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<algorithm>
#include<cstring>
#include<cctype>
#include<cstdio>
#define rep(i,x,y) for(int i=x; i<=y; ++i)
#define repd(i,x,y) for(int i=x; i>=y; --i)

using namespace std;
const int N=405,M=100005;
bool fk[N],g[N][N];
int n,m,ans;
struct data{int u,v;} dat[M];

int getint()
{
	char ch;
	while(!isdigit(ch=getchar()));
	int x=ch-48;
	while(isdigit(ch=getchar())) x=x*10+ch-48;
	return x;
}

void solve(int x)
{
	g[x][x]=1;
	repd(i,m,1)
	{
		int u=dat[i].u,v=dat[i].v;
		if(g[x][u] && g[x][v]) {fk[x]=1; return;}
		if(!g[x][u] && !g[x][v]) continue;
		g[x][u]=g[x][v]=1;
	}
}

int main()
{
	n=getint(),m=getint();
	rep(i,1,m) dat[i].u=getint(),dat[i].v=getint();
	rep(i,1,n) solve(i);
	rep(i,1,n-1) if(!fk[i]) rep(j,i+1,n) if(!fk[j])
	{
		bool jdg=1;
		rep(k,1,n) if(g[i][k] && g[j][k]) jdg=0;
		ans+=jdg;
	}
	printf("%d\n",ans);
	return 0;
}