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 <iostream>
using namespace std;
#define ref(i,x,y)for(int i=x;i<=y;++i)
#define def(i,x,y)for(int i=x;i>=y;--i)
const int mod=1e9+7;
int n,m,K,ans,f[1<<20],A[401],B[401],C[401],t[21][21],g[1<<20][20],p[1<<20];
void inc(int&a,int b){a+=b;if(a>=mod)a-=mod;}
int main(){
	cin>>n>>m>>K;n--;
	ref(i,1,K){
		cin>>A[i]>>B[i]>>C[i],B[i]--;
		t[A[i]][++t[A[i]][0]]=i;
	}
	ref(i,0,(1<<n)-1){int I=i;
		ref(j,0,n-1)g[i][j]=I&-I,I=I^(I&-I);
		ref(j,0,n-1)if(i>>j&1)p[i]++;
	}
	f[0]=1;
	ref(i,1,m){
		ref(I,0,n-1){
			ref(j,0,(1<<n)-1)if((j>>n-1&1)&&p[j]==(I+1))
				inc(f[j],f[j^(1<<n-1)]);
			def(j,(1<<n)-1,0)if(g[j][I]){
				int G=g[j][I];
				if(G==(1<<n-1))continue;
				if(!(j&(G<<1)))inc(f[j],f[j^G^(G<<1)]);
			}
		}
		ref(j,0,(1<<n)-1)
			ref(I,1,t[i][0]){
				int id=t[i][I];
				if((j>>B[id]&1)!=C[id]){
					f[j]=0;break;
				}
			}
	}
	ref(i,0,(1<<n)-1)inc(ans,f[i]);
	cout<<ans<<endl;
}