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>
#define MAXN 410
#define MAXM 100010

int n,m;
int e[MAXM][2];
bool visit[MAXN],flag[MAXN],a[MAXN][MAXN];

bool gao(int x){
	memset(visit,0,sizeof visit);
	visit[x]=1;
	for(int i=m;i>=1;i--){
		if(visit[e[i][0]] && visit[e[i][1]]) return 0;
		else if(visit[e[i][0]]) visit[e[i][1]]=1;
		else if(visit[e[i][1]]) visit[e[i][0]]=1;
	}
	for(int i=1;i<=n;i++) a[x][i]=visit[i];
	return 1;
}

bool query(int x,int y){
	if(flag[x] || flag[y]) return 0;
	for(int i=1;i<=n;i++)
		if(a[x][i] && a[y][i]) return 0;
	return 1;
}

int main(){
#ifdef DEBUG
	freopen("E.in","r",stdin);
#endif
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++) scanf("%d%d",&e[i][0],&e[i][1]);
	for(int i=1;i<=n;i++) if(!gao(i)) flag[i]=1;
	int ans=0;
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++)
			if(query(i,j)) ans++;
	printf("%d\n",ans);
	return 0;
}
