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>
using namespace std;
typedef long long ll;
const int N=22,D=1<<19,P=1000000007;
int i,j,k,l,n,m,nm,K,ch,ff,ans;
int f[D];
int a[N][N];
void R(int &x) {
	ff=x=0;ch=getchar();
	while (ch<'0' || '9'<ch) { if (ch=='-') ff=1;ch=getchar();}
	while ('0'<=ch && ch<='9') x=x*10+ch-'0',ch=getchar();
	if (ff) x=-x;
}
int main() {
	R(n);R(m);R(K);
	n--;
	nm=1<<n;
	for (i=1;i<=m;i++)
		for (j=0;j<n;j++) a[i][j]=3;
	for (i=1;i<=K;i++) {
		int A,B,C;
		R(A);R(B);R(C);
		a[A][B-1]=1<<C;
	}
	f[0]=1;
	for (i=1;i<=m;i++) {
		for (j=0;j<n;j++) {
			for (k=nm-1;k>=0;k--) if (f[k]) {
				int F=f[k];
				f[k]=0;
				for (l=0;l<2;l++) if (a[i][j]&1<<l) {
					int t;
					if (k&1<<j) t=1;
					else t=0;
					if (t>l) continue;
					if (l>t) {
						int tt=k>>j<<j;
						if (tt) tt=k^(tt&-tt);
						else tt=k;
						tt|=1<<j;
						f[tt]=(f[tt]+F)%P;
					}
					else f[k]=(f[k]+F)%P;
				}
			}
		}
	}
	for (i=0;i<nm;i++) ans=(ans+f[i])%P;
	printf("%d\n",ans);
}