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 <algorithm>  
#include <iostream>  
#include <sstream>  
#include <string>  
#include <cstring>
#include <vector>  
#include <queue>  
#include <set>  
#include <map>  
#include <cstdio>  
#include <cstdlib>  
#include <cctype>  
#include <cmath>  
#include <list>  
#include <cassert>
#include <ctime>
#include <climits>
using namespace std;  

#define PB push_back  
#define MP make_pair  
#define SZ(v) ((int)(v).size())  
#define FOR(i,a,b) for(int i=(a);i<(b);++i)  
#define REP(i,n) FOR(i,0,n)  
#define FORE(i,a,b) for(int i=(a);i<=(b);++i)  
#define REPE(i,n) FORE(i,0,n)  
#define FORSZ(i,a,v) FOR(i,a,SZ(v))  
#define REPSZ(i,v) REP(i,SZ(v))  
typedef long long ll;
typedef unsigned long long ull;
ll gcd(ll a,ll b) { return b==0?a:gcd(b,a%b); }

const int MAXN=19;
const int MAXPATH=20;
const int MOD=1000000007;

int n,npath,nfixes;
int fix[MAXPATH][MAXN];

int dp[1<<MAXN];

int calcnmask(int mask,int j) {
	int suf=mask&((1<<j)-1),prf=mask^suf;
	if(prf==0) return suf^(1<<j);
	int lsb=prf&-prf;
	return mask^lsb^(1<<j);
}

void run() {
	scanf("%d%d%d",&n,&npath,&nfixes); --n;
	REP(i,npath) REP(j,n) fix[i][j]=-1;
	REP(i,nfixes) { int a,b,c; scanf("%d%d%d",&a,&b,&c); --a,--b; fix[a][b]=c; }
	memset(dp,0,sizeof(dp)); dp[0]=1;
	REP(i,npath) {
		REP(j,n) for(int mask=(1<<n)-1;mask>=0;--mask) {
			if(dp[mask]==0||((mask>>j)&1)==1) continue;
			int nmask=calcnmask(mask,j);
			if((dp[nmask]+=dp[mask])>=MOD) dp[nmask]-=MOD;
		}
		REP(j,n) if(fix[i][j]!=-1) REP(mask,1<<n) if(((mask>>j)&1)!=fix[i][j]) dp[mask]=0;
		//printf("%d:",i); REP(mask,1<<n) if(dp[mask]!=0) printf(" %x=%d",mask,dp[mask]); puts("");
	}
	int ret=0; REP(mask,1<<n) if((ret+=dp[mask])>=MOD) ret-=MOD; printf("%d\n",ret);
}

int main() {
	run();
	return 0;
}