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 "bits/stdc++.h"
using namespace std;
const int maxn = 100100;
int x[maxn], y[maxn];
set<int> group[maxn];
bool survive[maxn];
int M(){
    int n, m;
    if(!~scanf("%d %d", &n, &m)) return 0;
    for(int e = 0; e < m; e++)
        scanf("%d %d", x + e, y + e);
    for(int e = 1; e <= n; e++){
        group[e].clear();
        group[e].insert(e);
        survive[e] = true;
        for(int f = m-1; f >= 0; f--){
            int hasX = group[e].count(x[f]);
            int hasY = group[e].count(y[f]);
            if(hasX && hasY){
                survive[e] = false;
                break;
            } else if(hasX != hasY){
                group[e].insert(x[f]);
                group[e].insert(y[f]);
            }
        }
    }
    int pairs = 0;
    for(int e = 1; e <= n; e++)
        for(int f = e + 1; f <= n; f++){
            if(survive[e] && survive[f]){
                bool ok = true;
                for(int v : group[e])
                    if(group[f].count(v)) ok = false;
                if(ok) pairs++;
            }
        }
    printf("%d\n", pairs);
    return 1;
}
int main(){
    while(M());
    return 0;
}