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 <stdio.h>
const int MOD=1000000007;
int dp[2][1<<19],flp[1<<19][19],must[20][19];
inline void upd(int &a,int b)
{
    a+=b;
    if (a>=MOD)
        a-=MOD;
}
int main()
{
    int n,m,k;
    scanf("%d%d%d",&n,&m,&k);
    for (int i=0;i<(1<<(n-1));++i)
    {
        int lst=n-1;
        for (int j=n-2;j>=0;--j)
        {
            if ((i>>j)&1)
                lst=j;
            else
                flp[i][j]=lst==n-1?i^(1<<j):i^(1<<j)^(1<<lst);
        }
    }
    for (int i=0;i<k;++i)
    {
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        must[a-1][b-1]=c+1;
    }
    dp[0][0]=1;
    int p=0;
    for (int i=0;i<m;++i)
    {
        for (int j=0;j<n-1;++j)
        {
            for (int k=0;k<(1<<(n-1));++k)
                dp[p^1][k]=0;
            for (int k=0;k<(1<<(n-1));++k)
            {
                if (!dp[p][k])
                    continue;
                if (must[i][j]!=2&&!((k>>j)&1))
                    upd(dp[p^1][k],dp[p][k]);
                if (must[i][j]!=1)
                {
                    if ((k>>j)&1)
                        upd(dp[p^1][k],dp[p][k]);
                    else
                        upd(dp[p^1][flp[k][j]],dp[p][k]);
                }
            }
            p^=1;
        }
    }
    int ans=0;
    for (int i=0;i<(1<<(n-1));++i)
        upd(ans,dp[p][i]);
    printf("%d\n",ans);
    return 0;
}