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 <cstdio>
#include <cstring>

#define iter(i, n) for (int i = 1; i <= n; ++i)
#define iter0(i, n) for (int i = 0; i < n; ++i)

#define NR 22
#define SR (1 << 20)

const int mod = 1000000007;

int n, m, K;
int must[NR][NR], f[2][NR][SR];

inline void inc(int &x, int y) { x = (x + y) % mod; }
inline int remove_lowbit(int x) { return x ^ (x & -x); }

int main() {
	scanf("%d%d%d", &n, &m, &K); --n;
	int M = 1 << n;
	memset(must, -1, sizeof(must));
	iter(i, K) {
		int a, b, c;
		scanf("%d%d%d", &a, &b, &c);
		must[a - 1][b - 1] = c;
	}
	f[0][0][0] = 1;
	iter0(i, m) {
		bool d = i % 2;
		memset(f[!d], 0, sizeof(f[!d]));
		iter0(j, n) {
			bool _d = (d ^ (j + 1 == n));
			int _j = (j + 1) % n;
			iter0(s, M) if (f[d][j][s]) {
				if (!(s >> j & 1) && must[i][j] != 1)
					inc(f[_d][_j][s], f[d][j][s]);
				int tmp = (1 << j) - 1;
				int _s = remove_lowbit(s & ~tmp) ^ (1 << j) ^ (s & tmp);
				//printf("%d -%d-> %d\n", s, j, _s);
				if (must[i][j] != 0) inc(f[_d][_j][_s], f[d][j][s]);
				//printf("f[%d][%d][%d]=%d\n", i, j, s, f[d][j][s]);
			}
		}
	}
	int ans = 0;
	iter0(s, M) {
	//	printf("f[%d][0][%d]=%d\n", m, s, f[m%2][0][s]);
		inc(ans, f[m % 2][0][s]);
	}
	printf("%d\n", ans);
	return 0;
}