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 rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define forE(i,x) for(int i=head[x];i!=-1;i=ne[i])
using namespace std;
typedef long long i64;
typedef unsigned long long u64;
typedef unsigned u32;
typedef pair<int,int> pin;
#define mk(a,b) make_pair(a,b)
#define lowbit(x) ((x)&(-(x)))
#define sqr(a) ((a)*(a))
#define clr(a) (memset((a),0,sizeof(a)))
#define ls ((x)<<1)
#define rs (((x)<<1)|1)
#define mid (((l)+(r))>>1)
#define pb push_back
#define w1 first
#define w2 second
inline void read(int &x){
	x=0;int f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	x*=f;
}
inline void judge(){
	freopen("in.txt","r",stdin);
	freopen("out.txt","w",stdout);
}
/*******************************head*******************************/
const int maxn=20;
const int mod=1e9+7;
int f[2][1<<maxn];
int place[1<<maxn][maxn+6];
int n,m,c,col[maxn+5][maxn+5];
inline void add(int &x,int y){
	x=((x+y)>=mod?(x+y-mod):(x+y));
}
int main(){
	read(n);read(m);read(c);
	rep(i,1,m)rep(j,1,n)col[i][j]=3;
	rep(i,1,c){
		int x,y,z;read(x);read(y);read(z);
		col[x][y]=z+1;
	}
	f[0][0]=1;
	rep(i,0,(1<<(n-1))-1){
		int lst=n-1;
		per(j,n-2,0){
			if((1<<j)&i)lst=j;
			place[i][j]=lst;
		}
	}int st=1,ed=0;
	rep(i,1,m)rep(j,1,n-1){
		swap(st,ed);
		memset(f[ed],0,sizeof(f[ed]));
		rep(s,0,(1<<(n-1))-1)if(f[st][s]){
			rep(k,0,1){
				if(!((col[i][j])&(1<<k)))continue;
				int news=s;
				if((s&(1<<(j-1)))&&(k==0))continue;
				if((place[news][j-1]!=n-1)&&(k==1)){
					news&=((1<<(n-1))-1)^(1<<((place[news][j-1])));
				}
				news&=((1<<(n-1))-1)^(1<<((j-1)));
				news|=(k<<(j-1));
				add(f[ed][news],f[st][s]);
			}	
		}
	}int ans=0;
	rep(s,0,(1<<(n-1))-1)add(ans,f[ed][s]);
	cout<<ans<<endl;
	return 0;
}