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>
#define repu(i,x,y) for (int i=x; i<=y; ++i)
#define repd(i,x,y) for (int i=x; i>=y; --i)
using namespace std;

const int p=1000000007;
int n,m,k,s0[21],s1[21],f[21][530000],nxt[21][530000],ans;

int main()
{
    scanf("%d%d%d",&n,&m,&k),--n;
    while (k--)
    {
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        (c?s1[a]:s0[a])|=1<<(b-1);
    }
    repu(i,0,n-1)
        repu(j,0,(1<<n)-1)
            if (j&(1<<i))
            {
                nxt[i][j]=j^(1<<i);
                repu(k,i+1,n-1)
                    if (!(j&(1<<k)))
                    {
                        nxt[i][j]^=1<<k;
                        break;
                    }
            }
    f[0][0]=1;
    repu(i,1,m)
    {
        static int g[21][530000];
        repu(j,0,(1<<n)-1)
            g[n][j]=f[i-1][j];
        repd(j,n-1,0)
            repu(k,0,(1<<n)-1)
            {
                g[j][k]=g[j+1][k];
                if (k&(1<<j))
                    (g[j][k]+=g[j+1][nxt[j][k]])%=p;
            }
        repu(j,0,(1<<n)-1)
            if (!(s0[i]&j) && (s1[i]&j)==s1[i])
                f[i][j]=g[0][j];
    }
    repu(i,0,(1<<n)-1)
        (ans+=f[m][i])%=p;
    printf("%d\n",ans);
    return 0;
}