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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>

using namespace std;

const int MAXN = 20 + 1;
const int MAXM = 1 << 19;
const int MOD = 1000000007;

int n, m, k;
int lim[MAXN][MAXN];
int f[2][MAXM];

void add(int &x, int y)
{
	x = (x + y) % MOD;
}

int lowbit(int x)
{
	return x & (-x);
}

int main()
{
	ios::sync_with_stdio(false);
	// freopen("1.in", "r", stdin);
	// freopen("1.out", "w", stdout);

	cin >> n >> m >> k;
	memset(lim, -1, sizeof(lim));
	for(int i = 1; i <= k; i++)
	{
		int x, y, z;
		cin >> x >> y >> z;
		lim[x - 1][y - 1] = z;
	}
	int cur = 0, maxx = 1 << (n - 1);
	f[cur][0] = 1;
	for(int i = 0; i < m; i++)
		for(int j = 0; j < n - 1; j++)
		{
			cur ^= 1;
			memset(f[cur], 0, sizeof(f[cur]));
			if(lim[i][j] != 1)
			{
				for(int s = 0; s < maxx; s++)
					if(~(s >> j) & 1)
						add(f[cur][s], f[cur ^ 1][s]);
			}
			if(lim[i][j] != 0)
			{
				for(int s = 0; s < maxx; s++)
					if((s >> j) & 1)
						add(f[cur][s], f[cur ^ 1][s]);
					else
					{
						int t = lowbit(s & ~((1 << j) - 1));
						add(f[cur][s ^ (1 << j) ^ t], f[cur ^ 1][s]);
					}
			}
			// for(int s = 0; s < maxx; s++)
				// cout << i << ' ' << j << ' ' << s << ' ' << f[cur][s] << endl;
		}
	int ans = 0;
	for(int s = 0; s < maxx; s++)
		add(ans, f[cur][s]);
	cout << ans << endl;



	return 0;
}