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;
bool can[410][410];
int main()
{
	int n,m;scanf("%d%d",&n,&m);
	memset(can,true,sizeof(can));
	for (int i=1;i<=m;i++)
	{
		int x,y;scanf("%d%d",&x,&y);
		if (can[x][y])
		{
			can[x][y]=can[y][x]=0;
			for (int j=1;j<=n;j++)
			{
				if (!can[j][x]) can[j][y]=can[y][j]=0;
				if (!can[j][y]) can[j][x]=can[x][j]=0;
			}
		}
		else for (int j=1;j<=n;j++) can[j][x]=can[x][j]=can[j][y]=can[y][j]=0;
	}
	int ans=0;
	for (int i=1;i<=n;i++)
	for (int j=i+1;j<=n;j++) ans+=can[i][j];
	cout<<ans<<endl;
}