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 <bitset>
#include <algorithm>
#define rep(i,j,k) for (i=j;i<=k;i++)
#define down(i,j,k) for (i=j;i>=k;i--)
using namespace std;
const int N=405,M=1e5+5;
int n,m,i,j,ans,u[M],v[M],kill[N];
bitset<N> b[N];
int main()
{
//	freopen("turkey.in","r",stdin);
//	freopen("turkey.out","w",stdout);
	scanf("%d%d",&n,&m);
	rep(i,1,m) scanf("%d%d",&u[i],&v[i]);
	rep(i,1,n)
	{
		b[i][i]=1;
		down(j,m,1)
			if (b[i][u[j]] || b[i][v[j]])
			{
				if (b[i][u[j]] && b[i][v[j]])
				{
					kill[i]=1; break;
				}
				b[i][u[j]]=b[i][v[j]]=1;
			}
	}
	rep(i,1,n-1)
	if (!kill[i])
		rep(j,i+1,n)
			if (!kill[j] && (b[i]&b[j]).none()) ans++;
	printf("%d\n",ans);
	return 0;
}