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<bitset>
using namespace std;
const int N=440,M=100100;
int i,j,k,n,m,ch,ff,ans;
int x[M],y[M],fg[N];
bitset<N> A[N];
void R(int &x) {
	ff=x=0;ch=getchar();
	while (ch<'0' || '9'<ch) { if (ch=='-') ff=1;ch=getchar();}
	while ('0'<=ch && ch<='9') x=x*10+ch-'0',ch=getchar();
	if (ff) x=-x;
}
int main() {
	R(n);R(m);
	for (i=1;i<=m;i++) R(x[i]),R(y[i]);
	for (i=1;i<=n;i++) A[i][i]=1;
	for (i=m;i;i--) {
		for (j=1;j<=n;j++) if (A[j][x[i]]|A[j][y[i]]) {
			if (A[j][x[i]]&A[j][y[i]]) fg[j]=1;
			else A[j][x[i]]=A[j][y[i]]=1;
		}
	}
	for (i=1;i<=n;i++) if (!fg[i])
		for (j=i+1;j<=n;j++) if (!fg[j] && (A[i]&A[j]).count()==0) ans++;
	printf("%d\n",ans);
}