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 <bits/stdc++.h>

using namespace std;

const int md = 1000000007;

inline void add(int &a, int b) {
  a += b;
  if (a >= md) {
    a -= md;
  }
}

const int N = 1234567;

int need1[23], need2[23];
int go[N][23];
int put[N];
int f[N];

int main() {
  int n, m, k;
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 0; i < m; i++) {
    need1[i] = 0;
    need2[i] = 0;
  }
  while (k--) {
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    a--; b--;
    need1[a] |= (1 << b);
    need2[a] |= (1 << b) * c;
  }
  n--;
  for (int t = 0; t < (1 << n); t++) {
    for (int i = 0; i < n; i++) {
      go[t][i] = -1;
    }
    int id = 0;
    for (int i = 0; i < n; i++) {
      if (t & (1 << i)) {
        if (i > 0 && !(t & (1 << (i - 1)))) {
          go[t][id] = t ^ (1 << i) ^ (1 << (i - 1));
        }
        id++;
      }
    }
  }
  for (int t = 0; t < (1 << n); t++) {
    put[t] = -1;
    for (int i = n - 1; i >= 0; i--) {
      if (t & (1 << i)) {
        continue;
      }
      put[t] = t ^ (1 << i);
      break;
    }
  }
  for (int t = 0; t < (1 << n); t++) {
    f[t] = 0;
  }
  f[0] = 1;
  for (int i = 0; i < m; i++) {
    for (int t = 0; t < (1 << n) - 1; t++) {
      add(f[put[t]], f[t]);
    }
    for (int id = 0; id < n; id++) {
      for (int t = (1 << n) - 1; t >= 0; t--) {
        if (go[t][id] != -1) {
          add(f[go[t][id]], f[t]);
        }
      }
    }
    for (int t = 0; t < (1 << n); t++) {
      if ((t & need1[i]) != need2[i]) {
        f[t] = 0;
      }
    }
  }
  int ans = 0;
  for (int t = 0; t < (1 << n); t++) {
    add(ans, f[t]);
  }
  printf("%d\n", ans);
  return 0;
}