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:
//              +-- -- --++-- +-In the name of ALLAH-+ --++-- -- --+              \\

#include <bits/stdc++.h>

#define F first
#define S second
#define _sz(x) (int)x.size()

using namespace std ;
using ll = long long ;
using ld = long double ;
using pii = pair <int , int> ;

const int N = 20 + 2 , MASK = (1 << 20) + 20 , mod = 1e9 + 7 ;
int n , m , k , val[N][N] , dp[N][MASK] , go[N][MASK] ;

inline void add (int &a , int b) {
	a += b ;
	if (a >= mod) a -= mod ;
}

int main(){
	ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0) ;

	memset(val , -1 , sizeof val) ;

	cin >> n >> m >> k ;

	for (int i = 0 , a , b , c ; i < k ; i ++) {
		cin >> a >> b >> c ;
		a -- , b -- ;

		val[a][b] = c ;
	}

	n -- ;

	if (!n) return cout << "1\n" , 0 ;

	for (int mask = 0 ; mask < (1 << n) ; mask ++) {
		int last = -1 ;
		for (int i = 0 ; i < n ; i ++) {
			if (mask >> i & 1) last = i ;

			go[i][mask] = mask ^ (last == -1 ? 0 : (1 << last)) ;
		}
	}

	dp[n][0] = 1 ;

	for (int t = 0 ; t < m ; t ++) {
		for (int mask = 0 ; mask < (1 << n) ; mask ++) {
			dp[0][mask] = dp[n][mask] ;
			for (int i = 1 ; i <= n ; i ++) dp[i][mask] = 0 ;
		}

		for (int i = 0 ; i < n ; i ++) {
			for (int mask = 0 ; mask < (1 << n) ; mask ++) {
				if (val[t][i] != 1 && !(mask & (1 << n - i - 1))) {
					add(dp[i + 1][mask] , dp[i][mask]) ;			
				}
				if (val[t][i] != 0) {
					if (mask & (1 << n - i - 1)) {
						add(dp[i + 1][mask] , dp[i][mask]) ;
					}
					else {
						add(dp[i + 1][go[n - i - 1][mask] ^ (1 << n - i - 1)] , dp[i][mask]) ;
					}
				}
			}
		}
	}

	int ans = 0 ;
	for (int mask = 0 ; mask < (1 << n) ; mask ++) add(ans , dp[n][mask]) ;

	cout << ans << '\n' ;
}