Problem:
Score : 1400 points

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

Solution in Rust (1.42.0) (judge verdict: Accepted)

use proconio::marker::Usize1;

fn run() {
    proconio::input! {
        n: usize,
        m: usize,
        p: [(Usize1, Usize1); m],
    }
    let mut next = vec![vec![]; n];
    for (i, &(x, y)) in p.iter().enumerate() {
        next[x].push(i + 1);
        next[y].push(i + 1);
    }
    let mut ans = 0;
    for i in 1..n {
        for j in 0..i {
            let mut ok = 1;
            let mut used = vec![false; n];
            used[i] = true;
            used[j] = true;
            let mut stack = vec![(i, m), (j, m)];
            while let Some((v, k)) = stack.pop() {
                let k = match next[v].binary_search(&k) {
                    Ok(k) => next[v][k],
                    Err(k) => {
                        if k == 0 {
                            continue;
                        }
                        next[v][k - 1]
                    },
                };
                let (x, y) = p[k - 1];
                let z = x ^ y ^ v;
                if used[z] {
                    ok = 0;
                    break;
                }
                used[z] = true;
                stack.push((v, k - 1));
                stack.push((z, k - 1));
            }
            ans += ok;
        }
    }
    println!("{}", ans);
}

fn main() {
    run();
}