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<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int n,m,i,j,k,x[100005],y[100005],ans;
bool tag[405];
int main()
{
	scanf("%d%d",&n,&m);
	for(i=1;i<=m;++i)
		scanf("%d%d",x+i,y+i);
	for(i=1;i<n;++i)
		for(j=i+1;j<=n;++j)
		{
			for(k=1;k<=n;++k)
				tag[k]=false;
			tag[i]=tag[j]=true;
			for(k=m;k>=1;--k)
			{
				if(tag[x[k]]&&tag[y[k]])
					break;
				if(tag[x[k]]||tag[y[k]])
					tag[x[k]]=tag[y[k]]=true;
			}
			if(k<1)
				++ans;
		}
	printf("%d",ans);
	return 0;
}
