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
#include<bits/stdc++.h>
using namespace std;
const int N=21;
const int mod=1e9+7;
int gi() {
    int x=0,o=1;char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
    if(ch=='-') o=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
int n,m,k,c[N][N],f[2][N][1<<20],nxt[1<<20][N];
void inc(int &a,int b) {
    a+=b;if(a>=mod) a-=mod;
}
int main() {
    n=gi()-1,m=gi(),k=gi();
    memset(c,-1,sizeof(c));
    while(k--) {
        int x=gi(),y=gi();
        c[x][y]=gi()^1;
    }
    for(int i=0;i<(1<<n);i++) {
        k=-1;
        for(int j=n-1;j>=0;j--) {
            nxt[i][j]=k;
            if(i>>j&1) k=j;
        }
    }
    int now=0,lst=1;
    f[lst][n][0]=1;
    for(int i=1;i<=m;i++) {
        memset(f[now],0,sizeof(f[now]));
        memcpy(f[now][0],f[lst][n],sizeof(f[now][0]));
        for(int j=0;j<n;j++)
            for(int k=0;k<(1<<n);k++)
                if(f[now][j][k]) {
                    for(int t=(k>>j&1);t<2;t++)
                        if(c[i][j+1]!=t) {
                            if(t==(k>>j&1)) inc(f[now][j+1][k],f[now][j][k]);
                            else if(nxt[k][j]!=-1) inc(f[now][j+1][k+(1<<j)-(1<<nxt[k][j])],f[now][j][k]);
                            else inc(f[now][j+1][k+(1<<j)],f[now][j][k]);
                        }
                }
        swap(now,lst);
    }
    int ans=0;
    for(int i=0;i<(1<<n);i++) inc(ans,f[lst][n][i]);
    printf("%d\n",ans);
    return 0;
}
