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<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cassert>
#define ll long long
using namespace std;
inline int read(){
	int re=0,flag=1;char ch=getchar();
	while(ch>'9'||ch<'0'){
		if(ch=='-') flag=-1;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9') re=(re<<1)+(re<<3)+ch-'0',ch=getchar();
	return re*flag;
}
int n,m,l[100010],r[100010],s[510][510],ok[510];
int main(){
	n=read();m=read();int i,j,flag=0,k,ans=0;
	for(i=1;i<=m;i++) l[i]=read(),r[i]=read();
	for(i=1;i<=n;i++){
		ok[i]=1;
		s[i][i]=1;
		for(j=m;j>=1;j--){
			if(!s[i][l[j]]&&!s[i][r[j]]) continue;
			if(s[i][l[j]]&&s[i][r[j]]){
				ok[i]=0;break;
			}
			s[i][l[j]]=s[i][r[j]]=1;
		}
	}
	for(i=1;i<n;i++){
		if(!ok[i]) continue;
		for(j=i+1;j<=n;j++){
			if(!ok[j]) continue;
			flag=0;
			for(k=1;k<=n;k++) flag|=s[i][k]&s[j][k];
			ans+=(flag==0);
		}
	}
	cout<<ans;
}