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<bits/stdc++.h>
using namespace std;
typedef double db;
typedef long long ll;
#define w1 first
#define w2 second
#define ls (x<<1)
#define rs (x<<1|1)
#define pb push_back
#define mid ((l+r)>>1)
#define SZ(x) ((x).size())
#define All(x) (x).begin(),(x).end()
#define rep(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
#define rep2(i,a,b) for(int (i)=(a);(i)<(b);(i)++)
#define per(i,a,b) for(int (i)=(a);(i)>=(b);(i)--)
#define Rep(p,x) for(int (p)=head[(x)];(p);(p)=nxt[(p)])
template<class T>void read(T&num){
	num=0;T f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9')num=num*10+ch-'0',ch=getchar();
	num*=f;
}
const int maxn=21,mod=1000000007;
int n,m,k;
int must[maxn][maxn],F[maxn][1<<maxn],G[maxn][1<<maxn];
void upd(int &x,int y){
	x=(x+y)%mod;
}
int main(){
	memset(must,-1,sizeof must);
	read(n);n--;
	if(!n)return puts("1"),0;
	read(m);read(k);
	rep(i,1,k){
		int a,b,c;
		read(a);read(b);read(c);
		must[b][a]=c;
	}
	rep2(i,0,n)F[i][0]=1;
	rep(i,1,m)rep2(now,0,n){
		rep2(S,0,1<<n)G[now][S]=0;
		rep2(S,0,1<<n){
			int lst;
			if(!now)lst=n-1;
			else lst=now-1;
			if(S>>now&1){
				if(must[now+1][i]!=0)upd(G[now][S],F[lst][S]);
			}else{
				if(must[now+1][i]!=1)upd(G[now][S],F[lst][S]);
				if(must[now+1][i]!=0){
					int nxtS=S^(1<<now);
					rep2(k,now+1,n)if(S>>k&1){
						nxtS^=1<<k;
						break;
					}
					upd(G[now][nxtS],F[lst][S]);
				}
			}
		}
		rep2(S,0,1<<n)F[now][S]=G[now][S];
	}
	int ans=0;
	rep2(S,0,1<<n)upd(ans,F[n-1][S]);
	printf("%d\n",ans);
	return 0;
}