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 fo(i,a,b) for(i=a;i<=b;i++)
#define fd(i,a,b) for(i=a;i>=b;i--)
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
typedef long long ll;
using namespace std;
inline int read(){
	int n=0,f=1;char c;
	for(c=getchar();(c<'0'||c>'9')&&c!='-';c=getchar());
	if (c=='-') c=getchar(),f=-1;
	for(;c>='0'&&c<='9';c=getchar()) n=n*10+c-48;
	return n*f;
}
const int maxn=(1<<20)+5,mo=1e9+7;
int f[2][21][maxn],ne[maxn][21];
int i1,ans,i,j,n,m,k,s,t,w[25][25];
bool bz[25][25];
void mod(int &x){
	x-=(x>=mo)?mo:0;
}
int work(int s,int t){
	s>>=t,s&=-s;
	return s<<t;
}
int main(){
	n=read(),m=read(),k=read();
	n--;
	fo(i,1,k){
		int a=read(),b=read(),c=read();
		b--,bz[a][b]=1,w[a][b]=c;
	}
	s=(1<<n)-1;
	fo(k,0,s) 
	fo(i,0,n-1) if (!(k&(1<<i)))ne[k][i]=k+(1<<i)-work(k,i+1);
	f[1][0][0]=1;
	fo(i,1,m){
		i1^=1;
		fo(j,0,n-1){
			bool b=bz[i][j],x=w[i][j];
			int now=1<<j;
			fo(k,0,s) if (f[i1][j][k]){
				int sum=f[i1][j][k];
				fo(t,0,1){
					if (b&&t!=x) continue;
					int p=(k&now)>0;
					if (!t&&p) continue;
					if (t==p) mod(f[i1][j+1][k]+=sum);else
					mod(f[i1][j+1][ne[k][j]]+=sum);
				}
			}
		}
		memset(f[i1^1],0,sizeof(f[i1^1]));
		fo(k,0,s) f[i1^1][0][k]=f[i1][n][k];
	}i1^=1;
	fo(k,0,s) mod(ans+=f[i1][0][k]);
	printf("%d\n",ans);
}