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>
#define IL inline
#define ll long long
using namespace std;

IL int gi(){
	int data=0,fu=1; char ch=0; while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
	if(ch=='-')fu=-1,ch=getchar();
	while('0'<=ch&&ch<='9')data=(data<<1)+(data<<3)+(ch^48),ch=getchar(); return data*fu;
}

bitset<405>Need[405]; int n,m,a[100005],b[100005],Ans; bool vis[405],flag;
int main(){
	n=gi(); m=gi();
	for(int i=1;i<=m;++i)a[i]=gi(),b[i]=gi();
	for(int x=1;x<=n;++x)Need[x][x]=1;
	for(int x=1;x<=n;++x){
		for(int i=1;i<=n;++i)vis[i]=false; flag=true;
		vis[x]=true; 
		for(int i=m;i>=1;--i){
			if(vis[a[i]]&&vis[b[i]]){flag=false;break; }
			else if(vis[a[i]])Need[x][b[i]]=1,vis[b[i]]=true;
			else if(vis[b[i]])Need[x][a[i]]=1,vis[a[i]]=true;
		}
		if(!flag)Need[x].set();
	}
	for(int i=1;i<=n;++i)
		for(int j=i+1;j<=n;++j)if(!(Need[i]&Need[j]).count())++Ans; 
	cout<<Ans<<endl;
	return 0; 
}