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 <cstring>
#include <algorithm>
#include <cctype>
#include <vector>
using namespace std;
typedef long long int64;
inline int read(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 20, P = 1e9+7;
int n, m, all, d[N+5][N+5], f[2][1<<N|5], p = 1, q = 0, ans;
int main()
{
    n = read()-1, m = read(), all = (1<<n)-1;
    for(int i = read(); i; --i) 
    {
        int a = read(), b = read()-1, c = read();
        d[a][b] = c+1;
    }
    f[0][0] = 1;
    for(int i = 1, t, u; i <= m; ++i)
        for(int j = 0; j < n; ++j)
        {
            swap(p, q), t = 1<<j, u = all^((t<<1)-1);
            memset(f[q], 0, sizeof(int)*(all+1));
            for(int s = 0, r; s <= all; ++s)
                if(f[p][s])
                {
                    if(d[i][j] != 2&&(s&t) == 0) f[q][s] = (f[q][s]+f[p][s])%P;
                    if(d[i][j] != 1)
                    {
                        if(s&t) r = s;
                        else if((u&s) == 0) r = s|t;
                        else r = (s|t)^((u&s)&-(u&s));
                        f[q][r] = (f[q][r]+f[p][s])%P;
                    }
                }
        }
    for(int s = 0; s <= all; ++s) ans = (ans+f[q][s])%P;
    printf("%d\n", ans);
    return 0;
}