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 <iostream>
#include <cstdio>
#include <cstring>
#include <cassert>
typedef long long lint;
const int N=20;
const int mod=1e9+7;
int n,m,q,hbit[1<<N];
int limit[N+1][N+1],dp[N+1][1<<N];

inline int nxi(){
	int x=0;
	char c;
	while((c=getchar())>'9'||c<'0');
	while(x=x*10-48+c,(c=getchar())>='0'&&c<='9');
	return x;
}

void init(){
	for(int i=2; i<1<<n; ++i){
		hbit[i]=hbit[i>>1]+1;
	}
}

int main(){
#ifdef LOCAL
	//freopen("d.in","r",stdin);
#endif
	memset(limit,-1,sizeof(limit));
	n=nxi(),m=nxi(),q=nxi();
	--n;
	init();
	for(int i=1; i<=q; ++i){
		const int a=nxi(),b=nxi(),c=nxi();
		limit[a][b]=c;
	}
	dp[n][0]=1;
	for(int i=1; i<=m; ++i){
		memcpy(dp[0],dp[n],sizeof(dp[0]));
		for(int j=1; j<=n; ++j){
			memset(dp[j],0,sizeof(dp[j]));
		}
		for(int j=0; j<n; ++j){
			for(int k=0; k<1<<n; ++k){
				if(!dp[j][k]) continue;
				if((k>>j&1)==0&&limit[i][j+1]!=1){
					dp[j+1][k]=(dp[j+1][k]+dp[j][k])%mod;
				}
				if(limit[i][j+1]!=0){
					int lbit=((k>>j)&-(k>>j))<<j;
					int tgt=k^(1<<j)^lbit;
					dp[j+1][tgt]=(dp[j+1][tgt]+dp[j][k])%mod;
				}
			}
		}
	}
	lint ans=0;
	for(int i=0; i<1<<n; ++i){
		ans+=dp[n][i];
	}
	printf("%lld\n",ans%mod);
	return 0;
}