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<cstdio>
#include<cstring>
#include<algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,b,a) for(int i=b;i>=a;i--)
#define mset(a,x) memset(a,x,sizeof(a))
using namespace std;
typedef long long ll;
typedef double db;
char ch;
int read(){int n=0,p=1;for(ch=getchar();ch<'0' || '9'<ch;ch=getchar())if(ch=='-') p=-1;for(;'0'<=ch && ch<='9';ch=getchar()) n=n*10+ch-'0';return n*p;}
const int N=21,mo=1e9+7;
int _2[N],n,m,lim,c[N][N],f[2][N][524290],nxt[N][524290];
void upd(int &x,int y){x=(x+y)%mo;}
int main()
{
	_2[0]=1;fo(i,1,20) _2[i]=_2[i-1]<<1;
	int x,y,z;
	n=read(),m=read(),lim=read();--n;
	mset(c,-1);
	fo(i,1,lim) x=read(),y=read(),z=read(),c[x][y]=z;
	fo(s,0,_2[n]-1)
		fo(i,0,n-1) if(!(s&_2[i]))
			fo(j,i+1,n-1) if(s&_2[j]) {nxt[i][s]=j;break;}
	f[0][n][0]=1;
	z=1;
	for(int i=1;i<=m;i++,z^=1)
	{
		mset(f[z],0);
		fo(s,0,_2[n]-1) f[z][0][s]=f[z^1][n][s];
		fo(j,0,n-1)
			fo(s,0,_2[n]-1) if(f[z][j][s])
				if(s&_2[j])
				{
					if(c[i][j+1]==0) continue;
					upd(f[z][j+1][s],f[z][j][s]);
				}
				else
				{
					if(c[i][j+1]!=1)
					{
						upd(f[z][j+1][s],f[z][j][s]);
					}
					if(c[i][j+1]!=0)
					{
						int ps=nxt[j][s],ss=s|_2[j];
						if(ps) ss^=_2[ps];
						upd(f[z][j+1][ss],f[z][j][s]);
					}
				}
	}
	int ans=0;
	fo(s,0,_2[n]-1) upd(ans,f[z^1][n][s]);
	printf("%d",ans);
	return 0;
}