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(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
#define N 25
#define mo 1000000007
using namespace std;
int n,m,h,a[N][N],f[N][1048577],e[N],nxt[1048577][N],ans=0;
int main() {
	scanf("%d%d%d",&n,&m,&h);
	fo(i,1,h) { int x,y,z;scanf("%d%d%d",&x,&y,&z); a[x][y]=z+1; }
	e[0]=1;fo(i,1,n) e[i]=e[i-1]*2;
	fo(s,0,e[n]-1) { int y=-1; fd(j,n,1) { nxt[s][j]=y; if(s&e[j-1]) y=j; } }
	f[n][0]=1;
	fo(i,1,m){
		fo(s,0,e[n]-1) f[1][s]=f[n][s]; fo(j,2,n) fo(s,0,e[n]-1) f[j][s]=0;
		fo(j,1,n-1){
			fo(s,0,e[n]-1)
				if(f[j][s]){
					if(s&e[j-1]){ if(a[i][j]!=1) (f[j+1][s]+=f[j][s])%=mo; }
					else{
						if(a[i][j]!=2) (f[j+1][s]+=f[j][s])%=mo;
						if(a[i][j]!=1){ int s1=s; if(nxt[s][j]!=-1) s1-=e[nxt[s][j]-1]; s1+=e[j-1]; (f[j+1][s1]+=f[j][s])%=mo; }
					}
				}
		}
	}
	fo(s,0,e[n]-1) (ans+=f[n][s])%=mo;
	printf("%d\n",ans);
	return 0;
}