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=500;
bitset<maxn>S[maxn];
int n,m,x[100100],y[100100],dead[maxn];
int main()
{
//	freopen("c.in","r",stdin);
//	freopen("c.out","w",stdout);
	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]=1;
	for(int i=m;i>=1;i--)
	{
		int X=x[i],Y=y[i];
		for(int now=1;now<=n;now++)
		{
			if(S[now][X]&&S[now][Y]){
				dead[now]=1;
			}
			else if(S[now][X]&&!S[now][Y]){
				S[now][Y]=1;
			}
			else if(!S[now][X]&&S[now][Y]){
				S[now][X]=1;
			}
		}
	}
	int ans=0;
	for(int i=1;i<=n;i++)
	if(!dead[i])
		for(int j=i+1;j<=n;j++)
		{
			if(dead[j])continue;
			if((S[i]&S[j]).none())ans++;
		}
	cout<<ans;
}
 