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<algorithm>
#define fo(i,a,b) for(i=a;i<=b;i++)
#define fd(i,a,b) for(i=a;i>=b;i--)
using namespace std;
const int maxn=20+10,mo=1000000007;
int dp[2][maxn][(1<<21)+10],nxt[(1<<21)+10][maxn];
int bz[maxn][maxn];
int i,j,k,l,r,s,t,n,m,p,ans,now;
int main(){
	scanf("%d%d%d",&n,&m,&p);
	n--;
	fo(i,0,(1<<n)-1){
		t=-1;
		fd(j,n-1,0){
			if ((i&(1<<j))) t=j;
			nxt[i][j]=t;
		}
	}
	fo(i,1,m)
		fo(j,0,n-1)
			bz[i][j]=-1;
	fo(i,1,p){
		scanf("%d%d%d",&j,&k,&l);
		bz[j][k-1]=l;
	}
	dp[0][n][0]=1;
	now=1;
	fo(i,1,m){
		fo(s,0,(1<<n)-1) dp[now][0][s]=dp[1-now][n][s];
		fo(j,0,n)
			fo(s,0,(1<<n)-1)
				dp[1-now][j][s]=0;
		fo(j,0,n-1)
			fo(s,0,(1<<n)-1)
				if (dp[now][j][s]){
					fo(k,0,1){
						if (bz[i][j]!=-1&&bz[i][j]!=k) continue;
						t=(s&(1<<j));
						if (t) t=1;
						if (k==0&&t==1) continue;
						if (k==t) (dp[now][j+1][s]+=dp[now][j][s])%=mo;
						else{
							r=nxt[s][j];
							if (r==-1) (dp[now][j+1][s+(1<<j)]+=dp[now][j][s])%=mo;
							else (dp[now][j+1][s+(1<<j)-(1<<r)]+=dp[now][j][s])%=mo;
						}
					}
				}
		now=1-now;
	}
	fo(s,0,(1<<n)-1) (ans+=dp[1-now][n][s])%=mo;
	(ans+=mo)%=mo;
	printf("%d\n",ans);
}