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>
#define MOD 1000000007
int can[25][25], dp[25][1000005], nxt[1000005][25];
int main()
{
	// freopen("AGC017-F.in", "r", stdin);
	int n, m, lim;
	scanf("%d%d%d", &n, &m, &lim);
	for (int i = 0; i < m; i++)
	{
		for (int j = 0; j < n - 1; j++)
			can[i][j] = 3;
	}
	while (lim--)
	{
		int x, y, z; 
		scanf("%d%d%d", &x, &y, &z);
		can[--x][--y] = 1 << z;
	}
	for (int i = 0; i < 1 << n - 1; i++)
	{
		for (int j = 0; j < n - 1; j++)
		{
			if (i & (1 << j))
				continue;
			for (int k = j + 1; k < n - 1; k++)
			{
				if (i & (1 << k))
				{
					nxt[i][j] = i ^ (1 << j) ^ (1 << k);
					break;
				}
			}
			if (!nxt[i][j])
				nxt[i][j] = i ^ (1 << j);
		}
	}
	dp[n - 1][0] = 1;
	for (int i = 0; i < m; i++)
	{
		for (int j = 0; j < 1 << n - 1; j++)
		{
			dp[0][j] = dp[n - 1][j];
			dp[n - 1][j] = 0;
		}
		for (int j = 0; j < n - 1; j++)
		{
			for (int k = 0; k < 1 << n - 1; k++)
			{
				if (!dp[j][k])
					continue;
				if (can[i][j] & (1 << (k >> j & 1)))
					(dp[j + 1][k] += dp[j][k]) %= MOD;
				if ((can[i][j] & 2) && !(k & (1 << j)))
					(dp[j + 1][nxt[k][j]] += dp[j][k]) %= MOD;
				dp[j][k] = 0;
			}
		}
	}
	int ans = 0;
	for (int i = 0; i < 1 << n - 1; i++)
		(ans += dp[n - 1][i]) %= MOD;
	printf("%d\n", ans);
	return 0;
}