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>

int f[42][42];

int dp[1 << 20];

int const MOD = 1000000007;

void add(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 < m; i++) {
    for (int j = 0; j < n; j++) {
      f[i][j] = -1;
    }
  }
  for (int i = 0; i < k; i++) {
    int who, where, what;
    scanf("%d%d%d", &who, &where, &what);
    --who;
    --where;
    f[who][where] = what;
  }
  for (int i = 0; i < 1 << (n - 1); i++) dp[i] = 0;
  dp[0] = 1;
  for (int q = 0; q < m; q++) {
    for (int pos = 0; pos < n - 1; pos++) {
      for (int mask = 0; mask < 1 << (n - 1); mask++) {
        if (((mask >> pos) & 1) == 1) {
          continue;
        }
        int nmask = mask ^ (1 << pos);
        int npos = pos + 1;
        while (npos < n - 1 && ((nmask >> npos) & 1) == 0) {
          ++npos;
        }
        if (npos < n - 1) {
          nmask &= ~(1 << npos);
        }
        add(dp[nmask], dp[mask]);
      }
    }
    int ones = 0;
    int zeros = 0;
    for (int i = 0; i < n - 1; i++) {
      if (f[q][i] == 0) {
        zeros |= 1 << i;
      } else if (f[q][i] == 1) {
        ones |= 1 << i;
      }
    }
    for (int mask = 0; mask < 1 << (n - 1); mask++) {
      if ((mask & ones) != ones || (~mask & zeros) != zeros) {
        dp[mask] = 0;
      }
    }
    // printf("q = %d\n", q);
    // for (int mask = 0; mask < 1 << (n - 1); mask++) {
    //   printf(" mask = %d, val = %d\n", mask, dp[mask]);
    // }
  }
  int ans = 0;
  for (int i = 0; i < 1 << (n - 1); i++) add(ans, dp[i]);
  printf("%d\n", ans);
}