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>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<ctime>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
int dp[1048580];
int need1[21],need2[21];
int put[1048580],go[1048580][21];
int n,m,k;
int main()
{     int i,j,p,q,a,b,c;
      scanf("%d%d%d",&n,&m,&k);
      for(i=1;i<=k;i++){
      	scanf("%d%d%d",&a,&b,&c);
      	a--,b--;
      	need1[a]|=(1<<b);
      	need2[a]|=(1<<b)*c;
      }
      n--;
      memset(go,-1,sizeof(go));
      memset(put,-1,sizeof(put));
      for(i=0;i<(1<<n);i++){
      	 int num=0;
         for(j=0;j<n;j++)
            if(i&(1<<j)){
              if(j>0&&!(i&(1<<(j-1)))){
                go[i][num]=i^(1<<j)^(1<<(j-1));
              }
              num++;
            }
      }
      for(i=0;i<(1<<n);i++)
         for(j=n-1;j>=0;j--){
            if((i&(1<<j)))continue;
            put[i]=i^(1<<j);
            break;
		 }
      dp[0]=1;
      for(i=0;i<m;i++){
         for(j=0;j<(1<<n);j++)
		    if(put[j]){
              dp[put[j]]+=dp[j];
              dp[put[j]]%=1000000007;
         }
         for(p=0;p<n;p++)
            for(j=(1<<n)-1;j>=0;j--)
			   if(go[j][p]!=-1){
                 dp[go[j][p]]+=dp[j],
                 dp[go[j][p]]%=1000000007;
               }
         for(j=0;j<(1<<n);j++)
            if((j&need1[i])!=need2[i])
              dp[j]=0;
      }
      int ans=0;
      for(i=0;i<(1<<n);i++)
         ans+=dp[i],
         ans%=1000000007;
      printf("%d\n",ans%1000000007);
	  return 0;
}