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;

int n,m,col,ans;
int a[200010],b[200010],f[410],used[410];
bitset<512> g[410];

int check(int p)
{int i,j;
 
 col++;
 used[p]=col;
 for(i=m;i>=1;i--)
 {	if(used[a[i]]==col && used[b[i]]==col) return 0;
 	if(used[a[i]]==col)
 		used[b[i]]=col;
 	else if(used[b[i]]==col)
 		used[a[i]]=col;
 }
 for(i=1;i<=n;i++)
 	if(used[i]==col) g[p][i]=1;
 return 1;
}	

int main()
{int i,j,x,y,fx,fy;
 bitset<512> w;
 
 scanf("%d%d",&n,&m);
 for(i=1;i<=m;i++)
 	scanf("%d%d",&a[i],&b[i]);
 for(i=1;i<=n;i++)
 	f[i]=check(i);
 	
 //cout<<f[1]<<" "<<f[2]<<" "<<f[3]<<endl;	
 	
 for(i=1;i<=n;i++)
 	for(j=i+1;j<=n;j++)
 		if(f[i] && f[j])
		{	w=(g[i]&g[j]);
			ans+=w.none();
		}
 printf("%d\n",ans);
 return 0;
}