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:
#include <bits/stdc++.h>

#define For(i, j, k) for(int i = j; i <= k; i++)

using namespace std;

const int N = 21;
const int Mod = 1e9 + 7;

int n, m, q;
int dp[2][N][1 << 19];
bool ban[N][N][2];

void add(int &x, int y){
	x += y;
	if(x >= Mod) x -= Mod;
}

int main(){	
	scanf("%d%d%d", &n, &m, &q);
	while(q--){
		int a, b, c;
		scanf("%d%d%d", &a, &b, &c);
		ban[a][b][c ^ 1] = true;
	}

	int s = 0, P = (1 << (n - 1)) - 1;
	dp[0][0][0] = 1;
	For(i, 1, m){
		For(j, 0, n - 2){
			int msk = (1 << j) - 1;
			For(k, 0, P){
				int &val = dp[s][j][k];
				if(!val) continue;

				if(!(k & (1 << j)) && !ban[i][j + 1][0])
					add(dp[s][j + 1][k], val);
				if(!ban[i][j + 1][1]){
					if(k & (1 << j))
						add(dp[s][j + 1][k], val);
					else{
						int a = k & (P ^ msk), b = (k & msk) | (1 << j);
						a -= a & (-a);
						add(dp[s][j + 1][a | b], val);
					}
				}

				val = 0;
			}
			if(j == n - 2) For(k, 0, P) dp[s ^ 1][0][k] = dp[s][n - 1][k], dp[s][n - 1][k] = 0;
		}
		s ^= 1;
	}

	int ans = 0;
	For(i, 0, P) add(ans, dp[s][0][i]);
	printf("%d\n", ans);

	return 0;
}