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;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<int,P> P1;
typedef pair<P,P> P2;
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define mod 1000000007
#define fi first
#define sc second
#define rep(i,x) for(int i=0;i<x;i++)
#define repn(i,x) for(int i=1;i<=x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())
int n,m,q;
int pos[21][21];
ll dp[21][(1<<19)];
int mx[(1<<19)];
int main(){
	cin>>n>>m>>q;
	if(n == 1){
		cout<<1<<endl;
		return 0;
	}
	memset(pos,-1,sizeof(pos));
	rep(i,q){
		int a,b,c;
		cin>>a>>b>>c;
		pos[a][b] = c;
	}
	dp[1][0] = 1;
	for(int i=1;i<(1<<19);i++){
		for(int j=18;j>=0;j--){
			if(i >= (1<<j)){
				mx[i] = (1<<j);
				break;
			}
		}
	}
	repn(L,m){
		repn(i,n-1){
			rep(mask,(1<<(n-1))){
				if(dp[i][mask] == 0) continue;
				if(pos[L][i] != 1 && mask < (1<<(n-2))){
					int mask2 = mask%(1<<(n-2));
					mask2 *= 2;
					dp[i+1][mask2] += dp[i][mask];
				}
				if(pos[L][i] != 0){
					int mask2 = mask;
					if(mask >= (1<<(i-1))){
						mask2 = mask-mx[mask];
					}
					mask2 = mask2%(1<<(n-2))*2+1;
					dp[i+1][mask2] += dp[i][mask];
				}
			}
		}
		rep(mask,(1<<(n-1))){
			dp[1][mask] = dp[n][mask]%mod;
		}
		for(int i=2;i<=n;i++) rep(mask,(1<<(n-1))) dp[i][mask] = 0;
	}
	ll ans = 0;
	rep(mask,(1<<(n-1))) ans = (ans+dp[1][mask])%mod;
	cout<<ans<<endl;
}