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>
#define ll long long
#define bi(x) (1<<(x-1))
#define rep(i,j,k) for (i=j;i<=k;i++)
using namespace std;
const int N=20,mod=1e9+7;
int n,m,ans,k,i,cur,nxt,j,p,a,b,c,sn,lock[N+1][N+1];
int x,y,f[2][N+1][1<<N];
void updata(int &x,int y) { x+=y; if (x>=mod) x-=mod; }
int main()
{
//	freopen("zigzag.in","r",stdin);
//	freopen("zigzag.out","w",stdout);
	scanf("%d%d%d",&n,&m,&k);
	rep(i,1,k) {scanf("%d%d%d",&a,&b,&c); lock[a][b]=c+1; }
	f[0][0][0]=1;
	sn=(1<<n-1)-1;
	x=1; y=0;
	rep(i,1,m)
	{
		x^=1; y^=1;
		rep(j,0,n-1)
			rep(cur,0,sn) f[y][j][cur]=0;
		rep(j,0,n-1)
			rep(cur,0,sn)
			if (f[x][j][cur])
			{
				if (j==n-1) {
					if (i<m) updata(f[y][0][cur],f[x][j][cur]);
					else updata(ans,f[x][j][cur]);
				}
				if (lock[i][j+1]!=2 && !( cur&bi(j+1) )) 
					updata(f[x][j+1][cur],f[x][j][cur]);
				if (lock[i][j+1]!=1)
				{
					nxt=cur;
					if (!( cur&bi(j+1) )) {
						rep(p,j+2,n-1) if (cur&bi(p)) break;
						if (p<n) nxt^=bi(p);
						nxt|=bi(j+1);
					}
					updata(f[x][j+1][nxt],f[x][j][cur]);
				}
			}
	}
	printf("%d\n",ans);
	return 0;
}