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>
#define rep(i,n) for ((i)=1;(i)<=(n);(i)++)
#define per(i,n) for ((i)=n;(i)>=(1);(i)--)
using namespace std;
const int mod=1e9+7;
int n,m,lim[25][25],ans;
int dp[2][21][1<<20],nxt[1<<20][21];
int main()
{
	int i,j,k,l;
	cin>>n>>m>>k;n--;
	memset(lim,-1,sizeof(lim));
	while(k--)
	{
		int x,y,z;
		cin>>x>>y>>z;
		lim[x][y-1]=z;
	}
	for(i=0;i<(1<<n);i++)
	{
		int lst=-1;
		for(j=n-1;j>=0;j--)
		{
			if((i>>j)&1) lst=j;
			nxt[i][j]=lst;
		}
	}
	dp[0][n][0]=1;
	rep(i,m)
	{
		memset(dp[i&1],0,sizeof(dp[i&1]));
		memcpy(dp[i&1][0],dp[(i-1)&1][n],sizeof(dp[0][0]));
		for(j=0;j<n;j++)
		{
			for(k=0;k<(1<<n);k++) if(dp[i&1][j][k])
			{
//				cerr<<i<<' '<<j<<' '<<k<<' '<<dp[i&1][j][k]<<endl;
				for(l=0;l<=1;l++)
				{
					if(lim[i][j]!=-1&&lim[i][j]!=l) continue;
					if(((k>>j)&1)&&l==0) continue;
					if(((k>>j)&1)==l)
					{
						(dp[i&1][j+1][k]+=dp[i&1][j][k])%=mod;
					}
					else
					{
						int p=nxt[k][j];
						if(p==-1) (dp[i&1][j+1][k^(1<<j)]+=dp[i&1][j][k])%=mod;
						else (dp[i&1][j+1][k^(1<<j)^(1<<p)]+=dp[i&1][j][k])%=mod;
					}
				}
			}
		}
	}
	for(i=0;i<(1<<n);i++) (ans+=dp[m&1][n][i])%=mod;
	cout<<ans<<endl;
	return 0;
}