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<bitset>
#include<cctype>
#include<cstdio>
const int N=100007,M=407;
int a[N],b[N],d[N];std::bitset<M>f[M];
int read(){int x=0,c=getchar();while(isspace(c))c=getchar();while(isdigit(c))(x*=10)+=c&15,c=getchar();return x;}
int main()
{
    int n=read(),m=read(),ans=0;
    for(int i=1;i<=m;++i) a[i]=read(),b[i]=read();
    for(int i=1,j;i<=n;++i)
	for(f[i][i]=1,j=m;j;--j)
	{
	    if(f[i][a[j]]&&f[i][b[j]]) {d[i]=1;break;}
	    if(f[i][a[j]]&&!f[i][b[j]]) f[i][b[j]]=1;
	    else if(!f[i][a[j]]&&f[i][b[j]]) f[i][a[j]]=1;
	}
    for(int i=1;i<=n;++i) for(int j=i+1;j<=n;++j) if(!d[i]&&!d[j]&&(f[i]&f[j]).none()) ++ans;
    printf("%d",ans);
}
