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>
#define LL long long
using namespace std;
const LL mod = 1e9 + 7;
int n, m, K;
LL dp[21][1 << 19], pd[1 << 19][21];
int lim[21][21], pw2[21];
//inline void ADD(int &x,int &y){x = (x + y < mod) ? (x + y) : (x + y - mod);}
inline int lowbit(const int &i){ return i & (-i); }
#define add(x,y) ((x+y<mod)?(x+=y):(x+=y-mod))
void trans(int i)
{
	for (int s=0;s<(1<<n);s++)
	{
		memset(pd[s], 0, sizeof(pd[s][0]) * (n + 1));
		bool f = 1;
		for (int j=0;j<n;j++) if(~lim[i][j] && ((s >> j) & 1) != lim[i][j]){ f = 0; break; }
		pd[s][0] = (f ? dp[i - 1][s] : 0);
	}
	LL tmp;
	for (int i=0;i<n;i++) for (int s=0;s<(1<<n);s++) if(tmp = pd[s][i]) 
	{
		if(!((s >> i) & 1))
		{
			int t = pw2[n] ^ pw2[i + 1];
			t &= s;
			pd[s ^ lowbit(t) | (1 << i)][i + 1] += tmp;
		}
		pd[s][i + 1] += tmp;
	}
	for (int s=0;s<(1<<n);s++) dp[i][s] = pd[s][n] % mod;
}

int main()
{
	for (int i=0;i<=20;i++) pw2[i] = (1 << i) - 1;
	cin>>n>>m>>K;
	memset(lim, -1, sizeof lim);
	for (int i=0;i<K;i++){int x, y;cin>>x>>y;cin>>lim[x][y - 1];}
	n--;
	for (int s=0;s<(1<<n);s++)dp[0][s]=1;
	for (int i=1;i<=m;i++)trans(i);
	printf("%d\n", dp[m][(1 << n) - 1]);
}