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>
#include<algorithm>
#include<vector>
using namespace std;
#define rep(i,n) for(int i=0,_end_=(n); i<_end_;++i)
#define per(i,n) for(int i=(n)-1; i>=0 ; --i)
#define forn(i,l,r) for(int i=(l),_end_=(r); i<=_end_;++i)
using namespace std;
typedef long long ll;
const ll linf=1ll<<60;
const ll mod=1e9+7;
int n,m,K;
ll dp[21][1100005],pd[1100005][21];
int xjt[21][21],pw2[21];
inline void ADD(int &x,const int &y)
{
	x=(x+y<mod) ? (x+y) : (x+y-mod);
	return;
}
inline int lowbit(const int &i){ return i&(-i); }
void trans(int i)
{
	rep(s,1<<n)
	{
		memset(pd[s],0,sizeof(pd[s][0])*(n+1));
		bool f=1;
		rep(j,n) if(~xjt[i][j] && ((s>>j)&1) !=xjt[i][j]){ f=0; break; }
		pd[s][0]=(f ? dp[i-1][s] : 0);
	}
	ll tmp;
	rep(i,n) rep(s,1<<n) 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;
	}
	rep(s,1<<n) dp[i][s]=pd[s][n]%mod;
	return;
}

int main()
{
	rep(i,21) pw2[i]=(1<<i)-1;
	scanf("%d%d%d",&n,&m,&K);
	memset(xjt,-1,sizeof xjt);
	rep(i,K)
	{
		int x,y;scanf("%d%d",&x,&y);
		scanf("%d",&xjt[x][y-1]);
	}
	--n;
	rep(s,1<<n) dp[0][s]=1;
	forn(i,1,m) trans(i);
	printf("%d\n",dp[m][(1<<n)-1]);
	return 0;
}