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<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const int N=1000010,P=998244353;
inline int read(){
	int x=0,f=0,c=getchar();
	for(;c>'9'||c<'0';f=c=='-',c=getchar());
	for(;c>='0'&&c<='9';c=getchar())
	x=(x<<1)+(x<<3)+c-'0';return f?-x:x;
}
inline void write(ll x){
	if(x<0)putchar('-'),x=-x;
	if(x>9)write(x/10);
}
#include<bitset>
bitset<510>f[510];
int n,m,x[N],y[N],ans,del[N];
void bfs(int s,bitset<510>&f){
	f[s]=1;for(int i=m;i;i--){
		if(f[x[i]]&&f[y[i]]){del[s]=1;return;}
		if(f[x[i]])f[y[i]]=1;
		if(f[y[i]])f[x[i]]=1;
	}
}
int main(){
	//freopen("1.out","w",stdout);
	int i,j;
	n=read();m=read();
	for(i=1;i<=m;i++)x[i]=read(),y[i]=read();
	for(i=1;i<=n;i++)bfs(i,f[i]);
	for(i=1;i<=n;i++)
		for(j=i+1;j<=n;j++)
			if(!del[i]&&!del[j])ans+=(f[i]&f[j]).count()==0;
	printf("%d\n",ans);
	return 0;
}