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>
#define RI register int

using namespace std;

char buf[1000000], *p1 = buf, *p2 = buf;
inline char gc() {
  if (p1 == p2) p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin);
  return p1 == p2 ? EOF : *(p1++);
}
template <class T> inline void read(T &n) {
  n = 0; RI ch = gc();
  while (ch < '0' || ch > '9') ch = gc();
  while (ch >= '0' && ch <= '9') n = n * 10 + (ch ^ 48), ch = gc();
}

int const MAXN = 1 << 20, mod = 1e9 + 7;
int dp[2][MAXN], Limit[25][25];

inline int lowbit(int x) { return x & (-x); }
inline int Mo(int x) { return x >= mod ? x - mod : (x < 0 ? x + mod : x); }

int main() {
 // freopen("a.in", "r", stdin);
 // freopen("a.out", "w", stdout);
  int n, m, k;
  read(n), read(m), read(k); --n;
  for (RI i = 1, a, b, c; i <= k; ++i)
    read(a), read(b), read(c), Limit[a][b] = c + 1;
  int nxt = 0, cur = 1;
  int Max = (1 << n);
  dp[0][0] = 1;
  for (RI i = 1; i <= m; ++i) {
    for (RI j = 0; j < n; ++j) {
      swap(nxt, cur);
      memset(dp[nxt], 0, sizeof(dp[nxt]));
      int bit = 1 << j, lower = (Max - 1) ^ ((bit << 1) - 1);
      for (RI s = 0; s < Max; ++s) {
        if (Limit[i][j + 1] != 2)
          if (!(s & bit))
            dp[nxt][s] = Mo(dp[nxt][s] + dp[cur][s]);
        if (Limit[i][j + 1] != 1) {
          int nxts;
          if (s & bit) nxts = s;
          else if (!(s & lower)) nxts = s | bit;
          else nxts = (s | bit) ^ lowbit(s & lower);
          dp[nxt][nxts] = Mo(dp[nxt][nxts] + dp[cur][s]);
        }
      }
    }
  }
  int ans = 0;
  for (RI s = 0; s < Max; ++s) ans = Mo(ans + dp[nxt][s]);
  printf("%d\n", ans);
  return 0;
}