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 lowbit(x) ((x)&(-(x)))

const int N=22,Mod=1e9+7;

int n,m,k,nxt[N][N],dp[N][1<<20],Ml[N];

void Init() {
	int a,b,c;
	scanf("%d%d%d",&n,&m,&k);
	for (int i=1;i<=m;++i)
		for (int j=1;j<=n;++j) nxt[i][j]=-1;
	for (int i=1;i<=k;++i) {
		scanf("%d%d%d",&a,&b,&c);
		nxt[a][b]=c;
	}
}

void Add(int &a,int b) {
	a+=b;
	if (a>=Mod) a-=Mod;
}

void Solve() { 
	int Max=1<<n;
	Ml[n]=Max-1; for (int i=n-1;i>=0;--i) Ml[i]=Ml[i+1]^(1<<i);
	dp[1][0]=1;
	for (int i=1;i<=m;++i) {
		for (int j=1;j<n;++j) 
			for (int s=0;s<Max;++s) if (dp[j][s]) {
				if (nxt[i][j]!=1 && ((s>>(j-1))&1)==0) Add(dp[j+1][s],dp[j][s]);
				if (nxt[i][j]!=0) {
					if (((s>>(j-1))&1)==0) Add(dp[j+1][(s-(lowbit(s>>(j-1))<<(j-1)))|(1<<(j-1))],dp[j][s]);
					else Add(dp[j+1][s],dp[j][s]);
				}
			}
		for (int s=0;s<Max;++s) dp[1][s]=dp[n][s];
		for (int j=2;j<=n;++j)
			for (int s=0;s<Max;++s) dp[j][s]=0;
	}
	int ans=0; for (int s=0;s<Max;++s) Add(ans,dp[1][s]);
	printf("%d\n",ans);
}

int main() {
	Init();
	Solve();
	return 0;
}