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<algorithm>
#include<bits/stdc++.h>
#define LL long long
#define db double

using namespace std;
const int N=400+10,M=1e5+10;
int rd()
{
    int x=0,w=1;char ch=0;
    while(ch<'0'||ch>'9'){if(ch=='-') w=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+(ch^48);ch=getchar();}
    return x*w;
}
int n,m,e[M][2];
bitset<N> a[N]; 

int main()
{
	n=rd(),m=rd();
	for(int i=1;i<=m;++i) e[i][0]=rd(),e[i][1]=rd();
	for(int i=1;i<=n;++i)
	{
		a[i][i]=1;
		for(int j=m;j;--j)
		{
			int x=e[j][0],y=e[j][1];
			if(a[i][x]&a[i][y]) {a[i][i]=0;break;}
			if(a[i][x]) a[i][y]=1;
			else if(a[i][y]) a[i][x]=1;
		}
	}
	int ans=0;
	for(int i=1;i<=n;++i)
	{
		if(!a[i][i]) continue;
		for(int j=i+1;j<=n;++j)
		{
			if(!a[j][j]) continue;
			ans+=(bool)!(a[i]&a[j]).count();
		}
	}
	printf("%d\n",ans);
    return 0;
}