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>
#define ll long long
#define fir first
#define sec second
using namespace std;
int x[100010],y[100010],noans[410];
bitset<410> z,b[410];
int read()
{
	int x=0,flag=1;char c=getchar();
	while(!isdigit(c)) {if(c=='-') flag=-1;c=getchar();}
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x*flag;
}
int main()
{
	int n=read(),m=read();
	for(int i=1;i<=m;i++) x[i]=read(),y[i]=read();
	for(int i=1;i<=n;i++)
	{
		z.reset();
		z[i]=1;
		for(int j=m;j>=1;j--)
		{
			int ss=x[j],tt=y[j];
			if(z[ss]&&z[tt]) {noans[i]=1;break;}
			if(!z[ss]&&!z[tt]) continue;
			z[ss]=z[tt]=1;
		}
		b[i]=z;
	}
	int ans=0;
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++)
			if(!noans[i]&&!noans[j]) if(!(b[i]&b[j]).count()) ans++;
	cout<<ans;
	return 0;
}
