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 <cstring>

const int MOD = 1000000007;

const int N = 20 + 3;
const int M = 20 + 3;
const int K = (N - 1) * M;

const int T = 1 << 20;

int n, m, k;

int c[M][N];

int t;

int f[2][N][T];

int z = 0;

inline int lowbit(int n) {
	return n & -n;
}

inline int full(int j) {
	return (1 << (j)) - 1;
}

inline int func(int s, int j) {
	int top = s & ~full(j);
	int bottom = s & full(j);
	return top - lowbit(top) + (1 << j) + bottom;
}

inline int get(int s, int j) {
	return s & (1 << j);
}

void solve() {
	f[0][0][0] = 1;
	for (int _i = 0; _i < m; _i++) {
		int i = _i % 2;
		for (int j = 0; j < n; j++) {
			for (int s = 0; s < t; s++) {
				if (f[i][j][s] == 0) continue;
//				printf("%d %d %d : %d %d\n", _i, j, s, f[i][j][s], c[_i][j]);
				if (c[_i][j] != 1 && get(s, j) == 0) {
					f[i][j + 1][s] += f[i][j][s];
					f[i][j + 1][s] %= MOD;
				}
				if (c[_i][j] != 0) {
					f[i][j + 1][func(s, j)] += f[i][j][s];
					f[i][j + 1][func(s, j)] %= MOD;
				}
			}
		}
		memset(f[(i + 1) % 2], 0, sizeof f[(i + 1) % 2]);
		for (int s = 0; s < t; s++)
			f[(i + 1) % 2][0][s] = f[i][n][s];
	}
}

int main(int argc, char *argv[]) {
	memset(c, -1, sizeof c);
	scanf("%d%d%d", &n, &m, &k);
	n--;
	t = 1 << n;
	for (int i = 0; i < k; i++) {
		int a, b, c_;
		scanf("%d%d%d", &a, &b, &c_);
		c[a - 1][b - 1] = c_;
	}
	solve();
	int ans = 0;
	for (int s = 0; s < t; s++) {
		ans += f[(m - 1) % 2][n][s];
		ans %= MOD;
	}
	printf("%d\n", ans);
	return 0;
}