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>
using namespace std;
bool ban[410],vis[410][410];
int main(){
//	freopen("E.in","r",stdin);
	int n,m;
	scanf("%d%d",&n,&m);
	int s,t;
	while(m--)
	{
		scanf("%d%d",&s,&t);
		if(vis[s][t]||vis[t][s]){
			ban[s]=1,ban[t]=1;
			for(int i=1;i<=n;i++)
				vis[i][s]=1,vis[i][t]=1;
		}
		else{
			vis[s][t]=1;
			for(int i=1;i<=n;i++)
				if(i!=s&&i!=t)
					if(vis[s][i]||vis[t][i]||vis[i][s]||vis[i][t])
						vis[i][s]=vis[i][t]=1;
		}
	}
	int ans=0;
	for(int i=1;i<=n;i++)
		for(int j=1;j<i;j++)
			if(!vis[i][j]&&!vis[j][i])
				ans++;
	printf("%d\n",ans);
	return 0;
}
