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 <stdio.h>
#include <string.h>
#include <algorithm>
#include <bitset>
#define rep(i,st,ed) for (int i=st;i<=ed;++i)
#define drp(i,st,ed) for (int i=st;i>=ed;--i)

const int N=405;
const int M=100005;

std:: bitset <N> a[N];

int x[M],y[M],ans[N];

int read() {
	int x=0,v=1; char ch=getchar();
	for (;ch<'0'||ch>'9';v=(ch=='-')?(-1):(v),ch=getchar());
	for (;ch<='9'&&ch>='0';x=x*10+ch-'0',ch=getchar());
	return x*v;
}

int main(void) {
	int n=read(),m=read();
	rep(i,1,m) x[i]=read(),y[i]=read();
	rep(i,1,n) {
		a[i][i]=1;
		drp(j,m,1) {
			if (a[i][x[j]]&&a[i][y[j]]) {
				ans[i]=1; break;
			}
			if (a[i][x[j]]) a[i][y[j]]=1;
			else if (a[i][y[j]]) a[i][x[j]]=1;
		}
	}
	int prt=0;
	rep(i,1,n) if (!ans[i]) {
		rep(j,i+1,n) if (!ans[j]) {
			if (!(a[i]&a[j]).count()) prt++;
		}
	}
	printf("%d\n", prt);
	return 0;
}