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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma GCC optimize("O2")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,tune=native")
#include<bits/stdc++.h>
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define pb push_back
using namespace std;
using ll = long long;
//using arr = array;
using pi = pair<int, int>;
using vi = vector<int>;
using vvi = vector<vi>;
using vpi = vector<pi>;
using ld = long double;
const int maxn = 405, mod = 1e9 + 7;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int n, m;
vpi g[maxn];
int ok, vis[maxn];
void dfs(int v, int t) {
	vis[v] = 1;
	for(auto &i : g[v]) {
		if(i.second >= t) continue;
		if(vis[i.first]) ok = 0;
		if(ok) dfs(i.first, i.second);
	}
}
bool check(int u, int v) {
	memset(vis, 0, sizeof vis);ok = 1;
	dfs(u, 1<<30);
	if(ok) dfs(v, 1<<30);
	
	return ok;
}
int main() {
	cin.tie(0)->sync_with_stdio(0);
	cin >> n >> m;
	for(int f, t, i = 0; i < m; i++) {
		cin >> f >> t;
		g[f].pb({t, i});
		g[t].pb({f, i});
	}
	int ans = 0;
	for(int i = 1; i <= n; i++)
		for(int j = 1; j < i; j++)
			ans += check(i, j);
	cout << ans;
}