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 <bits/stdc++.h>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<60;
const double pi=acos(-1);
const double eps=1e-8;
const ll mod=1e9+7;
const int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};

int n,m,q;
vvi a;

int main(){
	cin>>n>>m>>q;
	n--;
	a=vvi(m,vi(n,-1));
	for(int i=0;i<q;i++){
		int A,B,C;
		cin>>A>>B>>C;
		a[A-1][B-1]=C;
	}
	int S=1<<n;
	vi dp(S),c(S);
	vvp b(S);
	dp[0]++;
	int sum=0;
	for(int i=1;i<S;i++){
		vi b_;
		if(i&1) b_.push_back(0);
		for(int j=0;j<n-1;j++) if(i&1<<(j+1)&&!(i&1<<j)) b_.push_back(j);
		c[i]=b_.size();
		b[i]=vp(1<<c[i]);
		for(int j=0;j<1<<c[i];j++){
			int T=i,t=-1;
			for(int k=0;k<c[i];k++) if(j&1<<k) T-=1<<b_[k],t*=-1;
			b[i][j]={T,t};
		}
	}
	for(int i=0;i<m;i++){
		vi DP(S);
		for(int j=0;j<S;j++){
			DP[j]=dp[j];
			for(int k=1;k<1<<c[j];k++){
				int T=b[j][k].first,t=b[j][k].second;
				if(t==1) (DP[j]+=DP[T])%=mod;
				else DP[j]=(DP[j]-DP[T]+mod)%mod;
			}
		}
		for(int j=0;j<S;j++) for(int k=0;k<n;k++) if(a[i][k]!=-1&&(j>>(n-k-1))%2!=a[i][k]) DP[j]=0;
		dp=DP;
	}
	int res=0;
	for(int i=0;i<S;i++) (res+=dp[i])%=mod;
	cout<<res<<endl;
}