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
#include<bits/stdc++.h>
using namespace std;

const int mo=1e9+7;

int n,m,k;
int g[21][21],dp[2][1<<19],S0[21][1<<19];

int main()
{
#ifdef h10
	freopen("F.in","r",stdin);
	freopen("F.out","w",stdout);
#endif
	int i,j,S,a,b,c;
	scanf("%d%d%d",&n,&m,&k);
	memset(g,-1,sizeof(g));
	for (i=1;i<=k;i++)
		scanf("%d%d%d",&a,&b,&c),g[a][b]=c;
	for (i=1;i<n;i++)
	for (S=0;S<1<<(n-1);S++)
	{
		S0[i][S]=S|(1<<(i-1));
		for (j=i;j<n-1;j++)
			if ((S0[i][S]>>j)&1) {S0[i][S]-=1<<j; break;}
	}

	int fk=1;
	dp[fk][0]=1;
	for (i=1;i<=m;i++)
	{
		for (j=1;j<n;j++)
		{
			memset(dp[fk^1],0,sizeof(dp[fk^1]));
			for (S=0;S<1<<(n-1);S++) if (dp[fk][S])
			{
				int tmp=dp[fk][S];
				if ((S>>(j-1))&1)
					(dp[fk^1][S]+=g[i][j]==0?0:tmp)%=mo;
				else
				{
					if (g[i][j]!=1)
						(dp[fk^1][S]+=tmp)%=mo;
					if (g[i][j]!=0)
						(dp[fk^1][S0[j][S]]+=tmp)%=mo;
				}
			}
			fk^=1;
		}
	}
	int Ans=0;
	for (S=0;S<1<<(n-1);S++)
		(Ans+=dp[fk][S])%=mo;
	printf("%d\n",Ans);
}
