now);
	}
	for(i=0;i<1<<N;i++)
		Ans=(Ans+F[lst][N][i])%Mod;
	printf("%d\n",Ans);
	return 0;
}