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).

<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

Solution (in CPP):

Solution:
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <algorithm>
#include <cassert>
#define rep(i, a, b) for (int i = (a), _ = (b); i <= _; ++ i)
#define per(i, a, b) for (int i = (a), _ = (b); i >= _; -- i)
#define For(i, a, b) for (int i = (a), _ = (b); i < _; ++ i)
#define ri rd<int>
using namespace std;
const int mod = 1e9 + 7;
const int maxN = 23;
const int maxS = (1 << 20) + 7;

inline int pls(int x, int y) {return (x + y) % mod;}
inline int mns(int x, int y) {return pls(x, mod - y);}
inline int mul(int x, int y) {return 1LL * x * y % mod;}
inline void Add(int &x, int y) {x = pls(x, y);}
inline void Mul(int &x, int y) {x = mul(x, y);}
inline void Minus(int &x, int y) {x = mns(x, y);}

template<class T> inline T rd() {
	bool f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = 0;
	T x = 0; for (; isdigit(c); c = getchar()) x = x * 10 + c - 48; return f ? x : -x;
}

int n, m, K;
int lim[maxN][maxN];
int f[maxN][maxS];

inline int lb(int x) {return x & -x;}

inline int chg(int x, int p) {
	int t = x & ((1 << p) - 1); x >>= p;
	if (lb(x)) x ^= lb(x);
	x ^= 1;
	return x << p | t;
}

int main() {

	n = ri() - 1, m = ri(), K = ri();
	memset(lim, -1, sizeof lim);
	rep (i, 1, K) {
		int x = ri(), y = ri(), z = ri();
		lim[x][y] = z;
	}

	int mask = (1 << n) - 1;
	f[0][0] = 1;
	rep (i, 1, m) {
		rep (j, 1, n) {
			memset(f[i], 0, sizeof f[i]);
			rep (s, 0, mask) if (f[i-1][s]) {
				if (s >> (j-1) & 1) {
					if (lim[i][j] == 0) continue;
					Add(f[i][s], f[i-1][s]);
				}
				else {
					if (lim[i][j] != 1) Add(f[i][s], f[i-1][s]);
					if (lim[i][j] != 0) Add(f[i][chg(s, j-1)], f[i-1][s]);
				}
			}
			rep (s, 0, mask) f[i-1][s] = f[i][s];
		}
	}

	int res = 0;
	rep (s, 0, mask) Add(res, f[m][s]);
	printf("%d\n", res);

	return 0;
}