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<cstdio>
#define neko 510
#define meko 100010
#define f(i,a,b) for(register int i=(a);i<=(b);i=-(~i))
#define rf(i,a,b) for(register int i=(a);i>=(b);i=~(-i))
int n,m,x[meko],y[meko],book[neko],now,ans;
int Must[neko][neko];
bool kill(int u)
{
	Must[u][u]=1;
	rf(i,m,1)
	{
		if(Must[u][x[i]]&&Must[u][y[i]])return 0;
		if(Must[u][x[i]]||Must[u][y[i]])Must[u][x[i]]=Must[u][y[i]]=1;
	}
	return 1;
}
int main()
{
	scanf("%d%d",&n,&m);
	f(i,1,m)scanf("%d%d",&x[i],&y[i]);
	f(i,1,n)book[i]=kill(i);
	f(i,1,n)if(book[i])
	 f(j,i+1,n)if(book[j])
	  {
		 now=1;
		 f(k,1,n)if(Must[i][k]&&Must[j][k]){now=0;break;}
		 ans+=now;
	  }return printf("%d\n",ans),0;
}