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<cmath>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<vector>
#define ll long long
#define N 200010
using namespace std;
inline int read()
{
	int neg=1,num=0;char c=getchar();
	for(;!isdigit(c);c=getchar())if(c=='-')neg=-1;
	for(;isdigit(c);c=getchar())num=(num<<1)+(num<<3)+c-'0';
	return neg*num;
}
int n,m,ans,u[N],v[N],f[410][410];
bool vis[410]; 
int main()
{
	n=read();m=read();
	for(int i=1;i<=m;i++)u[i]=read(),v[i]=read();
	for(int i=1;i<=n;i++)
	{
		f[i][i]=1;
		for(int j=m;j>=1;j--)
		{
			int x=f[i][u[j]],y=f[i][v[j]];
			if(x && y){vis[i]=1;break;}
			else if(x)f[i][v[j]]=1;
			else if(y)f[i][u[j]]=1;
		}
	}
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++)
		{
			if(vis[i] || vis[j])continue;
			ans++;
			for(int k=1;k<=n;k++)if(f[i][k] && f[j][k]){ans--;break;}
		}
	printf("%d",ans); 
	return 0;
}

