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<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,m,ans;
int a[100001],b[100001];
bool box[401][401],use[401];
int main()
{
	cin>>n>>m;
	for(int i=1;i<=m;i++) cin>>a[i]>>b[i];
	for(int i=1;i<=n;i++)
	{
		box[i][i]=1;
		for(int j=m;j>=1;j--)
		{
			bool x=box[i][a[j]],y=box[i][b[j]];
			if(x&&y) use[i]=true;//如果x,y此前已经在i的集合中i就不能选了 
			else if(x) box[i][b[j]]=1; 
			else if(y) box[i][a[j]]=1;
		}
	}
	int ans=0;
	for(int i=1;i<=n;i++)
	{
		if(use[i]) continue;
		for(int j=i+1;j<=n;j++)
		{
			if(use[j]) continue;
			bool flag=true;
			for(int k=1;k<=n;k++)
				if(box[i][k]&&box[j][k])//有交集不可取 
				{
					flag=0;
					break;
				}
			ans+=flag;
		}
	}
	cout<<ans; 
	return 0;
}