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;
   #define int long long
   #define PP pair<int,pair<int,int>>
const int MAX = 1000001;
const int MOD =1000000007;
const int INV2 = (MOD+1)/2;



     signed main(){
 int n,m;
 cin>>n>>m;
 bitset<400> b[n];
 for(int i=0;i<n;i++)b[i].set(i);
 int x,y;
 for(int i=0;i<m;i++){
   cin>>x>>y;x--;y--;
   if((b[x]&b[y]).none())b[x]|=b[y];
   else b[x]|=(~b[x]);
     
 

    b[y]=b[x];
  
 }
 int ans=0;
 for(int i=0;i<n;i++)for(int j=0;j<i;j++)if((b[i]&b[j]).none())ans++;
 cout<<ans<<endl;



}


        
