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
#include <bits/stdc++.h>
using namespace std;
#define mod 1000000007
typedef long long ll;
ll dp[21][1 << 21];
int must[21][21];
int mustbe[21][21];
int n, m, k;
ll ans;
int fix(int a, int b) // change first 1 after b to a 0
{
	for (int i = b+1; i <= 20; i++)
	{
		if (a & (1 << i))
		{
			a ^= (1 << i);
			return a;
		}
	}
	return a;
}
int main()
{
	scanf("%d%d%d", &n, &m, &k);
	for (int i = 0; i < k; i++)
	{
		int a, b, c;
		scanf("%d%d%d", &a, &b, &c);
		a--;
		b--;
		must[a][b] = 1;
		mustbe[a][b] = c;
	}
	dp[0][0] = 1;
	for (int i = 0; i < m; i++)
	{
		for (int j = 0; j < n; j++)
		{
			for (int k = 0; k < (1 << (n-1)); k++)
			{
				if (!dp[j][k]) continue;
				ll am = dp[j][k];
				dp[j][k] = 0;
				am %= mod;
				if (j == n-1)
				{
					if (i == m-1) ans += am;
					else
					{
						dp[0][k] += am;
					}
				}
				else
				{
					int prevmove = !!(k & (1 << j));
					if (must[i][j] && mustbe[i][j] < prevmove) continue;
					int newk = (k | (1 << j));
					if (newk != k) newk = fix(newk, j);
					if (!must[i][j] || mustbe[i][j])
					{
						dp[j+1][newk] += am;
					}
					if (!prevmove && (!must[i][j] || !mustbe[i][j]))
					{
						dp[j+1][k] += am;
					}
				}
			}
		}
	}
	ans %= mod;
	printf("%lld\n", ans);
}
