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<vector>
#include<cstring>
using namespace std;
const long long md=1e9+7;
int val[21][21];
long long dp[1<<20];
bool pd[1<<20];
int n,m,k;
int getnext(int p,int x){
	int st=p^(((1<<x)-1)&p);
	if(!st) return p|(1<<x);
	return (p^(st&(-st)))|(1<<x);
}
void solve(int pv){
	bool bo;
	for(int i=0;i<(1<<n);i++){
		bo=1;
		for(int j=0;j<n;j++){
			if(val[pv][j]==-1) continue;
			if(((i>>j)&1)!=val[pv][j]){
				bo=0;
				break;
			}
		}
		pd[i]=bo;
	}
	for(int i=0;i<n;i++){
		for(int j=0;j<(1<<n);j++){
			if((j&(1<<i))) continue;
			(dp[getnext(j,i)]+=dp[j])%=md;
		}
	}
	for(int i=0;i<(1<<n);i++)
		if(!pd[i]) dp[i]=0;
}
int main(){
//	freopen("F.in","r",stdin);
	scanf("%d%d%d",&n,&m,&k);
	n--;
	int A,B,C;
	for(int i=0;i<=m;i++)
		for(int j=0;j<=n;j++)
			val[i][j]=-1;
	for(int i=1;i<=k;i++){
		scanf("%d%d%d",&A,&B,&C);
		B--;
		if(val[A][B]!=-1){
			if(val[A][B]!=C){
				printf("0\n");
				return 0;
			}
		}
		val[A][B]=C;
	}
	dp[0]=1;
	long long ans=0;
	for(int i=1;i<=m;i++)
		solve(i);
	for(int i=0;i<(1<<n);i++)
		ans=(ans+dp[i])%md;
	printf("%lld\n",ans);
	return 0;
}