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 <iostream>
#include <cstdlib>
#include <cstdio>
#define N 22
#define MAX 1000000007
#define int long long
using namespace std;
int n,m,K,Map[N][N],dp[2][1<<N],ans;
int lowbit(int x){
	return x&(-x);
}
signed main(){
	int x,y,z;
	scanf("%lld%lld%lld",&n,&m,&K);n--;
	for(int i=1;i<=m;i++){
		for(int j=1;j<=n;j++){
			Map[i][j]=-1;
		}
	}
	for(int i=1;i<=K;i++){
		scanf("%lld%lld%lld",&x,&y,&z);
		Map[x][y]=z;
	}
	int cur=1,fin=0,All=(1<<n)-1;
	dp[cur][0]=1;
	for(int i=1;i<=m;i++){
		for(int j=1;j<=n;j++){
			int now=(1<<(j-1));
			swap(cur,fin);
			for(int s=0;s<=All;s++) dp[cur][s]=0;
			for(int s=0;s<=All;s++){
				if(Map[i][j]!=1 && !(s&now)){
					dp[cur][s]=(dp[cur][s]+dp[fin][s])%MAX;
				}
				if(Map[i][j]!=0){
					int x;
					if(s&now) x=s;
					else{
						int t=All-(now<<1)+1;
						t=s&t;
						t=lowbit(t);
						if(!t) x=s|now;
						else x=(s+now-t);
					}
					dp[cur][x]=(dp[cur][x]+dp[fin][s])%MAX;
				}
			}
		}
	}
	for(int i=0;i<=All;i++){
		ans=(ans+dp[cur][i])%MAX;
	}
	printf("%lld",ans);
	return 0;
}