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<stdio.h>
#include<string.h>

/*#### modular_arithmetic */
class modular_arithmetic{
public:
	int mod, range;

	modular_arithmetic(){
		mod = 1000000007;
	}
	//No specific settings.
	void set_mod(int _mod){mod = _mod;}
	int add(int x,int y){return (x+y)%mod;}
	int sub(int x,int y){return (x-y+mod)%mod;}
	int mul(int x,int y){return (int)((long long int)x*y%mod);}
	int pw(int x,int y){
		int r = 1;
		while(1){
			if(y&1)r=mul(r,x);
			if(y>>=1)x=mul(x,x);else
				break;
		}
		return r;
	}
};
/*####*/

modular_arithmetic md;
void upd(int&x,int y){
	x = md.add(x,y);
}

int c[22][22];
int N,M;
int dp[1<<20],dp2[1<<20];
int to[1<<20][20];

int main(){
	scanf("%d%d",&N,&M);
	memset(c,-1,sizeof(c));
	int q;
	for(scanf("%d",&q); q--;){
		int A,B,C;
		scanf("%d%d%d",&A,&B,&C);
		c[A-1][B-1] = C;
	}
	for(int mask=0; mask<(1<<(N-1)); mask++)
		for(int i=0; i<N-1; i++)
			if((mask&(1<<i))==0){
				int t = i;
				for(int j=i+1; j<N-1; j++)
					if(mask&(1<<j)){
						t = j;
						break;
					}
				if(t == i)
					to[mask][i] = mask|(1<<i);
				else
					to[mask][i] = (mask - (1<<t)) | (1<<i);
			}
	int res = 0;
	dp[0] = 1;
	for(int i=0; i<M; i++){
		for(int j=0; j<N-1; j++){
			memset(dp2,0,sizeof(dp2));
			for(int mask = 0, now; mask < (1<<(N-1)); mask++)
				if(now = dp[mask]){
					for(int k=0; k<2; k++)if(k == c[i][j] || c[i][j] == -1){
						if((bool)k == (bool)(mask&(1<<j))){
							upd(dp2[mask], now);
						}else
						if(k == 1 && (mask&(1<<j)) == 0){
							upd(dp2[to[mask][j]], now);
						}
					}
				}
			memcpy(dp,dp2,sizeof(dp));
		}
	}
	for(int mask = 0; mask < (1<<(N-1)); mask++)
		upd(res, dp2[mask]);
	printf("%d\n",res);
	return 0;
}