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;

const int maxn=1e5+10;

int n,m,u[maxn],v[maxn],Ans;
bitset<410> need[410];

int main()
{
#ifdef h10
	freopen("E.in","r",stdin);
	freopen("E.out","w",stdout);
#endif
	int i,j;
	bool flag;
	scanf("%d%d",&n,&m);
	for (i=1;i<=m;i++)
		scanf("%d%d",&u[i],&v[i]);

	for (i=1;i<=n;i++)
	{
		need[i][i]=1;
		flag=1;
		for (j=m;j>=1;j--)
		{
			if (need[i][u[j]]&&need[i][v[j]])
			{
				flag=0;
				break;
			}
			if (need[i][u[j]]||need[i][v[j]])
				need[i][u[j]]=need[i][v[j]]=1;
		}
		need[i][0]=!flag;
	}

	for (i=1;i<=n;i++) if (!need[i][0])
	for (j=i+1;j<=n;j++) if (!need[j][0])
		Ans+=!(need[i]&need[j]).count();
	printf("%d\n",Ans);
}
