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 N 1100000
#define mod 1000000007
using namespace std;
int n, m, k, bin[25], f[N], g[N][25], map[25][25], x, y, z, sum, LOG[N], ans, ff, L, last;
int main(){
	scanf("%d%d%d", &n, &m, &k);
	bin[0]=1; for(int i=1; i<=n; i++)bin[i]=bin[i-1]<<1;
	LOG[0]=-1; for(int i=1; i<=bin[n]; i++)LOG[i]=LOG[i>>1]+1;
	for(int i=1; i<=m; i++)
		for(int j=1; j<=n; j++)map[i][j]=2;
	for(int i=1; i<=k; i++){scanf("%d%d%d", &x, &y, &z); map[x][y]=z^1;}
	if(n==1){printf("1"); return 0;} n--;
	memset(g, 0, sizeof(g)); sum=1;
	for(int owo=1; owo<=m; owo++){
		memset(f, 0, sizeof(f));
		for(int s=0; s<=bin[n]-1; s++){
			ff=1;
			for(int i=1; i<=n; i++)if(((s&bin[n-i])?1:0)==map[owo][i]){ff=0; break;}
			if(!ff)continue;
			for(int i=1; i<=n; i++)if(!(s&bin[i-1])){
				f[s]+=g[((s+bin[i-1])>>(i-1))<<(i-1)][1]; if(f[s]>=mod)f[s]-=mod;
			}
			f[s]=sum-f[s]; if(f[s]<0)f[s]+=mod;
		}
		sum=0;
		for(int s=0; s<=bin[n]-1; s++){sum+=f[s]; if(sum>=mod)sum-=mod;}
		memset(g, 0, sizeof(g));
		for(int s=bin[n]-1; s; s--){
			L=n-LOG[s&(-s)]; g[s][L]=f[s];
			for(int i=1; i<=n-L; i++){g[s][L]+=g[s+bin[n-L-i]][L+i]; if(g[s][L]>=mod)g[s][L]-=mod;}
		}
		for(int s=1; s<=bin[n]-1; s++){
			L=n-LOG[s&(-s)]; last=n;
			for(int i=L-1; i; i--){
				g[s][i]=g[s][i+1];
				if(!(s&bin[n-L+i])){
					if(last==n)continue;
					g[s][i]+=g[s+bin[n-L+i]-bin[last]][i+1]; if(g[s][i]>=mod)g[s][i]-=mod;
				}else last=n-L+i;
			}
		}
	}
	ans=0; for(int i=0; i<=bin[n]-1; i++){ans+=f[i]; if(ans>=mod)ans-=mod;}
	printf("%d", ans);
	return 0;
}