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<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
#define fo(i,j,k) for(i=j;i<=k;i++)
#define fd(i,j,k) for(i=j;i>=k;i--)
#define cmax(a,b) (a=(a>b)?a:b)
#define cmin(a,b) (a=(a<b)?a:b)
typedef long long ll;
const int N=1e6+5,M=2e6+5,mo=1e9+7;
int f[2][(1<<20)+5],c[25][25],n,m,K,x,y,z,q1,q2,i,j,s,ts,ns,ds,ans;
int main()
{
    scanf("%d %d %d",&n,&m,&K);
    n--;
    fo(i,1,K)
    {
        scanf("%d %d %d",&x,&y,&z);
        c[x][y-1]=z+1;
    }
    f[0][0]=1;
    q1=0;
    q2=1;
    ts=1<<n;ts--;
    fo(i,1,m)
        fo(j,0,n-1)
        {
            fo(s,0,ts)
            if (f[q1][s])
            {
                //left
                if (c[i][j]!=2&&((s>>j)&1)==0)
                {
                    f[q2][s]+=f[q1][s];
                    if (f[q2][s]>=mo) f[q2][s]-=mo;
                }
                //right
                if (c[i][j]!=1)
                {
                    if ((s>>j)&1) ns=s;
                    else
                    {
                        ns=((-1)^(1<<j+1)-1)&s;
                        if (!ns) ds=0;
                        else ds=ns&(-ns);
                        ns=s^(1<<j)^ds;
                    }
                    f[q2][ns]+=f[q1][s];
                    if (f[q2][ns]>=mo) f[q2][ns]-=mo;
                }
            }
            fo(s,0,ts) f[q1][s]=0;
            q1^=1;
            q2^=1;
        }
    fo(s,0,ts) 
    {
        ans+=f[q1][s];
        if (ans>=mo) ans-=mo;
    }
    printf("%d\n",ans);
}