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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll ;
#define rep(i, a, b) for (int i = a ; i <= b; ++ i) 
const int N = 25, M = 6e5 + 5, mo = 1e9 + 7 ;
using namespace std ;

int n, m, T, A[N][N], g[N][M], f[2][N][M], nxt[M][N] ;
void upd(int &x, int y) { x = (x + y) % mo ; }
int main() {
	scanf("%d%d%d", &n, &m, &T) ;
	if (n == 1) { puts("1") ; return 0 ; }
	rep(i, 0, m - 1) rep(j, 0, n - 2) A[i][j] = - 1 ;
	for (int x, y, z ; T -- ; ) {
		scanf("%d%d%d", &x, &y, &z) ;
		A[x - 1][y - 1] = z ;
	} 
	int c = 0, st = (1 << (n - 1)) - 1 ;
	rep(i, 0, st) rep(j, 0, n - 2) {
		nxt[i][j] = - 1 ;
		rep(k, j + 1, n - 2) if ((i >> k) & 1) { nxt[i][j] = k ; break ; }
	}
	g[n - 2][0] = 1 ;
	rep(u, 0, m - 1) {
		c ^= 1 ;
		rep(i, 0, n - 1) rep(j, 0, st) f[c][i][j] = 0 ;
		rep(i, 0, st) upd(f[c][0][i], g[n - 2][i]) ;
		rep(i, 0, n - 2) rep(j, 0, st) g[i][j] = 0 ;
		rep(i, 0, n - 2) rep(j, 0, st) {
			upd(f[c][i + 1][j], f[c][i][j]) ;
			if ((j >> i) & 1) continue ;
			if (nxt[j][i] == - 1) upd(g[i][(j & ((1 << i) - 1)) | (1 << i)], f[c][i][j]) ;
				else upd(f[c][i + 1][j ^ (1 << i) ^ (1 << nxt[j][i])], f[c][i][j]) ;
		}
		rep(i, 0, st) upd(g[n - 2][i], f[c][n - 1][i]) ;
		rep(i, 0, n - 3) rep(j, 0, st) upd(g[i + 1][j], g[i][j]), upd(g[i + 1][j ^ (1 << (i + 1))], g[i][j]) ;
		rep(i, 0, st) rep(j, 0, n - 2) if (A[u][j] != - 1 && A[u][j] != ((i >> j) & 1)) g[n - 2][i] = 0 ;
	}
	int ans = 0 ;
	rep(i, 0, st) upd(ans, g[n - 2][i]) ;
	printf("%d\n", ans) ;
	return 0 ;
}