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>
using namespace std;

const int RLEN=1<<18|1;
inline char nc() {
	static char ibuf[RLEN],*ib,*ob;
	(ib==ob) && (ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob) ? -1 : *ib++;
}
inline int rd() {
	char ch=nc(); int i=0,f=1;
	while(!isdigit(ch)) {if(ch=='-')f=-1; ch=nc();}
	while(isdigit(ch)) {i=(i<<1)+(i<<3)+ch-'0'; ch=nc();}
	return i*f;
}

const int N=20, mod=1e9+7;
inline int add(int x,int y) {return (x+y>=mod) ? (x+y-mod) : (x+y);}
inline void up(int &x,int y) {x=add(x,y);}

int n,m,k,lim[N+20][N+20];
int bin[N+20],y,f[2][(1<<(N-1))+50],nxt[(1<<(N-1))+50][N];
inline int trans(int sta,int pos) {
	int p=nxt[sta][pos];
	sta^=bin[pos]; 
	if(p<n-1) sta^=bin[p];
	return sta;
}
inline void trans(int id) {
	for(int i=0;i<n-1;i++) {
		y^=1; memset(f[y],0,sizeof(f[y]));
		for(int pre=0;pre<bin[n-1];++pre) if(f[y^1][pre]) {
			if(!(pre&bin[i]) && (lim[id][i]!=2)) up(f[y][pre],f[y^1][pre]);
			if(lim[id][i]!=1) up(f[y][trans(pre,i)],f[y^1][pre]);
		}
	}
}

int main() {
	n=rd(), m=rd(), k=rd();
	if(n==1) return puts("1"),0;
	for(int i=0;i<=n;i++) bin[i]=1<<i;
	for(int i=0;i<bin[n-1];i++) 
		for(int j=0,z=0;j<n-1;j++) {
			while(z<n-1 && (z<j || (!(i&bin[z])))) ++z;
			nxt[i][j]=z;
		}
	for(int i=1;i<=k;i++) {
		int x=rd(), y=rd()-1;
		lim[x][y]=rd()+1;
	}
	f[y][0]=1;
	for(int i=1;i<=m;i++) trans(i);
	int ans=0;
	for(int i=0;i<bin[n-1];i++) up(ans,f[y][i]);
	cout<<ans<<'\n';
}