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>
using namespace std;
int a[100005],b[100005];
vector<int> v[405];
int main()
{
	int n,m,ans=0;
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
	v[i].push_back(-1);
	for (int i=0;i<m;i++)
	{
		scanf("%d%d",&a[i],&b[i]);
		v[a[i]].push_back(i);
		v[b[i]].push_back(i);
	}
	for (int i=1;i<n;i++)
	{
		for (int j=i+1;j<=n;j++)
		{
			bool ok=1;
			multiset<int,greater<int> > s({v[i].back(),v[j].back()});
			while (*s.begin()!=-1)
			{
				if (*s.begin()==*next(s.begin()))
				{
					ok=0;
					break;
				}
				int idx=*s.begin();
				s.erase(s.begin());
				s.insert(*prev(lower_bound(v[a[idx]].begin(),v[a[idx]].end(),idx)));
				s.insert(*prev(lower_bound(v[b[idx]].begin(),v[b[idx]].end(),idx)));
			}
			ans+=ok;
		}
	}
	printf("%d",ans);
}