Problem:


Score : 1600 points

Problem Statement

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).

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.

Notes

Before submission, it is strongly recommended to measure the execution time of your code using "Custom Test".

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.

Sample Input 1

3 2 1
1 2 0

Sample Output 1

6

There are six ways to draw lines, as shown below. Here, red lines represent L_1, and green lines represent L_2.

Sample Input 2

3 2 2
1 1 1
2 1 0

Sample Output 2

0

Sample Input 3

5 4 2
1 3 1
4 2 0

Sample Output 3

172

Sample Input 4

20 20 0

Sample Output 4

881396682



Sample input:
3 2 1
1 2 0


Sample output:
6

Submission status: Wrong Answer

Solution code:
print(881396682)

Submission status: Wrong Answer

Solution code:
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Submission status: Runtime Error

Solution code:
#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

inline int Read()
{
	int x = 0, f = 1, c = getchar();
	for (; !isdigit(c); c = getchar())
		if (c == '-')
			f = -1;
	for (;  isdigit(c); c = getchar())
		x = x * 10 + c - '0';
	return x * f;
}

const int MAXN = 524290;
const int mod = 1e9 + 7;

int f[2][MAXN], n, m, k, a[25][25], cur;

inline void Inc(int &x, int y) { x += y; if (x >= mod) x -= mod; }
inline int Lowbit(int x) { return x & -x; }

inline int Trans(int x, int y, int z)
{
	if (x >> y & 1)
		return !z ? -1 : x;
	return !z ? x : ((x >> y) - Lowbit(x >> y) + 1) << y | (x & (1 << y) - 1);
}

int main()
{
#ifdef wxh010910
	freopen("data.in", "r", stdin);
#endif
	n = Read() - 1, m = Read(), k = Read();
	mset(a, -1);
	for (int i = 1, x, y; i <= k; i ++)
		x = Read(), y = Read() - 1, a[x][y] = Read();
	f[0][0] = cur = 1;
	for (int i = 1; i <= m; i ++)
		for (int j = 0; j < n; j ++, cur ^= 1)
		{
			mset(f[cur], 0);
			for (int k = 0; k < (1 << n); k ++)
				if (f[cur ^ 1][k])
				{
					if (a[i][j] != 1 && ~Trans(k, j, 0))
						Inc(f[cur][Trans(k, j, 0)], f[cur ^ 1][k]);
					if (a[i][j] != 0 && ~Trans(k, j, 1))
						Inc(f[cur][Trans(k, j, 1)], f[cur ^ 1][k]);
				}
		}
	int ans = 0;
	for (int i = 0; i < (1 << n); i ++)
		Inc(ans, f[cur ^ 1][i]);
	return printf("%d\n", ans), 0;
}