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 <iostream>
#define llint long long
#define mod 1000000007

using namespace std;

llint n, m, k;
llint x[25][25];
int y[1<<19][20];
llint dp[21][1<<19];
llint dp2[20][1<<19];

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n >> m >> k;
	n--;
	
	for(int i = 1; i <= m; i++){
		for(int j = 0; j < n; j++){
			x[i][j] = -1;
		}
	}
	
	llint a, b, c;
	for(int i = 1; i <= k; i++){
		cin >> a >> b >> c;
		b--;
		x[a][b] = c;
	}
	
	llint N = 1<<n;
	for(int i = 0; i < N; i++){
		for(int j = 0; j < n; j++){
			if((i & (1<<j)) == 0) continue;
			int t = i & ~(1<<j);
			for(int k = j+1; k < n; k++){
				if((i & (1<<k)) == 0){
					t |= (1<<k);
					break;
				}
			}
			y[i][j] = t;
		}
	}
	
	dp[0][0] = 1;
	for(int i = 0; i < m; i++){
		for(int j = 0; j < n; j++){
			for(int k = 0; k < N; k++){
				dp2[j][k] = 0;
			}
		}
		for(int j = 0; j < N; j++) dp2[n][j] = dp[i][j];
		
		for(int j = n-1; j >= 0; j--){
			for(int k = 0; k < N; k++){
				dp2[j][k] = dp2[j+1][k];
				if(k & (1<<j)) dp2[j][k] += dp2[j+1][y[k][j]], dp2[j][k] %= mod;
			}
		}
		
		for(int j = 0; j < N; j++){
			dp[i+1][j] = dp2[0][j];
			for(int k = 0; k < n; k++){
				if(x[i+1][k] == -1) continue;
				if(((j>>k)&1) != x[i+1][k]){
					dp[i+1][j] = 0;
					break;
				}
			}
		}
	}
	
	llint ans = 0;
	for(int j = 0; j < N; j++) ans += dp[m][j], ans %= mod;
	cout << ans << endl;
	
	return 0;
}