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>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define sqr(x) ((x)*(x))
#define mp make_pair
#define uint unsigned
#define PI pair<int,int>
inline char gc(){
    static char buf[100000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
#define gc getchar
inline ll read(){
	ll x = 0; char ch = gc(); bool positive = 1;
	for (; !isdigit(ch); ch = gc())	if (ch == '-')	positive = 0;
	for (; isdigit(ch); ch = gc())	x = x * 10 + ch - '0';
	return positive ? x : -x;
}
inline void write(ll a){
    if(a<0){
    	a=-a; putchar('-');
	}
    if(a>=10)write(a/10);
    putchar('0'+a%10);
}
inline void writeln(ll a){write(a); puts("");}
inline void wri(ll a){write(a); putchar(' ');}
inline ull rnd(){
	return ((ull)rand()<<30^rand())<<4|rand()%4;
}
const int N=20,mod=1000000007;
#define lowbit(i) (i&-i)
int n,m,k,ans,ycl[20],dp[N+1][1<<N],tong[N][N];
inline void add(int &a,int b){
	a=a+b>=mod?a+b-mod:a+b;
}
signed main(){
	n=read(); m=read(); k=read(); n--;
	memset(tong,-1,sizeof(tong));
	for(int i=1;i<=k;i++){
		int a=read(),b=read(),c=read();
		tong[a-1][b-1]=c;
	}
	dp[0][0]=1; ycl[0]=(1<<n)-1;
	for(int i=1;i<n;i++)ycl[i]=ycl[i-1]^(1<<(i-1));
	for(int i=0;i<m;i++){
		for(int j=0;j<n;j++){
			for(int k=0;k<(1<<n);k++){
				if(tong[i][j]!=1&&((k>>j&1)==0)){
					add(dp[j+1][k],dp[j][k]);
				}
				if(tong[i][j]!=0){
					int t=k&ycl[j];// cout<<i<<" "<<j<<" "<<k<<" "<<(k^lowbit(t)^(1<<j))<<endl;
					add(dp[j+1][k^lowbit(t)^(1<<j)],dp[j][k]);
				}
			}
		}
		swap(dp[n],dp[0]);
		//for(int j=0;j<(1<<n);j++)cout<<i<<" "<<j<<" "<<dp[0][j]<<endl;
		for(int j=1;j<=n;j++)memset(dp[j],0,sizeof(dp[j]));
	}
	for(int i=0;i<(1<<n);i++)ans=(ans+dp[0][i])%mod;
	cout<<ans<<endl;
}