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)

#include <bits/stdc++.h>
using namespace std;

const int RLEN=1<<18|1;
inline char nc() {
	static char ibuf[RLEN],*ib,*ob;
	(ib==ob) && (ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob) ? -1 : *ib++;
}
inline int rd() {
	char ch=nc(); int i=0,f=1;
	while(!isdigit(ch)) {if(ch=='-')f=-1; ch=nc();}
	while(isdigit(ch)) {i=(i<<1)+(i<<3)+ch-'0'; ch=nc();}
	return i*f;
}

const int N=4e2+5;

int n,m;
bitset <N> sta[N],ban;
int main() {
	n=rd(), m=rd();
	for(int i=1;i<=m;i++) {
		int x=rd(), y=rd();
		sta[x]|=sta[y];
		sta[y]|=sta[x];
		for(int j=1;j<=n;j++) {
			int tag1=sta[j].test(x), tag2=sta[j].test(y);
			if(tag1) sta[j].set(y);
			if(tag2) sta[j].set(x);
		}
		sta[x].set(y); sta[y].set(x);
	}
	for(int i=1;i<=n;i++) 
		if(sta[i][i]) ban[i]=1;
	int cnt=0;
	for(int i=1;i<=n;i++) if(!ban[i]) {
		bitset <N> tp=sta[i];
		for(int j=1;j<=n;j++) tp.flip(j);
		cnt+=tp.count();
		tp&=ban;
		cnt-=tp.count();
	} cout<<(cnt-(n-ban.count()))/2<<'\n';
}