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 <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
template<typename T> inline void chkmin(T &a, const T &b) { a = a < b ? a : b; }
template<typename T> inline void chkmax(T &a, const T &b) { a = a > b ? a : b; }

const int MAXN = 25, MAXM = 1 << 19, MOD = 1e9 + 7;
int f[2][MAXM], g[2][MAXM], nxt[MAXN][MAXN], n, m, K;
inline void add(int &a, int b) { a += b - MOD, a += a >> 31 & MOD; }

int main() {
	scanf("%d%d%d", &n, &m, &K);
	memset(nxt, -1, sizeof(nxt));
	for (int i = 1; i <= K; i++) {
		int a, b, c;
		scanf("%d%d%d", &a, &b, &c);
		nxt[a][b - 1] = c;
	}
	f[0][0] = 1;
	for (int i = 1; i <= m; i++) {
		int *f1 = f[i & 1], *f0 = f[~i & 1];
		memcpy(g[1], f0, sizeof(g[1]));
		for (int j = 0; j < n - 1; j++) {
			int *g1 = g[j & 1], *g0 = g[~j & 1];
			memset(g1, 0, sizeof(g[0]));
			for (int k = 0; k < 1 << (n - 1); k++) if (g0[k]) {
				if (~k >> j & 1 && nxt[i][j] != 1) add(g1[k], g0[k]);
				if (nxt[i][j] != 0) {
					if (~k >> j & 1) {
						int t = k >> j;
						if (!t) { add(g1[k ^ 1 << j], g0[k]); continue; }
						add(g1[k ^ 1 << j ^ ((t & -t) << j)], g0[k]);
					} else add(g1[k], g0[k]);
				}
			}
		}
		memcpy(f1, g[n & 1], sizeof(g[0]));
	}
	int ans = 0;
	for (int i = 0; i < 1 << (n - 1); i++) add(ans, f[m & 1][i]);
	printf("%d\n", ans);
	return 0;
}