# AtCoder Grand Contest 017

## F - Zigzag

Time limit時間制限 : 4sec / Memory limitメモリ制限 : 256MB

### 問題文

• 折れ線 i は，N 個の点 (1, X_{i,1}), (2, X_{i,2}), ..., (N, X_{i,N}) をこの順で結んでいる．
• j=1, 2, ..., N-1 に対して，X_{i,j+1} = X_{i,j} または X_{i,j+1} = X_{i,j}+1 が成り立つ．

また，高橋君は，折れ線の曲がり方について K 個の条件を満たすように折れ線を描かなければなりません． i 番目の条件は (A_i, B_i, C_i) で表され，これは次を意味します：

• C_i=0 のときは，折れ線 A_i を描くとき，B_i 回目の移動においては，その時いる点のすぐ左下の点を選ぶ．
• C_i=1 のときは，折れ線 A_i を描くとき，B_i 回目の移動においては，その時いる点のすぐ右下の点を選ぶ．

すなわち，X_{A_i, {B_i}+1} = X_{A_i, B_i} + C_i を意味します．

### 制約

• 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 = 0,1
• (A_i, B_i) として同じ組は複数回与えられない．

### 入力

N M K
A_1 B_1 C_1
A_2 B_2 C_2
:
A_K B_K C_K


### 入力例 1

3 2 1
1 2 0


### 出力例 1

6


### 入力例 2

3 2 2
1 1 1
2 1 0


### 出力例 2

0


### 入力例 3

5 4 2
1 3 1
4 2 0


### 出力例 3

172


### 入力例 4

20 20 0


### 出力例 4

881396682


Score : 1600 points

### Problem Statement

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).

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.

### Notes

Before submission, it is strongly recommended to measure the execution time of your code using "Custom Test".

### 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.

### Sample Input 1

3 2 1
1 2 0


### Sample Output 1

6


There are six ways to draw lines, as shown below. Here, red lines represent L_1, and green lines represent L_2.

### Sample Input 2

3 2 2
1 1 1
2 1 0


### Sample Output 2

0


### Sample Input 3

5 4 2
1 3 1
4 2 0


### Sample Output 3

172


### Sample Input 4

20 20 0


### Sample Output 4

881396682