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 rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define REP(i,n) for(int i=(0);i<(n);i++)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;

template<class T> inline void read(T &x){
	int f=0;x=0;char ch=getchar();
	for(;!isdigit(ch);ch=getchar())f|=(ch=='-');
	for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';
	if(f)x=-x;
}

const int N=21,M=(1<<19)+1,mod=1e9+7;
int f[1<<N],R[1<<N],p[N][1<<N],zero[N],one[N];
int n,m,K,a,b,c,k,tt;

void add(int &x,int y){
	x=(x+y>=mod?x+y-mod:x+y);
}

#define doit(s) \
	if(f[s]&&(k=p[j][s])&&!(s>>(k-1)&1)) \
	add(f[s^(k<n?(1<<k):0)^(1<<(k-1))],f[s])

int main(){
	read(n),read(m),read(K),n--;
	rep(i,1,K){
		read(a),read(b),read(c);
		if(c) one[a]|=1<<(b-1);
		else zero[a]|=1<<(b-1);
	}
	REP(s,1<<n){
		int c=0;
		REP(i,n) if(s>>i&1) p[++c][s]=i;
		p[c+1][s]=n;
		R[s]=(s&1?R[s^1]|(1<<(n-1)):R[s>>1]>>1);
	}
	f[0]=1;
	int A=1<<(n-1),B=1<<n;
	rep(i,1,m){
		rep(j,1,n) REP(s,1<<n) doit(R[s]);
		REP(s,1<<n) if(f[s]){
			if((s&one[i])!=one[i]) f[s]=0;
			if((~s&zero[i])!=zero[i]) f[s]=0;
		}
	}
	int ans=0;
	REP(s,1<<n) add(ans,f[s]);
	cout<<ans<<endl;
	return 0;
}