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).

<image>

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>

const int N = 20, MOD = 1000000007;

int n, m, f[1 << N];

int mask[N][N];

bool init() {
  int k;
  scanf("%d%d%d", &n, &m, &k);
  --n;
  while (k--) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    mask[a - 1][b - 1] |= 1 << c;
  }
  for (int i = 0; i < m; ++i) if (std::count(mask[i], mask[i] + n, 3)) return false;
  return true;
}

void solve(int k) {
  int x = 0, y = 0;
  for (int i = 0; i < n; ++i) {
    if (mask[k][i] & 1) x |= 1 << i;
    if (mask[k][i] & 2) y |= 1 << i;
  }
  static int temp[1 << N];
  for (int i = 0; i < n; ++i) {
    memcpy(temp, f, sizeof temp);
    for (int s = 0; s < (1 << n); ++s) {
      if (!(s >> i & 1) || !f[s]) continue;
      int t = s ^ (1 << i);
      for (int j = i - 1; j >= 0; --j) if (t >> j & 1) break; else (temp[t | (1 << j)] += f[s]) %= MOD;
    }
    memcpy(f, temp, sizeof temp);
  }
  for (int i = 0; i < n; ++i) {
    memcpy(temp, f, sizeof temp);
    for (int s = 0; s < (1 << i); ++s) (temp[s | (1 << i)] += f[s]) %= MOD;
    memcpy(f, temp, sizeof temp);
  }
  for (int s = 0; s < (1 << n); ++s) if ((s & x) || (s & y) != y) f[s] = 0;
}

int main() {
  if (!init()) return puts("0"), 0;
  f[0] = 1;
  for (int i = 0; i < m; ++i) solve(i);
  int ans = 0;
  for (int s = 0; s < (1 << n); ++s) (ans += f[s]) %= MOD;
  printf("%d\n", (ans + MOD) % MOD);
  return 0;
}