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
/*
srt: 2017年06月19日 19時50分00秒
 《寄左省杜拾遗》岑参
联步趋丹陛，分曹限紫微。
晓随天仗入，暮惹御香归。
白发悲花落，青云羡鸟飞。
圣朝无阙事，自觉谏书稀。
end: 2017年06月19日 19時57分51秒
*/

#include<cstdio>
#include<bitset>
inline void read(int&r)
{
	register char c;r=0;
	do c=getchar();while(c<'0'||c>'9');
	do r=r*10+c-'0',c=getchar();while(c>='0'&&c<='9');
}

bool b[401];
std::bitset<401>a[401];
int x[100001],y[100001];
int main()
{
	int i,j,n,m,s=0;read(n),read(m);
	for(i=1;i<=m;i++)read(x[i]),read(y[i]);
	for(i=1;i<=n;i++)
		for(j=m,a[i][i]=true;j;j--)
			if(a[i][j[x]]&&a[i][j[y]])
			{ b[i]=true;break; }
			else if(a[i][j[x]])a[i][j[y]]=true;
			else if(a[i][j[y]])a[i][j[x]]=true;
	for(i=1;i<=n;i++)
		if(!b[i])
			for(j=1;j<i;j++)
				if(!b[j]&&(a[i]&a[j]).none())
					s++;
	printf("%d\n",s);
}