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 <cmath>
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <algorithm>
#define R register
#define IN inline
#define W while
#define gc getchar()
#define ll long long
#define MOD 1000000007
#define lbt(i) ((i) & (-(i)))
#define MX 21
int n, m, k, all, last = 1, cur = 0;
int dir[MX][MX], dp[2][1 << MX];
template <class T>
IN void in(T &x)
{
	x = 0; R char c = gc;
	for (; !isdigit(c); c = gc);
	for (;  isdigit(c); c = gc)
	x = (x << 1) + (x << 3) + c - 48;
}
IN void ad(int &x, R int v) {x += v; if (x >= MOD) x -= MOD;}
int main(void)
{
	int x, y, foo, bar, ans = 0;
	in(n), in(m), in(k), --n; all = (1 << n) - 1;
	for (R int i = 1; i <= k; ++i)
	in(x), in(y), in(foo), dir[x][y - 1] = foo + 1;
	dp[0][0] = 1;
	for (R int i = 1; i <= m; ++i)
	for (R int j = 0; j < n; ++j)
	{
		std::swap(last, cur); foo = 1 << j, bar = all ^ ((foo << 1) - 1);
		for (R int s = 0; s <= all; ++s) dp[cur][s] = 0;
		for (R int s = 0; s <= all; ++s)
		{
			if (dp[last][s])
			{
				if (dir[i][j] != 2 && (!(foo & s)))//go left
				ad(dp[cur][s], dp[last][s]);
				if (dir[i][j] != 1)//go right
				{
					if (foo & s) x = s;//must go right
					else
					{
						if (!(bar & s)) x = s | foo;//no more bit 1
						else x = (s | foo) ^ lbt(bar & s);//bit 1 exists in higher bits
					}
					ad(dp[cur][x], dp[last][s]);
				}
			}
		}
	}
	for (R int i = 0; i <= all; ++i) ad(ans, dp[cur][i]);
	printf("%d", ans);
}