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 N=45,mod=1e9+7;	
int n,m,K,p[N][N],f[2][20][1<<20],nxt[1<<20][20];
inline void add(int &x,int y){x+=y;if(x>=mod)x-=mod;}
int main(){
	cin>>n>>m>>K;
	int x,y,z,lim=(1<<n)-1;n--;
	memset(p,-1,sizeof(p));
	for(int i=1;i<=K;i++){
		scanf("%d%d%d",&x,&y,&z);
		p[x][y]=z;
	}
	for(int i=0;i<=lim;i++){
		for(int j=0;j<n;j++){
			if(!(i>>j&1)){
				for(int k=j+1;k<n;k++)
				if(i>>k&1){nxt[i][j]=i^(1<<k)^(1<<j);break;}
				if(!nxt[i][j])nxt[i][j]=i^(1<<j);
			}
		}
	}
	bool t=0;
	f[t^1][n][0]=1;
	for(int i=1;i<=m;i++,t^=1){
		for(int j=0;j<=lim;j++)f[t][0][j]=f[t^1][n][j],f[t^1][n][j]=0;
		for(int j=0;j<n;j++){
			for(int k=0;k<=lim;k++){
				if(!f[t][j][k])continue;
				if(p[i][j+1]!=(k>>j&1^1))add(f[t][j+1][k],f[t][j][k]);
				if(p[i][j+1] && !(k>>j&1))add(f[t][j+1][nxt[k][j]],f[t][j][k]);
				f[t][j][k]=0;
			}
		}
	}
	int ans=0;t^=1;
	for(int i=0;i<=lim;i++)add(ans,f[t][n][i]);
	cout<<ans<<endl;
	return 0;
}