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 mod=1000000007,N=25,N2=600001;
int p[N],q[N],n,m,K,trans[N2][N],f[N2],to[N2];
inline void reduce(int&x){x+=x>>31&mod;}

int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n>>m>>K;--n;
	for(;K --> 0;){
		int x,y,z;cin>>x>>y>>z;--x;--y;
		p[x]|=1<<y;q[x]|=z<<y;
	}
	memset(to,-1,sizeof to);memset(trans,-1,sizeof trans);
	for(int i=0;i<(1<<n);++i){
		int cnt=0;
		for(int j=0;j<n;++j)
			if(i>>j&1){
				if(j&&!(i>>(j-1)&1))trans[i][cnt]=i^(1<<j)^(1<<(j-1));
				++cnt;
			}
	}
	for(int i=0;i<(1<<n);++i)
		for(int j=0;j<n;++j)
			if(!(i>>j&1))to[i]=i^(1<<j);
	f[0]=1;
	for(int i=0;i<m;++i){
		for(int j=0;j<(1<<n);++j)
			if(~to[j])reduce(f[to[j]]+=f[j]-mod);
		for(int j=0;j<n;++j)
			for(int k=(1<<n)-1;~k;--k)
				if(~trans[k][j])reduce(f[trans[k][j]]+=f[k]-mod);
		for(int j=0;j<(1<<n);++j)
			if((j&p[i])!=q[i])f[j]=0;
	}
	int ans=0;
	for(int i=0;i<(1<<n);++i)reduce(ans+=f[i]-mod);
	cout<<ans<<'\n';
	return 0;
}