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 <cstdlib>
#include <algorithm>
#include <vector>
#include <cstring>
#include <queue>
#include <set>
#include <map>
#include <functional>
#include <cmath>
#include <string>
#define SIZE 25
#define BT (1<<21)
#define MOD 1000000007

using namespace std;
typedef long long int ll;
typedef pair <int,int> P;

ll dp[BT];
ll nxt[BT];
int id[SIZE][SIZE];
int n,m,K;

void solve()
{
	for(int S=0;S<1<<n;S++) nxt[S]=dp[S];
	for(int i=0;i<n;i++)
	{
		for(int S=0;S<1<<n;S++)
		{
			if(!(S>>i&1))
			{
				int to=i+1;
				while(to<n&&!(S>>to&1)) to++;
				if(to<n)
				{
					nxt[S+(1<<i)-(1<<to)]+=nxt[S];
				}
				else
				{
					nxt[S+(1<<i)]+=nxt[S];
				}
			}
		}
		for(int S=0;S<1<<n;S++) nxt[S]%=MOD;
	}
}
int main()
{
	scanf("%d %d %d",&n,&m,&K);n--;
	memset(id,-1,sizeof(id));
	for(int i=0;i<K;i++)
	{
		int a,b,c;
		scanf("%d %d %d",&a,&b,&c);a--,b--;
		id[a][b]=c;
	}
	memset(dp,0,sizeof(dp));
	dp[0]=1;
	for(int i=0;i<m;i++)
	{
		solve();
		for(int S=0;S<1<<n;S++)
		{
			bool up=true;
			for(int j=0;j<n;j++)
			{
				if(id[i][j]==0&&(S>>j&1)) up=false;
				else if(id[i][j]==1&&!(S>>j&1)) up=false;
			}
			if(!up) dp[S]=0;
			else dp[S]=nxt[S];
		}
	}
	ll ret=0;
	for(int S=0;S<1<<n;S++)
	{
		ret+=dp[S];
		if(ret>=MOD) ret-=MOD;
	}
	printf("%lld\n",ret);
	return 0;
}