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).

<image>

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 <cstring>
#include <algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

const int N=25,S=(1<<19)+5,Mo=1e9+7;

int n,m,k,a,b,c[N][N],nxt[S][N],f[2][N][S];

void inc(int &x,int y) {
	x=x+y;
	if (x>=Mo) x=x-Mo;
}

int main() {
	scanf("%d%d%d",&n,&m,&k);n--;
	fo(i,1,m) fo(j,0,n-1) c[i][j]=-1;
	fo(i,1,k) {
		scanf("%d%d",&a,&b);
		scanf("%d",&c[a][b-1]);
	}
	fo(s,0,(1<<n)-1) {
		nxt[s][n]=-1;
		fd(j,n-1,0)
			if (s&(1<<j)) nxt[s][j]=j;
			else nxt[s][j]=nxt[s][j+1];
	}
	f[0][n][0]=1;
	int p=0,q=1;
	fo(i,0,m-1) {
		memset(f[q],0,sizeof(f[q]));
		fo(s,0,(1<<n)-1) f[q][0][s]=f[p][n][s];
		fo(j,0,n-1) 
			fo(s,0,(1<<n)-1) 
				if (f[q][j][s]) {
					fo(x,0,1) {
						if (c[i+1][j]!=-1&&c[i+1][j]!=x) continue;
						if (!x&&(s&(1<<j))) continue;
						if (x==((s&(1<<j))>0)) inc(f[q][j+1][s],f[q][j][s]);
						else {
							if (nxt[s][j]==-1) inc(f[q][j+1][s+(1<<j)],f[q][j][s]);
							else inc(f[q][j+1][s+(1<<j)-(1<<nxt[s][j])],f[q][j][s]);
						}
					}
				}
		swap(p,q);
	}
	int ans=0;
	fo(s,0,(1<<n)-1) inc(ans,f[p][n][s]);
	printf("%d\n",ans);
	return 0;
}