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<stdio.h>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
#include<string.h>
#include<queue>
#include<bitset>
#define LL long long
using namespace std;
inline int read( )
{
  int sum=0;char c=getchar( );bool f=0;
  while(c<'0'||c>'9') {if(c=='-') f=1;c=getchar( );}
  while(c>='0'&&c<='9') {sum=sum*10+c-'0';c=getchar( );}
  if(f) return -sum;
  return sum;
}
const int N=405,M=100005;
int n,m,A[M],B[M];
bitset<N>R[N];bool can[N];
int main( )
{
  int i,j,ans=0;
  n=read( );m=read( );
  for(i=1;i<=m;i++) A[i]=read( ),B[i]=read( );
  for(i=1;i<=n;i++)
    {
      R[i][i]=1;can[i]=1;
      for(j=m;j>=1;j--)
	{
	  if(R[i][A[j]]&&R[i][B[j]]) {can[i]=0;break;}
	  if(R[i][A[j]]) R[i][B[j]]=1;
	  else if(R[i][B[j]]) R[i][A[j]]=1;
	}
    }
  for(i=1;i<=n;i++)
    if(can[i])
      for(j=i+1;j<=n;j++)
	if(can[j]&&(R[i]&R[j]).none( )) ans++;
  printf("%d",ans);
  return 0;
}
