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 <iostream>
#include <cstdio>
#include <cstring>
#include <bitset>
using namespace std;
int n,m,E[100005][2];
bitset<405> s[405],c;
bool vis[405];
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++) scanf("%d%d",&E[i][0],&E[i][1]);
	for(int i=n;i;i--){
		s[i][i]=1;
		for(int p=m;p;p--){
			int x=E[p][0],y=E[p][1];
			if(s[i][x]&&s[i][y]){
				vis[i]=1;break;
			}
			if(!s[i][x]&&!s[i][y]) continue;
			s[i][x]=s[i][y]=1;
		}
	}
	int ans=0;
	for(int i=1;i<=n;i++){
		for(int p=1;p<i;p++)if(!vis[i]&&!vis[p]){
			c=s[i]&s[p];
			if(!c.any()) ans++;
		}
	}
	printf("%d",ans);
	return 0;
}