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 <bits/stdc++.h>
#define lop(i,s,t) for(int i=s;i<(t);++i)
#define rep(i,s,t) for(int i=s;i<=(t);++i)
using namespace std;
const int N=21,P=1e9+7;
void upd(int &a,int b){ a=a+b>=P?a+b-P:a+b; }
int d[N][N][2],ans,n,m,q,f[2][1<<N],nxt[1<<N][N];
int main(int argc,char *argv[]){
	cin>>n>>m>>q,--n;
	rep(k,1,q){
		int a,b,c;
		cin>>a>>b>>c;
		d[a][b-1][!c]=1;
	}
	lop(s,0,(1<<n))lop(j,0,n)if((s>>j&1)==0){
		int k=j,t=s;
		for(;k<n&&!((s>>k)&1);++k); 
		if(k<n)t-=(1<<k);
		nxt[s][j]=t+(1<<j); 
	}
	f[n&1][0]=1;
	rep(i,1,m){
		if(n&1)memcpy(f[0],f[1],sizeof(f[0])); 
		lop(j,0,n){
			int p=j&1,c=!p;
			memset(f[c],0,sizeof(f[c])); 
			lop(s,0,(1<<n))rep(num,0,1)if(!d[i][j][num]){
				if(!(((s>>j&1)==1)&&!num))
					upd(f[c][(s>>j&1)==num?s:nxt[s][j]],f[p][s]); 
			}
		}
	}
	lop(s,0,(1<<n))upd(ans,f[n&1][s]); 
	cout<<ans<<endl;
	return 0;
}
