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 N = 20, mod = 1000000007;

void reduce(int &x) { x += x >> 31 & mod; }
int n, m, k, dp[2][1 << N - 1], ans, q[N][N];

int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::cin >> n >> m >> k;
	for (int i = 0; i < m; ++i)
		for (int j = 0; j < n; ++j) q[i][j] = -1;
	for (int i = 0, a, b, c; i < k; ++i)
		std::cin >> a >> b >> c, q[a - 1][b - 1] = c;
	int now = 0, lst = 1;
	dp[now][0] = 1;
	for (int t = 0; t < m; ++t) {
		for (int j = 0; j < n - 1; ++j) {
			std::swap(now, lst), std::memset(dp[now], 0, sizeof dp[0]);
			for (int k = 0; k < 1 << n - 1; ++k) if (int v = dp[lst][k]) {
				if (q[t][j] != 1 && !(k >> j & 1)) reduce(dp[now][k] += v - mod);
				if (q[t][j] != 0)
					if (k >> j & 1) reduce(dp[now][k] += v - mod);
					else {
						int p = k & -1 << j; p &= p - 1, p |= k & ~(-1 << j) | 1 << j;
						reduce(dp[now][p] += v - mod);
					}
			}
		}
	}
	for (int s = 0; s < 1 << n - 1; ++s)
		reduce(ans += dp[now][s] - mod);
	std::cout << ans << '\n';
	return 0;
}