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 <bits/stdc++.h>
#define B bitset<500>
using namespace std;
int n,m,x,cnt;
int lis[200001],st[200001],ed[200001];
bool die[2001];
B a[2001];
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{ 
	//	scanf("%d",&x);
		x=2;
		st[i]=cnt+1;
		for(int j=1;j<=x;j++)
			scanf("%d",&lis[++cnt]);
		ed[i]=cnt;
	} 
	for(int i=1;i<=n;i++)
	{
		a[i][i]=1;
		for(int j=m;j;j--)
		{
			int sum=0;
			for(int k=st[j];k<=ed[j];k++)
			if(a[i][lis[k]]) ++sum;
			if(sum>1)
			{
				die[i]=1;
				break;
			}
			else
			if(sum==1)
				for(int k=st[j];k<=ed[j];k++)
					a[i][lis[k]]=1;
		}
	}
	int ans=0;
	for(int i=1;i<n;i++)
	if(!die[i])
		for(int j=i+1;j<=n;j++)
		if(!die[j] && (a[i]&a[j]).none())
			++ans;
	printf("%d\n",ans);
	return 0;
}