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>
#include<bitset>
#define MAXN 400
#define MAXM 100000
using namespace std;
bitset<MAXN+5> vis[MAXN+5];
int n,m;
int a[MAXM+5],b[MAXM+5];
bool sp[MAXN+5];
int main()
{
	scanf("%d %d",&n,&m);
	for(int i=1;i<=m;i++)
		scanf("%d %d",&a[i],&b[i]);
	for(int i=1;i<=n;i++)
		for(int j=m;j>=1;j--)
		{
			vis[i][i]=1;
			if(vis[i][a[j]]==1&&vis[i][b[j]]==1)
			{
				sp[i]=true;
				break;
			}
			else if(vis[i][a[j]]==0&&vis[i][b[j]]==0)
				continue;
			else
				vis[i][a[j]]=vis[i][b[j]]=1;
		}
	int ans=0;
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++)
			if(sp[i]==false&&sp[j]==false&&(vis[i]&vis[j]).none())
				ans++;
	printf("%d\n",ans);
	return 0;
}