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<algorithm>
using namespace std;
#define _rep(i,s,t) for(register int i=s;i>=t;--i)
#define rep(i,s,t) for(register int i=s;i<=t;++i)
const int N=411,M=1e5+11;
int n,m,x,y,ans;
int a[M],b[M],p[N];
int f[N][N];
int main(){
	scanf("%d%d",&n,&m);
	rep(i,1,m)
		scanf("%d%d",a+i,b+i);
	rep(i,1,n){
		f[i][i]=1;
		_rep(j,m,1){
			x=a[j],y=b[j];
			if(!f[i][x]&&!f[i][y])continue;
			if(f[i][x]&&f[i][y]){
				p[i]=1;
				break;
			}
			f[i][x]=f[i][y]=1;
		}
	}
	rep(i,1,n)
		rep(j,i+1,n)
			if(!p[i]&&!p[j]){
				x=0;
				rep(k,1,n)
					if(f[i][k]&&f[j][k]){
						x=1;
						break;
					}
				if(!x)++ans;
			}
	printf("%d\n",ans);
    return 0;
}