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>
#define upd(a,b) a=add(a,b)
#define lb(x) ((x)&(-(x)))
using namespace std;
const int Mod=1e9+7;
int add(int a,const int b){a+=b;if(a>=Mod)a-=Mod;return a;}
const int N=23,V=(1<<20)+10;
int n,m,K,c=0,sx,ans=0;
int h[N][N],f[2][V];
int main()
{
	int i,j,ns,s1,s2,a,b;
	cin>>n>>m>>K;
	if(n==1)
	{
		puts("1");return 0;
	}
	n--;
	sx=1<<n;
	memset(h,-1,sizeof(h));
	for(i=1;i<=K;i++)
	{
		cin>>a>>b;b--;
		cin>>h[a][b];h[a][b]^=1;
	}
	f[0][0]=1;
	for(i=1;i<=m;i++)
	{
		for(j=0;j<n;j++)
		{
			memset(f[c^1],0,sizeof(f[c^1]));
			for(ns=0;ns<sx;ns++)
			{
				if(h[i][j]!=0)
				{
					if(!((ns>>j)&1))
					{
						upd(f[c^1][ns],f[c][ns]);
					}
				}
				if(h[i][j]!=1)
				{
					if((ns>>j)&1)
					{
						upd(f[c^1][ns],f[c][ns]);
					}
					else
					{
						s1=ns&((1<<j)-1);s1|=(1<<j);
						s2=ns>>j;s2^=lb(s2);s2>>=1;
						upd(f[c^1][s1|(s2<<(j+1))],f[c][ns]);
					}
				}
			}
			c^=1;
		}
	}
	for(ns=0;ns<sx;ns++) upd(ans,f[c][ns]);
	printf("%d",ans);
	return 0;
}