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>
#include <cstdio>
#include <algorithm>
#define N 21
#define ll long long
#define lb(x) ((x) & (-(x)))
#define mod 1000000007

using namespace std;

int n, m, k;
int c[N][N];
ll f[2][1 << N];

int main(){
	scanf("%d%d%d",&n, &m, &k);--n;
	for(int i = 1; i <= k; ++i){
		int x, y, z;scanf("%d%d%d", &x, &y, &z);
		c[x][y - 1] = z + 1;
	}
	f[0][0] = 1;
	int lst = 1, cur = 0, all = (1 << n) - 1;
	for(int i = 1; i <= m; ++i)
		for(int j = 0; j < n; ++j){
			swap(lst, cur);
			int now = 1 << j, bar = all ^ ((now << 1) - 1);
			for(int s = 0; s <= all; ++s) f[cur][s] = 0;
			for(int s = 0; s <= all; ++s){
				if(!f[lst][s]) continue;
				if(c[i][j] != 2 && (!(now & s))) (f[cur][s] += f[lst][s]) %= mod;
				if(c[i][j] != 1){
					int x;
					if(now & s) x = s;
					else{
						if(!(bar & s)) x = s | now;
						else x = (s | now) ^ lb(bar & s);
					}
					(f[cur][x] += f[lst][s]) %= mod;
				}
			}
		}
	ll ans = 0;
	for(int i = 0; i <= all; ++i) (ans += f[cur][i]) %= mod;
	cout << ans << endl;
	return 0;
}