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 <algorithm>

using namespace std;

int a[100010][2];
bool f[410][410],ok[410];

int main()
{
	//freopen("input.txt","r",stdin);
	//freopen("output.txt","w",stdout);
	int n,m;scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++) scanf("%d%d",&a[i][0],&a[i][1]);
	for (int i=1;i<=n;i++)
	{
		f[i][i]=1;ok[i]=1;
		for (int j=m;j;j--)
		{
			if (f[i][a[j][0]]&&f[i][a[j][1]]) {ok[i]=0;break;}
			if (f[i][a[j][0]]) f[i][a[j][1]]=1; else if (f[i][a[j][1]]) f[i][a[j][0]]=1;
		}
	}
	int ans=0;
	for (int i=1;i<=n;i++)
		if (ok[i])
			for (int j=i+1;j<=n;j++)
				if (ok[j])
				{
					bool t=1;
					for (int k=1;k<=n;k++) if (f[i][k]&&f[j][k]) t=0;
					ans+=t;
				}
	printf("%d\n",ans);
	return 0;
}
