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>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
using namespace std;
const int N=405;
const int M=100005;
int vis[405][405];
pair<int,int> xjt[M];
int gg[N];
int main()
{
	int n,m;scanf("%d%d",&n,&m);
	rep(i,1,m)scanf("%d%d",&xjt[i].first,&xjt[i].second);
	rep(i,1,n)
	{
		vis[i][i]=1;gg[i]=0;
		per(j,m,1)
		{
			if(vis[i][xjt[j].first] && vis[i][xjt[j].second]){gg[i]=1;break;}
			if(vis[i][xjt[j].first]) vis[i][xjt[j].second]=1;
			else if(vis[i][xjt[j].second]) vis[i][xjt[j].first]=1;
		}
	}
	int ans=0;
	rep(i,1,n)if(!gg[i])rep(j,i+1,n)if(!gg[j])
	{
		int fg=1;
		rep(k,1,n)if(vis[i][k] && vis[j][k]) {fg=0;break;}ans+=fg;
	}
	printf("%d\n",ans);
	return 0;
}