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>
using namespace std;
#define N 22
#define low(x) ((x)&(-(x)))
int go[N][N];
int dp[2][2][(1<<20)+3];
int n,m,k;
const int mod=1e9+7;
int main(){
    memset(go,-1,sizeof(go));
    scanf("%d%d%d",&n,&m,&k);
    for(int i=1,a,b,c;i<=k;i++){
        scanf("%d%d%d",&a,&b,&c);
        go[a][b-1]=c;
    }
    int now=0;
    dp[0][0][0]=1;
    for(int i=1;i<=m;i++){
        memset(dp[now^1],0,sizeof(dp[0]));
        int pos=0;
        for(int j=0;j<n-1;j++){
            memset(dp[now][pos^1],0,sizeof(dp[0][0]));
            for(int k=0,v;k<(1<<(n-1));k++){
                if((v=dp[now][pos][k])==0) continue;
                if(go[i][j]!=1&&((k>>j)&1)==0) (dp[now][pos^1][k]+=v)%=mod;
                if(go[i][j]!=0) (dp[now][pos^1][k^(low(k>>j)<<j)|(1<<j) ]+=v)%=mod;
            }
            pos^=1;
        }
        for(int k=0;k<(1<<(n-1));k++)
            dp[now^1][0][k]=dp[now][pos][k];
        now^=1;
    }
    long long ans=0;
    for(int k=0;k<(1<<(n-1));k++)
        ans+=dp[now][0][k];
    cout<<ans%mod<<endl;
    return 0;
}