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: Wrong Answer)

#include<cstdio>
#include<bitset>
#include<vector>
#include<algorithm>
using namespace std;
bitset<410>liv[410];
vector<int>v;
int n,m,ans,a[410],b[410];
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
		scanf("%d%d",a+i,b+i);
	for(int i=1;i<=n;i++){
		liv[i].set(i);
		bool ck=1;
		for(int j=m;j;j--){
			if(liv[i][a[j]]&&liv[i][b[j]]){
				ck=0;
				break;
			}
			if(liv[i][a[j]]||liv[i][b[j]]){
				liv[i].set(a[j]);
				liv[i].set(b[j]);
			}
		}
		if(ck)
			v.push_back(i);
	}
	for(int i=0;i<v.size();i++)
		for(int j=i+1;j<v.size();j++)
			ans+=((liv[v[i]]&liv[v[j]]).count()==0);
	printf("%d",ans);
	return 0;
}