Problem Statement

There are $N$ turkeys. We number them from $1$ through $N$.

$M$ men will visit here one by one. The $i$-th man to visit will take the following action:

• If both turkeys $x_i$ and $y_i$ are alive: selects one of them with equal probability, then eats it.
• If either turkey $x_i$ or $y_i$ is alive (but not both): eats the alive one.
• If neither turkey $x_i$ nor $y_i$ is alive: does nothing.

Find the number of pairs $(i,\\ j)$ ($1 ≤ i < j ≤ N$) such that the following condition is held:

• The probability of both turkeys $i$ and $j$ being alive after all the men took actions, is greater than $0$.

Constraints

• $2 ≤ N ≤ 400$
• $1 ≤ M ≤ 10^5$
• $1 ≤ x_i < y_i ≤ N$

Input

Input is given from Standard Input in the following format:

$N$ $M$ $x_1$ $y_1$ $x_2$ $y_2$ $:$ $x_M$ $y_M$

Output

Print the number of pairs $(i,\\ j)$ ($1 ≤ i < j ≤ N$) such that the condition is held.

Sample Input 1

3 1
1 2

Sample Output 1

2

$(i,\\ j) = (1,\\ 3), (2,\\ 3)$ satisfy the condition.

Sample Input 2

4 3
1 2
3 4
2 3

Sample Output 2

1

$(i,\\ j) = (1,\\ 4)$ satisfies the condition. Both turkeys $1$ and $4$ are alive if:

• The first man eats turkey $2$.
• The second man eats turkey $3$.
• The third man does nothing.

Sample Input 3

3 2
1 2
1 2

Sample Output 3

0

Sample Input 4

10 10
8 9
2 8
4 6
4 9
7 8
2 8
1 8
3 4
3 4
2 7

Sample Output 4

5