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<cstdio>
#include<cstring>
const int md=1e9+7,N=20;
inline void upd(int&a){a+=a>>31&md;}
int C,n,m,k,_0[N],_1[N],f[N+1][1<<N];
inline bool check(int id,int S){return(S&_0[id])==0&&(S&_1[id])==_1[id];}
inline int get(int S,int i){
	int T=S&((1<<i)-1);
	if(!T)return(1<<i)|S;
	return(S^(1<<(31-__builtin_clz(T))))|(1<<i);
}
int main(){
	scanf("%d%d%d",&n,&m,&k),--n;
	C=1<<n;
	while(k--){
		int A,B,C;
		scanf("%d%d%d",&A,&B,&C);
		B=n-B;
		(C?_1:_0)[A-1]|=1<<B;
	}
	for(int i=0;i<C;++i)f[n][i]=check(0,i);
	for(int t=1;t<m;++t){
		for(int i=n-1;~i;--i)
		for(int S=0;S<C;++S)if(int d=f[i+1][S]){
			//0
			if(!(_1[t]>>i&1)&&!(S>>i&1))upd(f[i][S]+=d-md);
			//1
			if(!(_0[t]>>i&1)){
				if(S>>i&1)upd(f[i][S]+=d-md);else{
					int T=get(S,i);
					upd(f[i][T]+=d-md);
				}
			}
		}
		memcpy(f[n],*f,sizeof*f);
		for(int i=0;i<n;++i)memset(f[i],0,sizeof*f);
	}
	int ans=0;
	for(int i=0;i<C;++i)upd(ans+=f[n][i]-md);
	printf("%d\n",ans);
	return 0;
}