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).

<image>

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>
#define ll long long
using namespace std;
const int p=1e9+7;
int mx,n,m,k,i,j,nt,pr,s,ns,ans,x,y,w,a[22][22],f[2][1200010];
int lowbit(int x){
	return x&(-x);
}
int main(){
	scanf("%d%d%d",&n,&m,&k);n--;
	for(i=1;i<=k;i++){
		scanf("%d%d%d",&x,&y,&w);
		a[x][y-1]=w+1;
	}
	mx=1<<n;
	f[0][0]=1;
	for(i=1;i<=m;i++){
		for(j=0;j<n;j++){
			nt^=1;pr=nt^1;
			memset(f[nt],0,sizeof(f[nt]));
			for(s=0;s<mx;s++)if(f[pr][s]){
				if((s&(1<<j))==0&&a[i][j]!=2)f[nt][s]=(f[nt][s]+f[pr][s])%p;
				if(a[i][j]!=1){
					if(s&(1<<j))ns=s;
					 else{
					 	int tmp=((-1)^((1<<(j+1))-1))&s,Ds=!tmp?0:lowbit(tmp);
                            ns=s^(1<<j)^Ds;
					 }
					 f[nt][ns]=(f[nt][ns]+f[pr][s])%p;
				}
			}
		}
	}
	for(s=0;s<mx;s++)ans=(ans+f[nt][s])%p;
	printf("%d",ans);
}