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
#include <bits/stdc++.h>

const int MAXN = 20;
const int mod = 1000000007;
typedef long long LL;
void reduce(int & x) { x += x >> 31 & mod; }
int mul(int a, int b) { return (LL) a * b % mod; }
int n, m, K;
int lim[MAXN], out[MAXN], f[2][1 << MAXN - 1];
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n >> m >> K; --n;
	for (int i = 1, a, b, c; i <= K; ++i) {
		std::cin >> a >> b >> c; --a, --b;
		lim[a] |= 1 << b, out[a] |= c << b;
	}
	int now = 1, lst = 0;
	f[now][0] = 1;
	const int U = 1 << n;
	for (int T = 0; T < m; ++T)
		for (int j = 0; j != n; ++j) {
			std::swap(lst, now);
			memset(f[now], 0, U << 2);
			for (int k = 0; k != U; ++k) if (int t = f[lst][k]) {
				if ((~lim[T] >> j & 1) || (out[T] >> j & 1) == 0) {
					if (~k >> j & 1) reduce(f[now][k] += t - mod);
				}
				if ((~lim[T] >> j & 1) || (out[T] >> j & 1) == 1) {
					if (k >> j & 1) reduce(f[now][k] += t - mod);
					else {
						int nx = k >> j ? __builtin_ctz(k >> j) + j : 30;
						int tx = (k & (1 << j) - 1) | (k >> nx + 1) << nx + 1 | 1 << j;
						reduce(f[now][tx] += t - mod);
					}
				}
			}
		}
	int ans = 0;
	for (int i = 0; i != U; ++i) reduce(ans += f[now][i] - mod);
	std::cout << ans << std::endl;
	return 0;
}
