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.

Examples

Input

3 1
1 2


Output

2


Input

4 3
1 2
3 4
2 3


Output

1


Input

3 2
1 2
1 2


Output

0


Input

10 10
8 9
2 8
4 6
4 9
7 8
2 8
1 8
3 4
3 4
2 7


Output

5
#include<bits/stdc++.h>
using namespace std;
using Int = long long;
//INSERT ABOVE HERE
Int E[444][444];
signed main(){
  Int n,m;
  cin>>n>>m;
  vector<Int> x(m),y(m);
  for(Int i=0;i<m;i++) cin>>x[i]>>y[i];
  reverse(x.begin(),x.end());
  reverse(y.begin(),y.end());
  memset(E,0,sizeof(E));
  for(Int i=0;i<m;i++){
    x[i]--;y[i]--;
    E[x[i]][y[i]]=E[y[i]][x[i]]=1;
  }
  
  vector<Int> lp(n,0);
  vector<vector<Int> > G(n);
  for(Int i=0;i<n;i++){
    vector<Int> used(n,0);
    used[i]=1;
    for(Int k=0;k<m;k++){
      if(!used[x[k]]&&!used[y[k]]) continue;
      lp[i]|=used[x[k]]&&used[y[k]];
      used[x[k]]=used[y[k]]=1;
    }
    G[i]=used;
  }
  
  Int ans=0;
  for(Int i=0;i<n;i++){
    if(lp[i]) continue;
    for(Int j=0;j<i;j++){
      if(E[i][j]||lp[j]) continue;
      Int flg=1;
      for(Int k=0;k<n;k++)
	flg&=!G[i][k]||!G[j][k];
      ans+=flg;
    }
  }
  cout<<ans<<endl;
  return 0;
}
