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<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<cmath>
#include<ctime>
#include<bitset>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<climits>
#include<complex>
#include<iostream>
#include<algorithm>
#define ll long long
#define lowbit(x) x&(-x)
#define _ %Mod
#define __ %=Mod
using namespace std;

const int maxn = 22;
const int maxm = 22;
const int maxk = 1100000;
const ll Mod = 1e9+7;

int n,m,K;
int ci[maxm][maxn];
ll f[2][maxk];

int main()
{
	scanf("%d%d%d",&n,&m,&K); n--;
	memset(ci,-1,sizeof ci);
	for(int i=1;i<=K;i++)
	{
		int x,y,z; scanf("%d%d%d",&x,&y,&z);
		ci[x][y-1]=z;
	}
	
	int al=1<<n;
	int now=0; f[0][0]=1ll;
	for(int i=1;i<=m;i++) for(int j=0;j<n;j++) 
	{
		for(int k=0;k<al;k++) if(f[now][k])
		{
			ll tmp=f[now][k];
			
			if(!(k>>j&1)&&ci[i][j]!=1) (f[!now][k]+=tmp)__;
			if(ci[i][j]==0) continue;
			if(k>>j&1) (f[!now][k]+=tmp)__;
			else
			{
				int lk=k&(1<<j+1)-1,nk=k^lk;
				nk=nk^lowbit(nk)^lk^1<<j;
				(f[!now][nk]+=tmp)__;
			}
		}
		memset(f[now],0,sizeof f[now]); now=!now;
	}
	ll ans=0;
	for(int k=0;k<al;k++) (ans+=f[now][k])__;
	printf("%lld\n",ans);
	
	return 0;
}