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>
#define ctz __builtin_ctz
#define EB emplace_back

typedef std::pair <int, int> pr;
typedef std::vector <pr> vector;
const int N = 24, N2 = 524354, mod = 1000000007;

int L, n, q, ALL;
int mask0[N], mask1[N], f[N2];
vector mdf[N];

inline void add(int &x, const int y) {x += y - mod, x += x >> 31 & mod;}

int main() {
	int i, j, c, S;
	scanf("%d%d%d", &L, &n, &q), ALL = ~(-1 << --L);
	for (; q; --q) scanf("%d%d%d", &i, &j, &c), (c ? mask1 : mask0)[--i] |= 1 << (L - j);
	for (i = 0; i <= ALL; ++i)
		for (c = 0, S = i & ((ALL & ~i) >> 1); S; S &= S - 1, ++c)
			 mdf[c].EB(i, i ^ ((S & -S) * 3));
	for (*f = 1, i = 0; i < n; ++i) {
		for (j = 0; j < ALL; ++j) add(f[j | (j + 1)], f[j]);
		for (c = 0; c < L; ++c) for (const pr e : mdf[c]) add(f[e.second], f[e.first]);
		for (j = 0; j <= ALL; ++j) if ((j & mask0[i]) | (~j & mask1[i])) f[j] = 0;
	}
	printf("%d\n", int(std::accumulate(f, f + (ALL + 1), 0ll) % mod));
	return 0;
}