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
//Heaplax
#include<bits/stdc++.h>
#define N 405
#define M 100005
#define LL long long
#define LOG(x) cerr<<#x<<" = "<<x<<endl
#define add_edge(u,v) nxt[++cnt]=head[u],head[u]=cnt,to[cnt]=v
#define open(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout)
char ch;bool fs;void re(int& x)
{
	while(ch=getchar(),ch<33);
	if(ch=='-')fs=1,x=0;else fs=0,x=ch-48;
	while(ch=getchar(),ch>33)x=x*10+ch-48;
	if(fs)x=-x;
}
using namespace std;
bitset<N> gg[N];
int n,m,x[M],y[M],d[N];
bool ggg[N];
int main()
{
	re(n),re(m);
	for(int i=1;i<=m;++i)
		re(x[i]),re(y[i]);
	for(int a=1;a<=n;++a)
	{
		gg[a][a]=1;
		for(int i=m;i;--i)
			if(gg[a][x[i]] && gg[a][y[i]])
			{
				ggg[a]=1;
				break;
			}
			else if(gg[a][x[i]])gg[a][y[i]]=1;
			else if(gg[a][y[i]])gg[a][x[i]]=1;
	}
	int ans=0;
	for(int a=1;a<=n;++a)if(!ggg[a])
		for(int b=a+1;b<=n;++b)if(!ggg[b])
			if((gg[a]&gg[b]).count()==0)
				++ans;
	printf("%d\n",ans);
}
