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: Memory Limit Exceeded)

use proconio::marker::Usize1;

fn run() {
    proconio::input! {
        n: usize,
        m: usize,
        p: [(Usize1, Usize1); m],
    }
    let mut next = vec![vec![0; m + 1]; n];
    for (i, &(x, y)) in p.iter().enumerate() {
        next[x][i + 1] = i + 1;
        next[y][i + 1] = i + 1;
    }
    for p in next.iter_mut() {
        for j in 1..=m {
            p[j] = std::cmp::max(p[j], p[j - 1]);
        }
    }
    let mut ans = 0;
    for i in 1..n {
        for j in 0..i {
            let mut ok = 1;
            let mut set = std::collections::BTreeSet::new();
            let mut h = std::collections::BinaryHeap::new();
            for &x in [i, j].iter() {
                let k = next[x][m];
                if k == 0 {
                    continue;
                }
                if set.contains(&k) {
                    ok = 0;
                    break;
                }
                set.insert(k);
                h.push(k);
            }
            while ok == 1 && !h.is_empty() {
                let k = h.pop().unwrap();
                let (x, y) = p[k - 1];
                for &x in [x, y].iter() {
                    let z = next[x][k - 1];
                    if z == 0 {
                        continue;
                    }
                    if set.contains(&z) {
                        ok = 0;
                        break;
                    }
                    set.insert(z);
                    h.push(z);
                }
            }
            ans += ok;
        }
    }
    println!("{}", ans);
}

#[allow(dead_code)]
fn naive() {
    proconio::input! {
        n: usize,
        m: usize,
        p: [(Usize1, Usize1); m],
    }
    let mut ans = 0;
    for i in 1..n {
        for j in 0..i {
            let mut dp = vec![false; n];
            dp[i] = true;
            dp[j] = true;
            let mut ok = 1;
            for &(x, y) in p.iter().rev() {
                if dp[x] && dp[y] {
                    ok = 0;
                    break;
                }
                if dp[x] || dp[y] {
                    dp[x] = true;
                    dp[y] = true;
                }
            }
            ans += ok;
        }
    }
    println!("{}", ans);
}

fn main() {
    run();
//    naive();
}