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 pb push_back
#define cmin(a,b) (a>b?a=b:a)
#define cmax(a,b) (a<b?a=b:a)
#define mem(a,k) memset(a,k,sizeof(a))
#define lop(i,s,t) for(int i=s;i<(t);++i)
#define rep(i,s,t) for(int i=s;i<=(t);++i)
#define dec(i,s,t) for(int i=s;i>=(t);--i)
using namespace std;

#define Pr(f,...) //fprintf(stderr,f,##__VA_ARGS__),fflush(stderr)

template<typename T>
void read(T &x){
	x=0; 
	char c;
	for(c=getchar();!isdigit(c);c=getchar()); 
	for(;isdigit(c);c=getchar())x=x*10+c-'0'; 
}

const int N=21,P=1e9+7;

int mul(int a,int b){ return 1ll*a*b%P; }
int sub(int a,int b){ a-=b; return a<0?a+P:a; }
int add(int a,int b){ a+=b; return a>=P?a-P:a; }
void upd(int &a,int b){ a=add(a,b); }

bool del[N][N][2];
int ans,n,m,q,f[N][1<<N],nxt[1<<N][N];

int main(int argc,char *argv[]){
#ifdef CURIOUSCAT
	freopen("dat.in","r",stdin);
	//freopen("my.out","w",stdout);
#endif
	read(n),read(m),read(q); 	
	--n;
	rep(k,1,q){
		int a,b,c;
		read(a),read(b),read(c); 
		del[a][b-1][!c]=1;
	}
	lop(s,0,(1<<n)){
		lop(j,0,n){
			if((s>>j&1)==0){
				int k=j,t=s;
				for(;k<n&&!((s>>k)&1);++k); 
				if(k<n)t-=(1<<k);
				nxt[s][j]=t+(1<<j); 
			}
		}
	}
	f[n][0]=1;
	rep(i,1,m){
		memcpy(f[0],f[n],sizeof(f[0])); 
		lop(j,0,n){
			int nj=j+1;
			mem(f[nj],0); 
			lop(s,0,(1<<n)){
				rep(num,0,1)if(!del[i][j][num]){
					if(((s>>j&1)==1)&&!num)continue;
					int t=(s>>j&1)==num?s:nxt[s][j];
					upd(f[nj][t],f[j][s]); 
				}
			}
		}
	}
	lop(s,0,(1<<n))ans=add(ans,f[n][s]); 
	printf("%d\n",ans);
	return 0;
}