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;
typedef long long LL;
LL read(){
	LL x=0,f=1;
	char ch=getchar();
	while (!isdigit(ch)&&ch!='-')
		ch=getchar();
	if (ch=='-')
		f=-1,ch=getchar();
	while (isdigit(ch))
		x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
	return x*f;
}
const int N=405,M=100005;
int n,m;
int x[M],y[M];
int vis[N];
int check(){
	for (int i=m;i>=1;i--){
		int a=x[i],b=y[i];
		if (vis[a]&&vis[b])
			return 0;
		if (!vis[a]&&!vis[b])
			continue;
		vis[a]=vis[b]=1;
	}
	return 1;
}
int main(){
	n=read(),m=read();
	for (int i=1;i<=m;i++)
		x[i]=read(),y[i]=read();
	int ans=0;
	for (int i=1;i<=n;i++)
		for (int j=i+1;j<=n;j++){
			memset(vis,0,sizeof vis);
			vis[i]=vis[j]=1;
			ans+=check();
		}
	cout << ans << endl;
	return 0;
}