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 <cctype>
#define repu(i,x,y) for (int i=x; i<=y; ++i)
using namespace std;

typedef unsigned long long ULL;
int n,m,ans;
bool flag[410][410],kill[410][410];

int getint()
{
    char ch;
    while (!isdigit(ch=getchar()));
    int x=ch-'0';
    for (; isdigit(ch=getchar()); x=x*10+ch-'0');
    return x;
}

void add(int u,int v)
{
    repu(i,1,n)
        if (flag[i][v])
            (u<i?kill[u][i]:kill[i][u])=1;
}

int main()
{
    scanf("%d%d",&n,&m);
    while (m--)
    {
        int u=getint(),v=getint();
        repu(i,1,n)
        {
            if (flag[v][i] && !flag[u][i])
                add(u,i);
            if (flag[u][i] && !flag[v][i])
                add(v,i);
        }
        add(u,v),add(v,u);
        repu(i,1,n)
            flag[u][i]|=flag[v][i],flag[v][i]|=flag[u][i];
        flag[u][v]=flag[v][u]=1;
        kill[u][u]|=kill[v][v],kill[v][v]|=kill[u][u];
    }
    repu(i,1,n)
        repu(j,i+1,n)
            ans+=!flag[i][j] && !flag[j][i] && !kill[i][j] && !kill[i][i] && !kill[j][j];
    printf("%d\n",ans);
    return 0;
}