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>
#define rep(i,st,ed) for (register int i=st;i<=ed;++i)
#define fill(x,t) memset(x,t,sizeof(x))

const int MOD=1e9+7;
const int N=200005;

int f[2][21][624288];
int lm[21][21],n;

int read() {
	int x=0,v=1; char ch=getchar();
	for (;ch<'0'||ch>'9';v=(ch=='-')?(-1):(v),ch=getchar());
	for (;ch<='9'&&ch>='0';x=x*10+ch-'0',ch=getchar());
	return x*v;
}

void upd(int &x,int v) {
	x+=v; (x>=MOD)?(x-=MOD):0;
}

int ch(int S,int x) {
	if ((S>>x)&1) return S;
	int w=((1<<n)-(1<<x))&S;
	if (w) S-=(w&-w);
	return S+(1<<x);
}

int main(void) {
	n=read()-1; int m=read(),k=read();
	rep(i,1,k) {
		int x=read(),y=read();
		lm[x][y]=read()+1;
	}
	f[1][0][0]=1;
	rep(i,1,m) {
		int now=i&1,nxt=!now;
		rep(j,0,n-1) {
			for (int S=0;S<(1<<n);++S) {
				if (!f[now][j][S]) continue;
				if (lm[i][j+1]==0) {
					if (!((S>>j)&1)) upd(f[now][j+1][S],f[now][j][S]);
					upd(f[now][j+1][ch(S,j)],f[now][j][S]);
				} else if (lm[i][j+1]==1) {
					if (!((S>>j)&1)) upd(f[now][j+1][S],f[now][j][S]);
				} else upd(f[now][j+1][ch(S,j)],f[now][j][S]);
			}
		}
		fill(f[nxt],0);
		for (int S=0;S<(1<<n);++S) upd(f[nxt][0][S],f[now][n][S]);
	}
	int ans=0;
	for (int S=0;S<(1<<n);++S) upd(ans,f[m&1][n][S]);
	printf("%d\n", ans);
	return 0;
}