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<bitset>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=505,maxm=100005;
int n,m,ans,x[maxm],y[maxm];bool pos[maxn];bool F[maxn][maxn];
inline int read()
{
	int ret=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-f;ch=getchar();}
	while(ch>='0'&&ch<='9'){ret=ret*10+ch-'0';ch=getchar();}
	return ret*f;
}
bool DFS(int stp,int s)
{
	if(!stp) return true;
	if(F[s][x[stp]]&&F[s][y[stp]]) return false;
	if(F[s][x[stp]]) F[s][y[stp]]=1;
	if(F[s][y[stp]]) F[s][x[stp]]=1;
	return DFS(stp-1,s);
}
int main()
{
	n=read();m=read();
	for(int i=1;i<=m;i++){x[i]=read();y[i]=read();}
	for(int i=1;i<=n;i++)
	{
		F[i][i]=1;
		pos[i]=DFS(m,i);
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=i+1;j<=n;j++)
		{
			if(pos[i]&&pos[j])
			{
				ans++;
				for(int k=1;k<=n;k++)
					if(F[i][k]&&F[j][k])
						{ans--;break;}
			}
		}
	}
	printf("%d\n",ans);
	return 0;
}