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 <iostream>
#include <cstring>
using namespace std;

const int N = 19;
const int mod = 1e9 + 7;

int G[N + 2][N], n, m, k;
int dp[1 << N | 5], swp[1 << N | 5];

inline void Inc(int &x, int y)  {
  if ((x += y) >= mod) x -= mod;
}

inline int lowbit(int x) { return x & (- x); }

int main() {
  memset(G, -1, sizeof G);
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1, x, y; i <= k; i ++) {
    scanf("%d%d", &x, &y);
    scanf("%d", &G[x][y - 1]);
  }
  n --;
  int U = 1 << n;
  int *x = dp, *y = swp;
  x[0] = 1;
  for (int i = 1; i <= m; i ++) {
    for (int j = 0; j < n; j ++) {
      memset(y, 0, sizeof (int) * U);
      for (int k = U - 1; ~ k; k --)
        if (x[k]) {
          Inc(y[k], x[k]);
          if (!((k >> j) & 1)) Inc(y[k - lowbit(k & (U - (2 << j))) + (1 << j)], x[k]);
        }
      swap(x, y);
    }
    int key = 0, save = 0;
    for (int j = 0; j < n; j ++) {
      if (~ G[i][j]) key |= G[i][j] << j;
      else save |= 1 << j;
    }
    for (int j = 0; j < U; j ++)
      if (((j ^ key) | save) != save) x[j] = 0;
  }
  int ans = 0;
  for (int i = 0; i < U; i ++) Inc(ans, x[i]);
  printf("%d\n", ans);
  return 0;
}