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 <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN=1100005,MOD=1000000007;
int n,m,K,a[MAXN],b[MAXN],c[MAXN],tr[25][MAXN][2];
int x[MAXN],y[25][MAXN];
bool f[MAXN];
int dig(int x,int b){return (x>>b)&1;}
inline void Add(int &a,int b){a+=b;(a>=MOD)?a-=MOD:0;}
int main(){
	scanf("%d%d%d",&n,&m,&K);
	n--;
	for(int i=0;i<K;i++)scanf("%d%d%d",&a[i],&b[i],&c[i]),b[i]--;
	for(int k=0;k<(1<<n);k++)
		for(int i=0;i<n;i++){
			if(dig(k,i))tr[i][k][0]=-1;else tr[i][k][0]=k;
			if(dig(k,i))tr[i][k][1]=k;
			else{
				tr[i][k][1]=k^(1<<i);
				for(int j=i+1;j<n;j++)if(dig(k,j)){
					tr[i][k][1]^=1<<j;
					break;
				}
			}
		}
	x[0]=1;
	for(int i=0;i<m;i++){
		memset(y,0,sizeof(y));
		for(int k=0;k<(1<<n);k++)y[0][k]=x[k];
		for(int j=0;j<n;j++)
			for(int k=0;k<(1<<n);k++){
				tr[j][k][0]!=-1?Add(y[j+1][k],y[j][k]):void(0);
				Add(y[j+1][tr[j][k][1]],y[j][k]);
			}
		memset(f,1,sizeof(f));
		for(int j=0;j<K;j++)if(a[j]==i+1){
			for(int k=0;k<(1<<n);k++)(dig(k,b[j])!=c[j])?f[k]=0:0;
		}
		for(int k=0;k<(1<<n);k++)x[k]=f[k]?y[n][k]:0;
	}
	int ans=0;
	for(int k=0;k<(1<<n);k++)Add(ans,x[k]);
	printf("%d\n",ans);
	return 0;
}