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<bits/stdc++.h>
using namespace std;
const int mod=1000000007;
int n,m,k,dp[1<<21],lim1[21][21],lim2[21][21],to[1<<21],ans;
vector<int> g[1<<21];
int main() {
	scanf("%d%d%d",&n,&m,&k);
	n--;
	for(int i=1;i<=k;i++) {
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		b=n-b;
		lim1[a][b]=1;
		lim2[a][b]=c;
	}
	for(int i=0;i<(1<<n);i++) {
		for(int j=1;j<n;j++) {
			if (i&(1<<j)) continue;
			if (!(i&(1<<(j-1)))) continue;
			int tmpi=i^(1<<j)^(1<<(j-1));
			g[i].push_back(tmpi);
		}
	}
//	for(int i=0;i<(1<<n);i++) {
//		printf("%d: ",i);
//		for(int j=0;j<(int)g[i].size();j++) printf("%d ",g[i][j]);
//		puts("");
//	}
	for(int i=0;i<(1<<n);i++) {
		to[i]=(1<<n);
		for(int j=0;j<n;j++) {
			if (i&(1<<j)) continue;
			int tmpi=i^(1<<j);
			to[i]=tmpi;
			break;
		}
//		printf("%d: %d\n",i,to[i]);
	}
	dp[0]=1;
	for(int i=1;i<=m;i++) {
		for(int j=0;j<(1<<n);j++) {
			dp[to[j]]=(dp[to[j]]+dp[j])%mod;
		}
		for(int l=0;l<n;l++) {
			for(int j=0;j<(1<<n);j++) {
				if ((int)g[j].size()>l) dp[g[j][l]]=(dp[g[j][l]]+dp[j])%mod;
			}
		}
		for(int j=0;j<n;j++) {
			if (!lim1[i][j]) continue;
			for(int l=0;l<(1<<n);l++) {
				if ((l&(1<<j) ? 1 : 0)!=lim2[i][j]) {
					dp[l]=0;
				}
			}
		}
//		for(int j=0;j<(1<<n);j++) printf("%d%c",dp[j]," \n"[j+1==(1<<n)]);
	}
	for(int i=0;i<(1<<n);i++) {
		ans=(ans+dp[i])%mod;
	}
	printf("%d\n",ans);
	return 0;
}