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
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<set>
#include<bitset>
#include<map>

#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)

using namespace std;

typedef long long LL;
typedef double db;

int get(){
	char ch;
	while(ch=getchar(),(ch<'0'||ch>'9')&&ch!='-');
	if (ch=='-'){
		int s=0;
		while(ch=getchar(),ch>='0'&&ch<='9')s=s*10+ch-'0';
		return -s;
	}
	int s=ch-'0';
	while(ch=getchar(),ch>='0'&&ch<='9')s=s*10+ch-'0';
	return s;
}

const int MAXN = 21;
const int LEN = 1048600;
const int mo = 1e+9+7;

int f[MAXN][LEN];
int N;
int n,m,k;
int ty[MAXN][MAXN];
int tr[LEN][MAXN];

inline int add(int x,int y){return x+y>=mo?x+y-mo:x+y;}

int main(){
	n=get();m=get();k=get();
	fo(i,1,k){
		int a=get(),b=get(),c=get();
		ty[a][n-b]=c+1;
	}
	N=1<<(n-1);
	fo(i,0,N-1){
		int w=-1;
		fo(j,0,n-2){
			if ((i&(1<<j))>0)w=j;
			if (w>-1)tr[i][j+1]=i^(1<<w)^(1<<j);else tr[i][j+1]=i^(1<<j);
		}
	}
	f[1][0]=1;
	fo(p,1,m){
		fo(i,0,N-1)f[n][i]=f[1][i];
		fo(x,1,n-1)fo(i,0,N-1)f[x][i]=0;
		fd(x,n-1,1){
			fo(i,0,N-1)
			if (f[x+1][i]){
				if (ty[p][x]==0){
					if ((i&(1<<(x-1)))==0)f[x][i]=add(f[x][i],f[x+1][i]);
					f[x][tr[i][x]]=add(f[x][tr[i][x]],f[x+1][i]);
				}
				if (ty[p][x]==1&&(i&(1<<(x-1)))==0)f[x][i]=add(f[x][i],f[x+1][i]);
				if (ty[p][x]==2)f[x][tr[i][x]]=add(f[x][tr[i][x]],f[x+1][i]);
			}
		}
	}
	int ans=0;
	fo(i,0,N-1)ans=add(ans,f[1][i]);
	printf("%d\n",ans);
	return 0;
}