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<iostream>
#include<cstdio>
#include<bitset>
#define MN 100000
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
int n,m,a[MN+5],b[MN+5],num=0,ans=0;
bitset<400*399/2+5> fail,s[405];
int main()
{
	n=read();m=read();
	for(int i=1;i<=n;++i) for(int j=i+1;j<=n;++j)
		++num,s[i][num]=1,s[j][num]=1;
	for(int i=1;i<=m;++i) a[i]=read(),b[i]=read();
	for(int i=m;i;--i)
	{
		fail|=s[a[i]]&s[b[i]];
		s[a[i]]|=s[b[i]];s[b[i]]|=s[a[i]];
	}
	for(int i=1;i<=num;++i) if(!fail[i]) ++ans;
	cout<<ans; 
	return 0;
}

