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>
inline int read()
{
	int x;char c;
	while((c=getchar())<'0'||c>'9');
	for(x=c-'0';(c=getchar())>='0'&&c<='9';)x=x*10+c-'0';
	return x;
}
#define MN 400
#define MM 100000
int x[MM+5],y[MM+5],u[MN+5],g[MN+5][MN+5];
int main()
{
	int n,m,i,j,k,ans=0;
	n=read();m=read();
	for(i=1;i<=m;++i)x[i]=read(),y[i]=read();
	for(i=1;i<=n;++i)
	{
		g[i][i]=1;
		for(j=m;j;--j)
		{
			if(g[i][x[j]]&&g[i][y[j]])u[i]=1;
			g[i][x[j]]|=g[i][y[j]];
			g[i][y[j]]|=g[i][x[j]];
		}
		if(!u[i])for(j=1;j<i;++j)if(!u[j])
			{for(k=1;k<=n;++k)if(g[i][k]&&g[j][k])break;if(k>n)++ans;}
	}
	printf("%d",ans);
}