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>
#include<algorithm>
#define LL long long
#define fo(i,j,k) for(int i=j;i<=k;i++)
#define fd(i,j,k) for(int i=j;i>=k;i--)
#define fr(i,j) for(int i=beg[j];i;i=nex[i])
using namespace std;
int const mn=20+3,ms=(1<<19)+3,mo=1e9+7;
int n,m,K,a[mn][mn],f[2][ms],nx[ms][mn];
inline int mod(int x){return (x>=mo)?x-mo:x;}
int main(){
	scanf("%d%d%d",&n,&m,&K);
	fo(i,1,K){
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		a[x][y]=z+1;
	}
	int lim=(1<<(n-1))-1,tg=0;
	fd(s,lim,0)fd(i,n-1,1)nx[s][i]=((s>>(i-1))&1)?i:nx[s][i+1];
	f[tg][0]=1;
	fo(i,1,m)fo(j,1,n-1){
		fo(s,0,lim)if(f[tg][s]){
			int tmp=(nx[s][j])?(1<<(nx[s][j]-1)):0,
				tm2=s+(1<<(j-1))-tmp;
			if((a[i][j]!=1)&&(!(s&(1<<(j-1)))))
				f[!tg][tm2]=mod(f[!tg][tm2]+f[tg][s]);
			if(a[i][j]+((s>>j-1)&1)!=2)
				f[!tg][s]=mod(f[!tg][s]+f[tg][s]);
		}
		memset(f[tg],0,sizeof(f[tg]));
		tg=!tg;
	}
	int ans=0;
	fo(s,0,lim)ans=mod(ans+f[tg][s]);
	printf("%d",ans);
	return 0;
}