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<iostream>
#include<cstdio>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<cassert>
#define PB push_back
#define MP make_pair
#define sz(v) (in((v).size()))
#define forn(i,n) for(in i=0;i<(n);++i)
#define forv(i,v) forn(i,sz(v))
#define fors(i,s) for(auto i=(s).begin();i!=(s).end();++i)
#define all(v) (v).begin(),(v).end()
using namespace std;
typedef int in;
typedef long long ll;
typedef vector<in> VI;
typedef vector<VI> VVI;
const in mdl=1000000007;
in p2(in a){
  return 1<<a;
}
vector<ll> wys;
in n;
void goup(){
  in pn=p2(n);
  in pi;
  forn(i,n){
    pi=p2(i);
    in pl=p2(n-1-i);
    in pr=p2(i);
    in msk;
    in nxt;
    forn(msk,pn){
      if(msk&pi)
	continue;
      nxt=i;
      while(nxt<n && !(msk&p2(nxt)))
	++nxt;
      if(nxt<n)
	wys[(msk|pi)^p2(nxt)]+=wys[msk];
      else
	wys[msk|pi]+=wys[msk];
    }
  }
}

VI condmembers;
VI condvalue;
int main(){
  ios::sync_with_stdio(0);
  cin.tie(0);
  in m,k;
  cin>>n>>m>>k;
  --n;
  wys.resize(p2(n));
  wys[0]=1;
  condmembers.resize(m);
  condvalue.resize(m);
  in a,b,c;
  forn(z,k){
    cin>>a>>b>>c;
    --a;
    --b;
    condmembers[a]|=p2(b);
    condvalue[a]|=c*p2(b);
  }
  forn(z,m){
    goup();
    forn(msk,p2(n)){
      if((msk&condmembers[z])!=condvalue[z])
	wys[msk]=0;
      wys[msk]%=mdl;
    }
  }
  ll sm=0;
  forn(i,p2(n))
    sm+=wys[i];
  sm%=mdl;
  if(sm<0)
    sm+=mdl;
  cout<<sm<<endl;
  return 0;
}