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<cstring>
#define maxn 25
#define maxt 2097160
#define mod 1000000007
int n,m,k,dp[2][maxn][maxt],nxt[maxt][maxn],bz[maxn][maxn],ans;
int main()
{
	scanf("%d%d%d",&n,&m,&k); n--;
	for(int i=0;i<(1<<n);i++)
		for(int j=n-1,t=-1;j>=0;j--)
		{
			if(i&(1<<j)) t=j;
			nxt[i][j]=t;
		}
	for(int i=1;i<=m;i++)
		for(int j=0;j<n;j++) bz[i][j]=-1;
	for(int i=1;i<=k;i++)
	{
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		bz[x][y-1]=z;
	}
	int now=1; dp[0][n][0]=1;
	for(int i=1;i<=m;i++,now^=1)
	{
		for(int s=0;s<(1<<n);s++) dp[now][0][s]=dp[now^1][n][s];
		//memset(dp[now^1],0,sizeof(dp[now^1]));
		for(int j=0;j<=n;j++)
			for(int s=0;s<(1<<n);s++) dp[now^1][j][s]=0;
		for(int j=0;j<n;j++)
			for(int s=0;s<(1<<n);s++)
				if(dp[now][j][s])
					for(int k=0;k<2;k++)
					{
						if(bz[i][j]!=-1&&bz[i][j]!=k) continue;
						int t=(s>>j)&1;
						if(!k&&t) continue;
						if(k==t) dp[now][j+1][s]=(dp[now][j+1][s]+dp[now][j][s])%mod;
						else
						{
							int r=nxt[s][j];
							if(r==-1) dp[now][j+1][s+(1<<j)]=(dp[now][j+1][s+(1<<j)]+dp[now][j][s])%mod;
							else dp[now][j+1][s+(1<<j)-(1<<r)]=(dp[now][j+1][s+(1<<j)-(1<<r)]+dp[now][j][s])%mod;
						}
					}
	}
	for(int s=0;s<(1<<n);s++) ans=(ans+dp[now^1][n][s])%mod;
	printf("%d\n",(ans+mod)%mod);
}