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).

<image>

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 <algorithm>
#include <cstring>
#include <ctime>

#define Rep(i, n) for (int i = 1; i <= n; i ++)
#define Rep0(i, n) for (int i = 0; i <= n; i ++)
#define RepG(i, x) for (int i = head[x]; i; i = edge[i].next)
#define v edge[i].to
#define mp(a, b) make_pair(a, b)

using namespace std;

typedef long long LL;
const int N = 21;
const int N0 = 1050010;
const int mod = 1e9 + 7;

int l[N], r[N];
int f[2][N][N0];

void upd(int &a, int b) { a = (a + b) % mod;}

int main()
{
	//freopen("f.in", "r", stdin);
	int n, m, k;
	scanf("%d%d%d", &n, &m, &k);
	Rep(i, k) {
		int a, b, c;
		scanf("%d%d%d", &a, &b, &c);
		if (!c) l[a] |= (1 << (b - 1));
		else r[a] |= (1 << (b - 1));
	}
	
	f[1][1][0] = 1;
	Rep(i, m){
		Rep(j, n) Rep0(s, (1 << n) - 1) f[(i + 1) & 1][j][s] = 0;
		Rep(j, n) Rep0(s, (1 << n) - 1){
			int i0 = (i & 1);
			if (!f[i0][j][s]) continue;
			if (j == n){
				if ((l[i] & s) || (r[i] & s) != r[i]) continue;
				upd(f[(i + 1) & 1][1][s], f[i0][j][s]);
			}
			else {
				upd(f[i0][j + 1][s], f[i0][j][s]);
				if (!(s & (1 << (j - 1)))){
					int tmp = s >> (j - 1), ts = s;
					tmp = (tmp & (-tmp)) << (j - 1);
					ts -= tmp, ts += 1 << (j - 1);
					upd(f[i0][j + 1][ts], f[i0][j][s]);
				}
			}
		}
	}
	
	int ans = 0;
	Rep0(i, (1 << n) - 1) upd(ans, f[(m + 1) & 1][1][i]);
	
	printf("%d\n", ans);
	
	return 0;
}
/*
3 2 1
1 2 0

5 4 2
1 3 1
4 2 0
*/