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<cctype>
#include<bitset>
inline int getint() {
	register char ch;
	while(!isdigit(ch=getchar()));
	register int x=ch^'0';
	while(isdigit(ch=getchar())) x=(((x<<2)+x)<<1)+(ch^'0');
	return x;
}
const int N=401,M=1e5+1;
bool die[N];
int x[M],y[M];
std::bitset<N> b[N];
int main() {
	const int n=getint(),m=getint();
	for(register int i=1;i<=m;i++) {
		x[i]=getint(),y[i]=getint();
	}
	int ans=0;
	for(register int i=1;i<=n;i++) {
		b[i][i]=true;
		for(register int j=m;j>=1;j--) {
			if(b[i][x[j]]&&b[i][y[j]]) b[i].set();
			if(b[i][x[j]]) b[i][y[j]]=true;
			if(b[i][y[j]]) b[i][x[j]]=true;
		}
		for(register int j=1;j<i;j++) {
			if((b[i]&b[j]).count()==0) ans++;
		}
	}
	printf("%d\n",ans);
	return 0;
}