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<cstdio>
#include<algorithm>
using namespace std;

const int MAXN=405;
const int MAXM=1e5+5;

int n,m;
int S[MAXN][MAXN],alive[MAXN];
pair<int,int> a[MAXM];

int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++) scanf("%d%d",&a[i].first,&a[i].second);
    for(int i=1;i<=n;i++){
        alive[i]=1;
        S[i][i]=1;
        for(int j=m;j>=1;j--){
            int x=a[j].first,y=a[j].second;
            if(!S[i][x]&&!S[i][y]) continue;
            if(S[i][x]&&S[i][y]){
                alive[i]=0;
                break;
            }
            S[i][x]=S[i][y]=1;
        }
    }
    int ans=0;
    for(int i=1;i<=n;i++)
        for(int j=i+1;j<=n;j++){
            if(!alive[i]||!alive[j]) continue;
            bool flag=1;
            for(int k=1;k<=n;k++)
                if(S[i][k]&&S[j][k]){
                    flag=0;
                    break;
                }
            if(flag) ans++;
        }
    printf("%d",ans);
}