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<bits/stdc++.h>
#define maxn 100005
#define inf (1<<30)
#define mod 1000000007
using namespace std;
typedef long long LL;
int read()
{
	char c=getchar();int f=1,sum=0;
	while(c<'0' || c>'9'){if(c=='-') f=-1;c=getchar();}
	while(c>='0' && c<='9'){sum=sum*10+c-'0';c=getchar();}
	return sum*f;
}
int n,m,k,ans;
int nex[1<<21][25],lim[25][25];
void init()
{
	n=read()-1;m=read();k=read();
	for(int i=0;i<(1<<n);i++)
	{
		int pos=-1;
		for(int j=n-1;j>=0;j--)
		{
			if(i&(1<<j)) pos=j;
			nex[i][j]=pos;
		}
	}
	memset(lim,-1,sizeof(lim));
	for(int i=1;i<=k;i++)
	{
		int a=read(),b=read(),c=read();
		lim[a][b-1]=c;	
	}
}
int dp[2][25][(1<<21)];
int main()
{
	init();
	dp[0][n][0]=1;
	int now=1;
	for(int i=1;i<=m;i++)
	{
		for(int sta=0;sta<(1<<n);sta++) dp[now][0][sta]=dp[1-now][n][sta];
		for(int j=0;j<=n;j++)
        for(int sta=0;sta<(1<<n);sta++)
        dp[1-now][j][sta]=0;
        for(int j=0;j<n;j++)
        for(int sta=0;sta<(1<<n);sta++)
        if(dp[now][j][sta])
		{
            for(int k=0;k<=1;k++)
			{
            	if(lim[i][j]!=-1 && lim[i][j]!=k) continue;
            	int t=(bool)(sta&(1<<j));
                if(k==0 && t) continue;
                if(k==t) (dp[now][j+1][sta]+=dp[now][j][sta])%=mod;
                else
				{
                    int r=nex[sta][j];
                    if (r==-1) (dp[now][j+1][sta+(1<<j)]+=dp[now][j][sta])%=mod;
                    else (dp[now][j+1][sta+(1<<j)-(1<<r)]+=dp[now][j][sta])%=mod;
                }
            }
        }
        now=1-now;
	}
	for(int sta=0;sta<(1<<n);sta++)
	ans=(ans+dp[1-now][n][sta])%mod;
    printf("%d\n",ans);
	return 0;
}