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>

int x[100005],y[100005],n,m,ans;

int ok[100005],eat[405];
bool gun[405];

int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;++i)
	scanf("%d%d",&x[i],&y[i]);

    for(int i=1;i<=n;++i)
    {
	for(int j=m;j;--j)
	    if(eat[x[j]]==i)
	    {
		if(eat[y[j]]==i)eat[i]=i;
		eat[y[j]]=i,ok[j]=i;
	    }
	    else if(eat[y[j]]==i)eat[x[j]]=i,ok[j]=i;
	    else if(x[j]==i)eat[y[j]]=i,ok[j]=i;
	    else if(y[j]==i)eat[x[j]]=i,ok[j]=i;
	for(int j=1;j<=m;++j)
	    if(ok[j]!=i)
		if(eat[x[j]]==i)eat[y[j]]=i;
		else if(eat[y[j]]==i)eat[x[j]]=i;
	if(eat[i]==i){gun[i]=1;continue;}
	for(int j=1;j<i;++j)
	    if(eat[j]!=i&&!gun[j])++ans;
    }

    printf("%d\n",ans);
    
    return 0;
}