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<bits/stdc++.h>
using namespace std;
#define MAXN	25
#define MAXS	1048576
#define P	1000000007
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
bool limited[MAXN][MAXN], type[MAXN][MAXN];
int ans[2][MAXN][MAXS];
void update(int &x, int y) {
	x = (x + y) % P;
}
int func(int s, int tmp) {
	int t = (tmp - 1) & s;
	s &= ~(tmp - 1);
	s ^= s & -s;
	s ^= tmp;
	return s | t;
}
int main() {
	int n, m, k;
	read(n), read(m), read(k);
	for (int i = 1; i <= k; i++) {
		int x, y, z;
		read(x), read(y), read(z);
		limited[x - 1][y - 1] = true;
		type[x - 1][y - 1] = z;
	}
	ans[0][0][0] = 1;
	int goal = (1 << n - 1) - 1;
	for (int i = 0, now = 0, dest = 1; i < m; i++, now ^= 1, dest ^= 1) {
		memset(ans[dest], 0, sizeof(ans[dest]));
		for (int j = 0; j < n - 1; j++) {
			int tmp = 1 << j;
			for (int s = 0; s <= goal; s++) {
				if ((!limited[i][j] || !type[i][j]) && ((tmp & s) == 0))
					update(ans[now][j + 1][s], ans[now][j][s]);
				if (!limited[i][j] || type[i][j]) {
					if (tmp & s) update(ans[now][j + 1][s], ans[now][j][s]);
					else update(ans[now][j + 1][func(s, tmp)], ans[now][j][s]);
				}
			}
		}
		for (int s = 0; s <= goal; s++)
			update(ans[dest][0][s], ans[now][n - 1][s]);
	}
	int finalans = 0;
	for (int i = 0; i <= goal; i++)
		update(finalans, ans[m & 1][0][i]);
	cout << finalans << endl;
	return 0;
}