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

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