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>
#include<cstring>
#include<bitset>
using namespace std;
const int M=100005;
bitset<405> bt[405];
int n,m,ans,x[M],y[M],die[405];
int read()
{
   int x=0,f=1;char ch=getchar();
   while (ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=getchar();}
   while (ch>='0'&&ch<='9') x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
   return x*f;
}
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++)
	{
		bt[i][i]=1;
		for (int j=m;j>=1;j--)
		{
			if (bt[i][x[j]]&&bt[i][y[j]]) die[i]=1;
			if (bt[i][x[j]]||bt[i][y[j]]) bt[i][x[j]]=bt[i][y[j]]=1;
		}
	}
	for (int i=1;i<n;i++)
	  if (!die[i])
	  for (int j=i+1;j<=n;j++)
	    if (!die[j]&&(bt[i]&bt[j]).none()) ans++;
	printf("%d\n",ans);
   return 0;
}
