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>
#define maxn 405
#define maxm 100005
int n,m,x[maxm],y[maxm],ans;
bool s[maxn][maxn],fl[maxn];
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++)
	{
		s[i][i]=true;
		for(int j=m;j;j--)
		{
			if(s[i][x[j]]&&s[i][y[j]]) fl[i]=true;
			else if(s[i][x[j]]) s[i][y[j]]=true;
			else if(s[i][y[j]]) s[i][x[j]]=true;
		//	else s[i][x[j]]=s[i][y[j]]=true;
		}
	}
	for(int i=1;i<=n;i++)
		if(!fl[i])
			for(int j=i+1;j<=n;j++)
				if(!fl[j])
				{
					bool f=true;
					for(int k=1;k<=n&&f;k++)
						if(s[i][k]&&s[j][k]) f=false;
					ans+=f;
				}
	printf("%d\n",ans);
}