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 MAX=1e5+5;
const int N=403;
int x[MAX];
int y[MAX];
int arr[N][N];
int fl[N];

int main()
{
	int n,m;
	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++)
	{
		arr[i][i]=1;
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=m;j>=1;j--)
		{
			int a=x[j];
			int b=y[j];
			if(arr[i][a]&&arr[i][b])
			{	
				fl[i]=1;
			}
			else if(arr[i][a])
			{
				arr[i][b]=1;
			}
			else if(arr[i][b])
			{
				arr[i][a]=1;
			}
		}
	}
	int ans=0;
	for(int i=1;i<=n;i++)
	{
		for(int j=i+1;j<=n;j++)
		{
			if(fl[i]||fl[j])
			{
				continue;
			}
			int ok=1;
			for(int k=1;k<=n;k++)
			{
				if(arr[i][k]&&arr[j][k])
				{
					ok=0;
				}
			}
			ans+=ok;
		}
	}
	printf("%d\n",ans);
}