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>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dwn(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline int read() {
    int x=0,f=1;char c=getchar();
    for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
    for(;isdigit(c);c=getchar()) x=x*10+c-'0';
    return x*f;
}
typedef long long ll;
const int maxn=21;
const int maxm=1<<20;
const int mod=1000000007;
int f[2][maxm],trans[maxm][maxn];
int g[maxn][maxn];
int n,m,k;
int main() {
	n=read();m=read();k=read();
	memset(g,-1,sizeof(g));
	rep(i,1,k) {
		int a=read()-1,b=read()-1;
		g[a][b]=read();
	}
	rep(S,0,(1<<n)-1) {
		rep(i,0,n-1) if(!(S>>i&1)) {
			int k=n;
			rep(j,i+1,n-1) if(S>>j&1) {k=j;break;}
			if(k==n) trans[S][i]=S^(1<<i);
			else trans[S][i]=S^(1<<i)^(1<<k);
		}
	}
	int cur=0;f[0][0]=1;
	rep(i,0,m-1) {
		rep(j,0,n-1) {
			cur^=1;
			memset(f[cur],0,sizeof(f[cur]));
			int ok1=(g[i][j]<0||g[i][j]==1),ok2=g[i][j]<=0;
			rep(S,0,(1<<n)-1) if(f[cur^1][S]) {
				if(ok1) {
					if(S>>j&1) (f[cur][S]+=f[cur^1][S])%=mod;
					else (f[cur][trans[S][j]]+=f[cur^1][S])%=mod;
				}
				if(ok2) {
					if(!(S>>j&1)) (f[cur][S]+=f[cur^1][S])%=mod;
				}
			}
		}
		cur^=1;
	}
	int ans=0;
	rep(S,0,(1<<n)-1) (ans+=f[cur][S])%=mod;
	printf("%d\n",ans);
  	return 0;
}